Class 11 Engineering Maths - Straight Lines

Find inclination (in degrees) of a line perpendicular to x-axis.

Line perpendicular to

$x-$ axis is the

$y-$ axis.
Inclination of

$y-$ axis with respect to

$x-$ axis is

${90}^{0}$ .
Hence, inclination of a line which is perpendicular to x-axis is

${ 90 }^{ 0 }$ .

If P is the mid point of seg AB and AB $=7$ cm, then AP is $3.5$ cm
If true then enter $1$ and if false then enter $0$ .

A midpoint divides a line segment into equal parts.
Here P is the midpoint of line segment AB dividing it into two equal parts namely AP & PB.
Hence AP=half of 7 cm =3.5cm

Write the slope of the line whose inclination is $0^{\circ}$ .

Slope of line with an inclination

$\theta = tan \theta$ .

Hence, slope of line, with inclination

$\theta = {0}^0$ is tan

${0}^0 = 0$

Find inclination (in degrees) of a line parallel to x-axis.

Inclination of

$x-$ axis is

${ 0 }^{ 0 }$ . Slope of two or more parallel lines is the same or equal to each other.
Hence, inclination of a line which is parallel to

$x-$ axis

${ 0 }^{o}$ .

Find inclination (in degrees) of a line parallel to $y$ -axis.

Inclination of

$y-$ axis is

${ 90 }^{ 0 }$ . Slope of two or more parallel lines is the same or equal to each other.
Hence, inclination of a line which is parallel to

$y$ -axis is

${ 90 }^{ 0 }$ .

Find inclination (in degrees) of a line perpendicular to y-axis.

Line perpendicular to $y-$ axis is the $x-$ axis.
Inclination of $x-$ axis is ${ 0 }^{ 0 }$ .
Hence, inclination of a line which is perpendicular to y-axis is ${ 0 }^{ 0 }$ .

Find the inclination of the line whose slope is 1

Slope of a line

$m = \tan { \theta }$ , where

$\theta$ is the inclination of the line with the x - axis.
Given slope

$= 1$

$\tan { \theta } = 1$
We know that

$\tan { { 45 }^{ 0 } } = 1$

$=> \theta = \quad { 45 }^{ 0 }$
Inclination

$= \quad { 45 }^{ 0 }$

Find the inclination of the line whose slope is $0$ .

$\textbf{Step 1: Find angle of line with x-axis}$

$\text{Slope of a line } m = \tan { \theta }$ ,

$\text{where } \theta$

$\text{is the inclination of the line with the x - axis.}$

$\text{Given slope} = 0$

$\tan { \theta } = 0$

$\text{We know that } \tan { { 0 }^\circ } = 0$

$\Rightarrow\theta = \quad { 0 }^{ \circ}$

$\text{Inclination} = \quad { 0 }^{ \circ }$

$\textbf{Hence, Inclination of line is 0 }^\circ.$

Consider the following population and year graph, find the slope of the line $AB$ and using it, find what will be the population in the year $2010$ ?

Since line

$AB$ passes through point

$A(1985,92)$ and

$B(1995,97)$ , its slope is

$\displaystyle \frac{97-92}{1995-1985}= \frac{5}{10}= \frac{1}{2}$ .
Now, let

$y$ be the population in the year

$2010$ . Then according to the given graph line

$AB$ must pass through point

$C(2010,y)$ .

$\therefore$ slope of

$AB$ = slope of

$BC$

$\displaystyle\Rightarrow \frac{1}{2}= \frac{y-97}{2010-1995}$

$\displaystyle\Rightarrow \frac{1}{2}= \frac{y-97}{15}$

$\displaystyle\Rightarrow \frac{15}{2}= y-97$

$\displaystyle\Rightarrow \ y-97= 7.5$

$\displaystyle\Rightarrow \ y= 7.5+97= 104.5$
Thus, the slope of line

$AB$ is

$\displaystyle\frac{1}{2}$ , while in the year

$2010$ the population will be

$104.5$ crores.

Find the coordinates of the foot of perpendicular from the point $(-1, 3)$ to the line
$3x - 4y - 16\displaystyle = 0$

Let

$(a,b)$ be the coordinates of the foot of the perpendicular from the point

$(-1,3)$ to the line

$3x -4y - 16\displaystyle = 0$
Slope of the line joining

$(-1,3)$ and

$(a,b)$ , is

$\displaystyle m_{1}= \frac{b-3}{a+1}$
Slope of the line

$3x -4y - 16\displaystyle = 0$ or

$\displaystyle y= \frac{3}{4}x-4,m_{2}= \frac{3}{4}$
Since these two lines are perpendicular,

$\displaystyle m_{1}m_{2}= -1$

$\displaystyle \therefore \left \{ \frac{b-3}{a+1} \right \}\cdot \left \{ \frac{3}{4} \right \}= -1$

$\displaystyle \Rightarrow \frac{3b-9}{4a+4}= -1$

$\displaystyle \Rightarrow 3b-9= 4a-4$

$\displaystyle \Rightarrow 4a+3b= 5$ ...(1)
Point(a,b) lies on line

$\displaystyle 3x-4y= 16$

$\displaystyle \therefore 3a-4b= 16$ ....(2)
On solving equation (1) and (2) we obtain

$\displaystyle a= \frac{68}{25}$ and

$b= -\cfrac{49}{25}$
Thus, the required coordinates of the foot of the perpendicular are

$\displaystyle \left \{ \frac{68}{25},\frac{49}{25} \right \}$

Find the point on the X-axis which is equidistant from $(2, -5)$ and $(-2, 9)$ .

Let the required point on the X-axis be

$P(x,0)$

As given

$P$ is equidistant from

$A(2,-5)$ and

$B(-2,9)$

$\therefore PA=PB$

$\Rightarrow \sqrt{(x-2)^2+(0-(-5))^2}=\sqrt{(x+2)^2+(0-9)^2}$

$\Rightarrow (x-2)^2+25=(x+2)^2+81$

$\Rightarrow x^2-4x+4+25=x^2+4x+4+81$

$\Rightarrow -8x=85-29$

$\Rightarrow -8x=56$

$\Rightarrow x=\dfrac{56}{-8}=-7$

$\therefore$ Required point on X-axis

$=(-7,0)$

Determine the equation of the line in the given graph.

We may take the equation in the form of

$y=ax+b$ .

Observe that the straight-line passes through

$(5,0)$ .

This means that

$y=0$ when

$x=5$ .

But the substitution

$y=0, x=5$ in the equation gives

$0=5a+b$ .

Thus

$b=-5a$
We also observe that the line passes through

$(0,5)$ ; that is

$y=5$ , whenever

$x=0$ .

This gives

$5=0+b$ or

$b=5\implies a=-1$ .

Therefore, the equation of the line is

$y=-x+5$ or

$x+y=5$

Draw the graph of $y=3x+5$ .

Given different values for

$x$ and get values for

$y$ . Tabulate them.
Let us trace these on a graph paper. Fix

$x$ and

$y$ axes and locate the points

$(x,y)=(0,5), (1,8), (2,11), (3,14), (-1,2), (-2,-1)$

We see that all these points lie on a straight line. Use a straight edge and join all these points by a straight line segment.

Hence, we get the graph of

$y=3x+5$ .

Determine the equation of the line in the given graph.

Recall that the general equation of a straight line is

$y=ax+b$ , where

$a,b$ are real constants. The given graph shows that

$a\ne 0$ . (If

$a=0$ , then

$y=b$ represents a straight line parallel to

$y$ -axis).

$x=0$ , then

$y=b$ . Thus

$(0,b)$ is a point on the straight line.
Similarly, taking

$y=0$ , you get

$ax+b=0$ or

$x=-\cfrac{b}{a}$ . Thus

$(-\cfrac{b}{a},0)$ is also a point on the line.

Looking at the graph, we see that it cuts

$y$ -axis at

$(0,4)$ and the x-axis at

$(3,0)$ .

So, we conclude that

$(0,b)=(0,4)$ and

$\left(-\cfrac{b}{a},0\right)=(3,0)$
Thus,

$b=4$ and

$-\cfrac{b}{a}=3$ , which gives

$a=-\cfrac{4}{3}$ .

Hence, the equation of the given graph is

$y=-\cfrac{4}{3}x+4$ .

This may also be written as

$3y+4x=12$

Determine the equation of the line in graph.

Given points of the line

$\left( 3,8 \right) \left( -3,-3 \right)$

$\therefore$ Slope =

$\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\dfrac { -3-8 }{ -3-3 } =\dfrac { -11 }{ -6 } =\dfrac { 11 }{ 6 }$

Equation of the line in standard form is

${ y-y }_{ 1 }=m\left( x-{ x }_{ 1 } \right)$

Substituting the value of

$m$ and

$(x,y) = (3,8)$ in the above equation:

$y-8=\dfrac { 11 }{ 6 } \left( x-3 \right)$

$6y-48=11x-33$

$6y=11x+15$

Find the area of a triangle whose vertices are $A(3, 4), B(-1, 2),$ and $C(2, 3)$ .

Area of triangle with vertices

$(x_1, y_1), (x_2, y_2)$ and

$(x_3, y_3)$ is given by

$A$ :

$A = \dfrac{1}{2} \begin{vmatrix} (x_{1}-x_{2}) & (x_{2}-x_{3}) \\ (y_{1}-y_{2}) & (y_{2}-y_{3}) \end{vmatrix}$

$= \dfrac{1}{2} \begin{vmatrix} (3+1) & (-1-2) \\ (4-2) &(2-3) \end{vmatrix}$

$= \dfrac{1}{2} [-4-(-6)]$

$\Rightarrow \boxed{A = 1}$

For the angle in standard position if the Initial arm rotates $130^o$ in anticlockwise direction, then state the quadrant in which terminal arm lies. (Draw the figure and write the answer).

The initial arm,

$X$ axis, if rotated by

$130^0$ in anti-clockwise direction, then it will cross the positive

$Y$ axis.

Hence, the terminal arm lies in the

$II$ quadrant.

Determine the equation of the line in graph.

Points of the given line are

$(12,-3), (-4,2)$

slope(m) will be=

$\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\dfrac { 2-\left( -3 \right) }{ -4-12 } =\dfrac { 5 }{ -16 } =-\dfrac { 5 }{ 16 }$

Standard equation of the line is

${ y-y }_{ 1 }=m\left( x-{ x }_{ 1 } \right)$

Substituting the value of

$m$ and point

$(12 , -3)$ in the equation we get:

$y-\left( -3 \right) =-\dfrac { 5 }{ 16 } \left( x-12 \right)$

$y+3=-\dfrac { 5 }{ 16 } \left( x-12 \right)$

$16y+48=-5x+60$

$16y=-5x+60-48$

$16y=-5x+12$

Find the slope of the line passing through the points $G(-4, 5)$ and $H(-2, 1)$

$G\equiv (-4, 5) = (x_{1}, y_{1})$
and

$H\equiv (-2, 1) = (x_{2}, y_{2})$
Then, Slope of line

$GH = \dfrac {y_{1} - y_{2}}{x_{1} - x_{2}}$

$= \dfrac {5 - 1}{-4 + 2} = \dfrac {4}{-2}$

$= -2$

Find the slope of the line having its inclination ${60}^{o}$ .

Inclination of the line is

${60}^{o}$

$\therefore$ Slope

$= \tan{\theta} = \tan{{60}^{o}} = \sqrt{3}$

$\therefore$ Slope

$= \sqrt{3}$

Find the slope of line passing through the point $P\left(1, -1\right)$ and $Q\left(-2, 5\right)$ .

Given :

$P\equiv \left( 1,-1 \right) =\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and

$Q\equiv \left( -2,5 \right) =\left( { x }_{ 2 },{ y }_{ 2 } \right)$
Then, Slope of line

$PQ=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$

$=\dfrac { 5-\left( -1 \right) }{ -2-\left( -1 \right) }$

$=\dfrac { 5+1 }{ -3 }$

$=\dfrac { 6 }{ -3 }$

$=-2$

Find the slope of the line passing through the points $M(4, 0)$ and $N(-2, -3)$ .

$M\equiv (4, 0) = (x_{1}, y_{1})$
and

$N\equiv (-2, -3) = (x_{2}, y_{2})$
Then, slope of line

$MN = \dfrac {y_{1} - y_{2}}{x_{1} - x_{2}}$

$= \dfrac {0 - (-3)}{4 - (-2)}$

$= \dfrac {3}{6} = \dfrac {1}{2}$

Find the area of the triangle, whose vertices are $(-5, -1), (3, -5), (5, 2).$

Given:

$(-5, -1)\equiv(x_1,y_1), (3, -5)\equiv(x_2,y_2), (5, 2)\equiv(x_3,y_3).$

Area of the triangle

$=\dfrac 12|{x}_{1}({y}_{2}-{y}_{3})+{x}_{2}({y}_{3}-{y}_{1})+{x}_{3}({y}_{1}-{y}_{2})|$

$=\cfrac 12|(-5)(-5-2)+3(2-(-1))+5(-1-(-5))|$ square units

$=\cfrac 12|35+9+20|$ square units

$=\cfrac 12\times64$ square units

$=32$ square units

Find the area of the triangle, whose vertices are $(2, 0), (1, 2), (-1, 6)$ . What do you observe?

Apply the formula for area of triangle whose co-ordinates of vertices are known.

Area of the triangle

$= \begin{vmatrix}2 & 0 &1\\1&2&1\\-1&6&1 \end{vmatrix}$

$=|2(2-6)-0(1+1)+1(6+2)|$

$=|(-8)+0+8|$

$=0$

We know that if area of triangle is zero, then the three points will be collinear.

So, the above three points are collinear.

Find the area of the triangles (in sq. units) formed by the points $(-2, 3), (7, 5)$ and $(3, -5)$ .

Area of the triangle

$=\dfrac12\times|{x}_{1}({y}_{2}-{y}_{3})+{x}_{2}({y}_{3}-{y}_{1})+{x}_{3}({y}_{1}-{y}_{2})|$

$=\dfrac12\times|-2(5-(-5))+7(-5-3)+3(3-5)|$

$=\dfrac12\times|(-20)+(-56)+(-6)|$

$=\dfrac12\times|-82|$

$=41$ square units.

Show that $S(4, 3)$ is the circum-centre of the triangle joining the points $A(9, 3), B(7, -1)$ and $C(1, -1)$ .

$SA = \sqrt{(9 - 4)^2 + (3 - 3)^2} = \sqrt{25} = 5$

$SB = \sqrt{(7 - 4)^2 + (-1 - 3)^2} = \sqrt{25} = 5$

$SC = \sqrt{(1 - 4)^2 + (-1 -3)^2} = \sqrt{25} = 5$

$\therefore SA = SB = SC$ .
It is known that the circum-centre is equidistant from all the vertices of a triangle.
Since, S is equidistant from all the three vertices, it is thee circum-centre of the triangle ABC.

Find the area of the triangle formed by joining the midpoints of the sides of a triangle whose vertices are $(0,-1),(2,1)$ and $(0,3)$ .

Mid Points of the sides of the triangle with the given vertices can be calculated using mid point formula

which is

$(1,0) , (1,2) , (0,1)$

Area

$= 1/2*[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$

$= 1/2 * [0(1-3) + 2(3+1) + 0(-1-1)]$

$= 4 sq.units$

Find the area of triangle $\triangle ABC$ having vertices $A(4, 2),B(3, 9)$ and $C(10, 10)$ .

$\triangle ABC$ has vertices

$A(4,2),B(3,9)$ and

$C(10,10)$ .
The area of a triangle formed by joining the points

$\left( { x }_{ 1 },{ y }_{ 1 } \right) ,\left( { x }_{ 2 },{ y }_{ 2 } \right) ,\left( { x }_{ 3 },{ y }_{ 3 } \right)$ is

$\cfrac { 1 }{ 2 } \left| { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right|$

$\therefore$ Area of

$\triangle ABC$

$\cfrac { 1 }{ 2 } \left| 4(9-10)+3(10-2)+10(2-9) \right|$

$=\cfrac { 1 }{ 2 } \left| -4+24-70 \right| =\cfrac { 50 }{ 2 } =25$

Hence, area of triangle

$ABC$ is

$25$ .

Find the slope of the straight line passing through the points $(3,-2)$ and $(-1,4)$ .

Slope of the straight line passing through the points

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and

$\left( { x }_{ 2 },{ y }_{ 2 } \right)$ is given by

$m=\cfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$
Slope of the straight line passing through the points

$(3,-2)$ and

$(-1,4)$ is

$m=\cfrac { 4+2 }{ -1-3 } =-\cfrac { 3 }{ 2 }$

Show that the following point taken in order form the vertices of a rhombus.
(0, 0), (3, 4), (0, 8) and (-3, 4)

Consider the given points.

$A(0, 0), B(3, 4), C(0, 8), D(-3, 4)$

Distance of

$AB$ ,

$AB=\sqrt {(3-0)^2+ (4-0)^2}$

$AB=\sqrt {(3)^2+ (4)^2}$

$AB=\sqrt {9+16}=\sqrt {25}=5$

Similarly,

$BC=\sqrt {(0-3)^2+ (8-4)^2}$

$BC=\sqrt {(-3)^2+ (4)^2}$

$BC=\sqrt {9+16}=5$

Similarly,

$CD=\sqrt {(-3-0)^2+ (4-8)^2}$

$CD=\sqrt {(-3)^2+ (-4)^2}$

$CD=\sqrt {9+16}=5$

Similarly,

$DA=\sqrt {(0+3)^2+ (0-4)^2}$

$DA=\sqrt {(3)^2+ (-4)^2}$

$DA=\sqrt {9+16}=5$

Therefore,

$AB=BC=CD=AD$

Now,

$AC=\sqrt {(0-0)^2+ (8-0)^2}$

$AC=\sqrt {(0)^2+ (8)^2}$

$AC=\sqrt {64}=8$

Similarly,

$BD=\sqrt {(-3-0)^2+ (4-8)^2}$

$BD=\sqrt {(-3)^2+ (-4)^2}$

$BD=\sqrt {9+16}=5$

So,

$AC\ne BD$

As all the sides are equal and diagonals are not equal.

Its shows that the following vertices are of rhombus.

Hence, this is the answer.

Show that the three points $(4, 2), (7, 5)\ and\ (9, 7)$ lie on a straight line.

Let the points be A(4, 2), B(7, 5) and C(9, 7). By the distance formula

$AB^2 = (4 - 7)^2 + (2- 5)^2 = (-3)^2 + (-3)^2 = 9 + 9 = 18$

$BC^2 = (9-7)^2 + (7 - 5)^2 = 2^2 + 2^2 = 4 + 4 = 8$

$CA^2 = (9 - 4)^2 + (7 - 2)^2 = 5^2 + 5^2 = 25+ 25 = 50$

So,

$AB = \sqrt{18} = \sqrt{9 \times 2} = 3 \sqrt 2; BC = \sqrt 8 = \sqrt{4 \times 2} = 2 \sqrt 2; CA = \sqrt{50} = \sqrt{25 \times 2} = 5 \sqrt 2$

This gives

$AB + BC = 3\sqrt 2 + 2 \sqrt 2 = 5 \sqrt 2 = AC$ . Hence the points A, B and C are collinear.

Find the distance between $A(a+b,b-a)$ and $B(a-b,a+b)$ .

$A({x}_{1},{y}_{1})=A(a+b,b-a)$ and

$B({x}_{2},{y}_{2})=B(a-b,a+b)$

$|AB|^{ 2 }={ \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-{ y }_{ 2 } \right) }^{ 2 }$

$={ \left[ (a+b)-(a-b) \right] }^{ 2 }+{ \left[ (b-a)-(a+b) \right] }^{ 2 }$

$={ (2b) }^{ 2 }-{ (-2a) }^{ 2 }$

$=4{ b }^{ 2 }-4{ a }^{ 2 }=4\left( { a }^{ 2 }-{ b }^{ 2 } \right)$

$\therefore AB=\sqrt { 4\left( { a }^{ 2 }-{ b }^{ 2 } \right) } =2\sqrt { \left( { a }^{ 2 }-{ b }^{ 2 } \right) }$

Find the angle of inclination(in degrees) of the line passing through the points $(1,2)$ and $(2,3)$

Consider the given points.

$(1, 2)$ and

$(2, 3)$

We know that the slope of the line which is passing through the two points

$m=\dfrac{y_2-y_1}{x_2-x_1}$

So,

$m=\dfrac{3-2}{2-1}$

$m=\dfrac{1}{1}=1$

Since,

$\tan\theta=m$

Therefore,

$\tan\theta=1$

$\tan\theta=\tan45^0$

$\theta=45^0$

Hence, this is the answer.

Prove that the points (-2, -1), (1, 0), (4, 3), and (1, 2) are at the vertices of a parallelogram.

We have the points

$P (-2 , -1), Q (1 ,0), R (4 , 3)$ and

$S (1 , 2)$
We know the property the of parallelogram that diagonals of parallelogram bisect each other.
Let us find out mid-point of line joining

$P$ and

$R$ and line joining

$Q$ and

$S$
(i) Mid-point

$M$ of diagonal

$PR$

$M\left(\dfrac{-2+4}{2} , \dfrac{-1+3}{2}\right)$

$\Rightarrow M(1, 1)$

(ii) Mid-point

$M'$ of diagonal

$QS$

$M' \left(\dfrac{1+1}{2} , \dfrac{0+2}{2}\right)$

$\Rightarrow M' (1, 1)$

From (i) & (ii)
Mid-points

$M$ &

$M'$ are identical

$\Rightarrow$ Diagonals of the figure

$PQRS$ bisect each other and this property is enough to prove that it is a parallelogram.

Although we can also check by distance formula i.e.

$d = \sqrt{(a - c)^{2} +(b - d)^{2}}$

$PQ = RS$

$SP = QR$

A line is of length 10 and one end is at the point (2, -3); if the abscissa of the other end be 10, prove that its ordinate must be 3 or -9.

Distance $d$ between two points $P(a, b)$ and $Q(c ,d)$ is given by

$d = \sqrt{(a - c)^{2} +(b - d)^{2} }$

Given that :

Length of line i.e $d = 10$

Abscissa of second point is $10$

And let us assume the ordinate of second point be $y$

Points $(2 , -3)$ & $(10 , y)$

Now apply distance formula

$10 = \sqrt{(2 - 10)^{2} +(-3 - y)^{2} }$

Squaring both sides

$\Rightarrow 100 = ( - 8)^{2} +(-3 - y)^{2}$

$\Rightarrow 100 = 64+(-3 - y)^{2}$

$\Rightarrow 100-64 = (-3 - y)^{2}$

$\Rightarrow 36 = (-3 - y)^{2}$

Taking square root both sides

$\Rightarrow \pm6 = (-3 - y)$

or $\Rightarrow \pm6 = (3 + y)$ (taking - sign common on both sides)

$\Rightarrow \pm6 -3 = y$

Taking both cases

$y= +6-3 = 3$

$y= -6-3 = -9$

Prove (by showing that the area of the triangle formed by them is zero) that the following sets of three points are in a straight line:
$(1,4), (3, -2),$ and $(-3,16)$

Area of triangle having angular points
$(x_{1} , y_{1} )$ $(x_{2} , y_{2})$ $(x_{3} , y_{3})$ is given by the formula

$\triangle=\dfrac{1}{2}\left | x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3} \right |$

$\triangle=\dfrac{1}{2}\left | 1\times(-2) +3\times16+(-3)\times4-3\times4-(-3)(-2)-1\times16 \right |$

$\Rightarrow \triangle=0$

$\Rightarrow$ All three points are in a straight line.

Find the areas of the triangles the whose coordinates of the points are respectively.
(5, 2), (-9, -3) and (-3, -5)

Area of triangle having angular points

$(x_{1} , y_{1} ),$

$(x_{2} , y_{2})$ and

$(x_{3} , y_{3})$ is given by the formula

$\triangle=\dfrac{1}{2}\left | x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3} \right |$

$\triangle=\dfrac{1}{2}\left | 5\times -3 +-9\times -5+-3\times2 -(-9)\times2-(-3)\times(-3)-5\times(-5) \right |$

$\Rightarrow \triangle=\dfrac{1}{2}\left |-15+45-6+18-9+25 \right |$

$\Rightarrow \triangle= \dfrac{1}{2}\times 58 = 29$

Prove that the point $\left ( -\frac{1}{14}, \frac{39}{14} \right )$ is the centre of the circle circumscribing the triangle whose angular points are (1, 1), (2, 3), and ( - 2, 2).

Given that points of triangle

$(1 , 1) , (2 , 3), (-2 , 2)$

Given circumcentre

$\left(\dfrac{-1}{14} , \dfrac{39}{14}\right)$

By distance formula we can check the point as circumcentre

$\sqrt{\left(1+\dfrac{1}{14}\right)^{2}+\left(1-\dfrac{39}{14}\right)^{2}} = \sqrt{\left(2+\dfrac{1}{14}\right)^{2}+\left(3-\dfrac{39}{14}\right)^{2}}=\sqrt{\left(-2+\dfrac{1}{14}\right)^{2}+\left(2-\dfrac{39}{14}\right)^{2}} = \dfrac{\sqrt{850}}{14}$
Then, circumradius

$R= \dfrac{\sqrt{850}}{14}$
Hence proved

Find the areas of the triangles the whose coordinates of the points are respectively.
(0, 4), (3, 6) and ( - 8, - 2)

Area of triangle having angular points

$(x_{1} , y_{1} )$

$(x_{2} , y_{2})$

$(x_{3} , y_{3})$ is given by the formula

$\triangle=\dfrac{1}{2}\left |x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right |$

Area of the triangle whose angular points are (0, 4), (3, 6) and ( - 8, - 2) ia

$\triangle=\dfrac{1}{2}\left | 0\times(6-(-2))+3\times(-2-4)+(-8)(4-6) \right |$

$\Rightarrow \triangle=\dfrac{1}{2}\left | -18+16 \right |$

$\Rightarrow \triangle= 1$

$(a, c + a), (a, c)$ and $(-a, c - a).$

Area of triangle having angular points

$(x_{1} , y_{1} )$

$(x_{2} , y_{2})$

$(x_{3} , y_{3})$ is given by the formula

$\triangle=\dfrac{1}{2}\left | x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3} \right |$

The given points are

$(a, c + a), (a, c)$ and

$(-a, c - a)$
Then,

$\triangle=\dfrac{1}{2}\left | a\times c+a\times (c-a)+-a\times (c+a)-a\times (c+a)-(-a)\times c-a\times (c-a) \right |$

$\Rightarrow \triangle = \dfrac{1}{2}\left | -2a^{2} \right |$

$\Rightarrow \triangle= a^{2}$

Lay down in a figure the positions of the points (1, -3) and (- 2,1), and prove that the distance between them is 5.

We can calculate distance between points by Pythagoras theorem
As shown in the figure

$d= \sqrt{4^{2}+3^{2}}$

$\Rightarrow d= 5 units$

Also calculating by distance formula
Points are

$A(-2, 1), B(1 , -3)$

$d=\sqrt{(-2-1)^{2}+(1-(-3))^{2}}$

$d=\sqrt{(3)^{2}+(4)^{2}}$

$d=\sqrt{25}$

$\Rightarrow d= 5 units$

Find the areas of the triangles the whose coordinates of the points are respectively.
(1, 3), ( - 7, 6) and (5, - 1)

Area of triangle having angular points $(x_{1} , y_{1} )$ $(x_{2} , y_{2})$ $(x_{3} , y_{3})$ is given by the formula

$\triangle=\dfrac{1}{2}\left | x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3} \right |$

$\triangle = \dfrac{1}{2}\left | (x_{1}- x_{3}) y_{2}+(x_{2}- x_{1}) y_{3}+(x_{3}- x_{2}) y_{1} \right |$

Then, area of the triangle the coordinates of whose angular points are respectively (1, 3), ( - 7, 6) and (5, - 1) is

$\triangle = \dfrac{1}{2}\left | (1-5) 6+(-7-1)(-1)+(5-(-7)) 3 \right |$

$\Rightarrow \triangle = \dfrac{1}{2}\left | -24+8+36 \right |$

$\Rightarrow \triangle =10$

Prove that the straight line x + y = 1 touches the parabola $y = x - x^2 .$

straight line

$x+y=1$
parabola

$y=x-x^{2}$

putting value of y in parabola eqn

$1-x=x-x^{2}$

$x^{2}-2x+1=0$

$(x-1)^{2}=0$

$x=1$
we get only one value of x
it means given straight line touches the parabola

Find the slope of the line joining the points $(4, -8)$ and $(5, -2)$ .

Let

$A(4,-8)$ and

$B(5,-2)$ be two points.

Slope of the line

$=\cfrac{y_2-y_1}{x_2-x_1}$

$=\cfrac{-2-(-8)}{5-4}$

$=\cfrac{-2+8}{1}$

$=6$

Find the perpendicular distance of the point $(-1,1)$ from the line $12(x+6)=5(y-2)$ .

The given equation of the line is

$12(x+6)=5(y-2)$ .

$12x+72=5y-10$

$12x-5y+82=0$

On comparing this equation with general equation of line

$Ax+By+C=0$ , we get,

$A=12, B=-5, C=82$

We know, perpendicular distance (

$d$ ) of a line

$Ax+By+C=0$ from a point

$(x_{1},y_{1})$ is given by

$d=\dfrac{|Ax_{1}+By_{1}+C|}{\sqrt{A^2+B^2}}$

Therefore, the distance of the point

$(-1,1)$ from given line is :

$=\dfrac{|12(-1)+(-5)(1)+82|}{\sqrt{(12)^2+(-5)^2}}=\dfrac{65}{13}=5$ units

$(1, 30^o), (2, 60^o)$ and $(3, 90^o)$

Given polar co-ordinates are

(i)

$(1 , 30^{\circ})$

Then,

$r = 1$ and

$\theta = \dfrac{\pi}{6}$

$x = rcos{\theta}= 1\times cos\left(\dfrac{\pi}{6}\right) =\dfrac{\sqrt{3}}{2}$

$y = rcos{\theta}= 1\times sin\left(\dfrac{\pi}{6}\right) =\dfrac{1}{2}$

$A \left(\dfrac{\sqrt{3}}{2} , \dfrac{1}{2}\right)$

(ii)

$(2 , 60^{\circ})$

Then,

$r = 2$ and

$\theta = \dfrac{\pi}{3}$

$x = rcos{\theta}= 2\times cos\left(\dfrac{\pi}{3}\right) =1$

$y = rcos{\theta}= 2\times sin\left(\dfrac{\pi}{3}\right) =\sqrt{3}$

$B (1 , \sqrt{3})$

(iii)

$(3 , 90^{\circ})$

Then,

$r = 1$ and

$\theta = \dfrac{\pi}{2}$

$x = rcos{\theta}= 3\times cos(\dfrac{\pi}{2}) =0$

$y = rcos{\theta}= 3\times sin(\dfrac{\pi}{2}) =3$

$C (0 , 3)$

Now we have points

$A \left(\dfrac{\sqrt{3}}{2} , \dfrac{1}{2}\right)$ ,

$B (1 , \sqrt{3})$ and

$C (0 , 3)$

Then, area of triangle is given by the formula

$\triangle=\dfrac{1}{2}\left | x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3} \right |$

$A(\triangle ABC)=\dfrac{1}{2}\left | \dfrac{3}{2}+ 3+0-\dfrac{1}{2} -0-\dfrac{3\sqrt{3}}{2} \right |$

$\Rightarrow A(\triangle ABC)= \dfrac{8-3\sqrt{3}}{4}$

$\left ( -a, \dfrac{\pi}{6} \right ), \left ( a, \dfrac{\pi}{2} \right )$ , and $\left ( -2a, -\dfrac{2\pi}{2} \right )$

Given polar co-ordinates

$(i)$

$\left ( -a, \frac{\pi}{6} \right )$

Then,

$r = -a$ and

$\theta = \dfrac{\pi}{6}$

$x = r \cos {\theta}= -a\times \cos \left(\dfrac{\pi}{6}\right) =\dfrac{-\sqrt{3}a}{2}$

$y =r \sin {\theta}= -a\times \sin \left(\dfrac{\pi}{6}\right) =\dfrac{-a}{2}$

$A \left(\dfrac{-\sqrt{3}a}{2} , \dfrac{-a}{2}\right)$

$\left(ii\right)$

$\left(-2a , -\pi \right)$

Then,

$r = -2a$ and

$\theta =- \pi$

$x = r\cos {\theta}= -2a\times \cos \left(-\pi\right) =2a$

$y = r\sin {\theta}= -2a\times \sin \left(-\pi\right) = 0$

$B \left(2a , 0\right)$

$\left(iii\right)$

$\left(a , \dfrac{\pi}{2}\right)$

Then,

$r = a$ and

$\theta = \dfrac{\pi}{2}$

$x = r\cos {\theta}= a\times \cos \left(\dfrac{\pi}{2}\right) =0$

$y = r\sin {\theta}= a\times \sin \left(\dfrac{\pi}{2}\right) =a$

$C \left(0 , a\right)$

Now we have points

$A \left(\dfrac{-\sqrt{3}a}{2} , \dfrac{-a}{2}\right)$ ,

$B \left(2a , 0\right)$ and

$C \left(0 , a\right)$

Area of triangle having angular points

$\left(x_{1} , y_{1} \right)$

$\left(x_{2} , y_{2}\right)$

$\left(x_{3} , y_{3}\right)$ is given by the formula

$\triangle=\dfrac{1}{2}\left | x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3} \right |$

Then,

$A(\triangle ABC)=\dfrac{1}{2}\left | 0+2a^{2}+0-\left(-a^{2}\right)-0+\dfrac{\sqrt{3}a^{2}}{2} \right |$

$\Rightarrow A(\triangle ABC)= a^{2} \left(\dfrac{6+\sqrt{3}}{4} \right)$

Find the equation of the straight line which passes through the point $(2,-3)$ and the point of intersection of the lines $x+y+4=0$ and $3x-y-8=0$ .

Any line through the intersection of the lines

$x+y+4=0$ and

$3x-y-8=0$ has the equation :

$(x+y+4)+\lambda (3x-y-8)=0$

This will pass through

$(2,-3)$ if

$(2-3+4)+\lambda (6+3-8)=0$

$3+\lambda =0$

$\lambda =-3$

Putting the value of

$\lambda$ , we get the equation of the required line as

$2x-y-7=0$ .

Show that the line joining the points $A(1, -1, 2)$ and $B(3, 4, -2)$ is perpendicular to the line joining the points $C(0, 3, 2)$ and $D(3, 5, 6)$ .

1298813_870258_ans_abf4981a91754ed68407f3209a751c96.jpg

$(2, 30^o)$ and $(4, 120^o)$

(i)

$(2, 30^{\circ}$ )
Then,

$r= 2$ and

$\theta = 30^{\circ}$

$x= rcos(30^{\circ})$ =

$2\times \dfrac{\sqrt{3}}{2}$ =

$\sqrt{3}$

$y= rsin(30^{\circ})= 2\times \dfrac{1}{2} = 1$

(ii)

$(4, 120^{\circ}$ )
Then,

$r= 4$ and

$\theta = 120^{\circ}$

$x= rcos(120^{\circ}) =-4sin(30^{\circ})=-2$

$y= rsin(120^{\circ})=4cos(30^{\circ}) = 2\sqrt{3}$

Now applying distance formula

$d =\sqrt{(\sqrt{3}+2)^{2}+(1-2\sqrt{3})^{2}}$

$\Rightarrow d=\sqrt{20}$

$\Rightarrow d= 2\sqrt{5}$

If be the origin, and if the coordinates of any two points $P_1$ and $P_2$ respectively $(x_1, y_2)$ and $(x_2, y_2)$ , prove that
$OP_1 \cdot OP_2\cdot cos P_1OP_2=x_1x_2+y_1y_2$

$cosP_{1}OP_{2}=\dfrac{(OP_{1})^{2}+(OP_{2})^{2}-(P_{1}P_{2})^{2} }{2(OP_{1})(OP_{2})}$

We have points

$P_{1} (x_{1} , y_{1})$ ,

$P_{2} (x_{2} , y_{2})$ and

$O (0 , 0)$
Then,

$OP_{1}=\sqrt{x_{1}^{2}+y_{1}^{2}}$

$OP_{2}=\sqrt{x_{2}^{2}+y_{2}^{2}}$

$P_{1}P_{2}=\sqrt{(x_{1}-x_{2})^{2} +(y_{1}-y_{2})^{2} }$

Substituting the above results in the given equation, we get

$OP_{1}.OP_{2}.cos(P_{1}OP_{2})=\dfrac{(x_{1}^{2}+y_{1}^{2})+(x_{2}^{2}+y_{2}^{2})-[(x_{1}-x_{2})^{2}+( y_{1}-y_{2})^{2} ]}{2}=x_{1}x_{2}+y_{1}y_{2}$

Hence proved

Find the area of the triangle formed by the lines
$y^{2}-9\ xy+18x^{2}=0$ and $y=9$

The sides of the triangle are

$y-3x=0$ and

$y-6x=0$ and

$y=9$ .

Solving in pairs, its vertices are
$$(0,0); \left( \dfrac { 3 }{ 2 } ,9 \right) ;\left(
3,9 \right)$$

or $$\triangle =\dfrac{ 1 }{ 2 }(x_{ 1 }y_{ 2 }-x_{ 2 }y_{
1 })=\dfrac{ 27 }{ 4 }sq.$$ units.

If the point $(x,y)$ is equidistant from the points $(a+b,b-a)$ and $(a-b,a+b),$ prove that $bx=ay$

Let P(x,y), Q(a+b,b-a) and R(a-b,a+b) be the given points. Then,PQ=PR

$\Rightarrow \sqrt { { \left\{ x-\left( a+b \right) \right\} }^{ 2 }+{ \left\{ y-\left( b-a \right) \right\} }^{ 2 } } =\sqrt { { \left\{ x-\left( a-b \right) \right\} }^{ 2 }+{ \left\{ y-\left( a+b \right) \right\} }^{ 2 } }$

$\Rightarrow { \left\{ x-\left( a+b \right) \right\} }^{ 2 }+{ \left\{ y-\left( b-a \right) \right\} }^{ 2 }={ \left\{ x-\left( a-b \right) \right\} }^{ 2 }+{ \left\{ y-\left( a+b \right) \right\} }^{ 2 }$

$\Rightarrow { x }^{ 2 }-2x\left( a+b \right) +{ \left( a+b \right) }^{ 2 }+{ y }^{ 2 }-2y\left( b-a \right) +{ \left( b-a \right) }^{ 2 }$

$={ x }^{ 2 }+{ \left( a-b \right) }^{ 2 }-2x\left( a-b \right) +{ y }^{ 2 }-2y\left( a+b \right) +({ a+b) }^{ 2 }$

$\Rightarrow -2x\left( a+b \right) -2y\left( b-a \right) =-2x\left( a-b \right) -2y\left( a+b \right)$

$\Rightarrow ax+bx+by-ay=ax-bx+ay+by$

$\Rightarrow 2bx=2ay\Rightarrow bx=ay$

REMARK-We know that a point which is equidistant from point P and Q lies on the
perpendicular bisector of PQ. Therefore, bx=ay is the equation of the perpendicular
bisector of PQ

If the area of triangle formed by points $A(x,y), B(1,2)$ and $C(2,1)$ is $68$ units then show that $x+y=15$ .

Let the given point is

$A(x,y),B(1,2),C(2,1).$

Area of a triangle

$ABC =\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$6=\dfrac{1}{2}[x(2-1)+1(1-y)+2(y-2)]$

$12 = x + 1 - y + 2y - 4$

$x + y - 3 = 12$

$x+y=15$

Hence, proved.

Find the angle which the straight line $y = \sqrt 3 x - 4$ makes with y-axis.

We have,

$y=\sqrt{3}x-4$

Slope =

$m=\sqrt{3}$

$tan$

$\theta$ =

$\sqrt{3}$

$\theta = 60^{o}$

This is the angle made by the line with the x-axis.

Therefore, the angle made by the line with the y-axis is

$90-60=30^{o}$

Find the area of a triangle whose vertices are (3,8), (-4,2) and (5,-1).

Let A(3,8), B(-4,2) and C(5,-1) be the vertices of the given

$\triangle$ ABC. Then,

$(x_{1}=3,y_{1}=8),(x_{2}=-4,y_{2}=2),(x_{3}=5,y_{3}=-1)$

Area of

$\triangle$ ABC =

$\dfrac{1}{2}|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|$

$=\dfrac{1}{2}|3(3)-4(-9)+5(6)|$

$=\dfrac{1}{2}|9+36+30|=\dfrac{75}{2}=37.5 \ sq. units$

Find the acute angle between the lines $2x-y+3=0$ and $x-3y+2=0$ .

Let the slope of the lines

$m_1=2$ and

$m_2=\frac{1}{3}$

And

$tan{\theta}=|\dfrac{m_1-m_2}{1+m_1m_2}|$

$=\left|\dfrac{\dfrac{1}{3}-2}{1+\dfrac{1}{3}.2} \right|=1$

$\theta=45^{\circ}$

Find the perpendicular distance of $(0,0)$ from $3x+4y=12$ .

Given the point is

$(0,0)$ and the line

$3x+4y=12$ ......(1).

Now the perpendicular distance of

$(0,0)$ to (1) is

$\left|\dfrac{0.3+0.4-12}{\sqrt{3^2+4^2}}\right|=\dfrac{12}{5}=2.4$ units.

How many equilateral triangles of side 2a with one vertex at origin and side along the x-axis is possible.

Four equilateral triangles are possible in the four quadrants.

$A$ is a point on $x-$ axis with abscissa $-8$ and $B$ is point on $y-$ axis with coordinate $15$ . Find distance $AB$ .

$A=(-8,0),B=(0,15)$

$AB=\sqrt{(-8,-0)^{2}+(0-15)^{2}}$

$=\sqrt{64+225}=17 units$

Find the length of sides of a triangle whose vertices are $(1, -1), (0, 4)$ and $(-5, 3)$ .

Vertices are

$(1, -1), (0, 4)$ and

$(-5, 3)$ .

$AB=\sqrt {(0-1)^2+(4+1)^2}=\sqrt{1+25}=\sqrt {26}$

$BC=\sqrt {(0-5)^2+(4-3)^2}=\sqrt{25+1}=\sqrt {26}$

$AC=\sqrt {(1+5)^2+(-1-3)^2}=\sqrt{36+16}=\sqrt {52}$

Find the distance of a point P(x,y) from the origin.

The distance of p(x, y) from the origin.

Solution: let p(x, y) be a point in 'x, y' coordinate system.

By distance formula

$OP=\sqrt{(x-0)^2+(y-0)^2}$

${OP=\sqrt{x^2+y^2}}$ .

1252952_1055003_ans_d9406f4a48684022a686075a2edbae04.png

Determine the ratio in which the point $P(3, 5)$ divides the join of $A(1, 3)$ & $B(7, 9)$ .

(Refer to Image)

$\Rightarrow P \Rightarrow \left(\cfrac {7k+1}{k+1},\cfrac {9k+3}{k+1}\right)=(3,5)$

$\Rightarrow 7k+1=3(k+1)$

$\Rightarrow k=\cfrac {1}{2}$

$\therefore$ k divides in

$1:2$ .

1210355_1095397_ans_80c77d4a1f8e4d3fa21c6096edd9a623.JPG

Find the area of rhombus if its vertices are $(3, 0), (4, 5), (-1, 4)$ and $(-2, -1)$ taken in order.
[Hint: Area of rhombus $=\dfrac{1}{2}\times$ product of its diagonals.]

We know that,

Area of Rhombus

$= \dfrac{1}{2}$ (Product of its diagonals)

Given co-ordinates of rhombus are

$(3,0),(4,5),(-1,4)$ and

$(-2,-1)$

Diagonal

$1 = \sqrt{(3+1)^{2} + (0-4)^{2}} = 4\sqrt2$

Diagonal

$2 = \sqrt{(4+2)^{2}+(5+1)^{2}} = 6\sqrt2$

$\therefore$ Area of rhombus

$=\dfrac{1}{2}( 4\sqrt2 \times 6\sqrt2)$

$= 24$ unit

$^{2}$

If the point $A(-2,1),\ B(a,b)$ and $C(4,-1)$ are collinear of $a-b=1$ . Find the value of $a$ of $b$ .

Given that points

$(-2,1), (a,b)$ and

$(4,-1)$ are collinear.

Using the point-slope formula,

$\dfrac{b-1}{a-(-2)} = \dfrac{-1-1}{4-(-2)}$

$\Rightarrow \dfrac{b-1}{a+2} = \dfrac{-1}{3}$

$\therefore 3b-3 = -a-2$

$a+3b=1$

Also,

Given that

$a-b=1$

Using both the equations, we get

$a = 1 , b = 0$

The length of line PQ is 10 units and the coordinates of P are $\left( {2, - 3} \right)$ ; calculate the coordinates of point Q, if its abscissa is 10.

The length of line pl is 10 units

and the coordinate, p are (2,-3)

;calculate the coordinate of point

l, if abscissa is 10.

$\Rightarrow$ pl = 10 units

$10 = \sqrt{(y-y_{1})^{2} + (x_{2}-x_{1})^{2}}$

$10 = \sqrt{(-3-y_{1})^{2}+ (2-10)^{2}}$

$10^{2} = (3+y_{1})^{2}+64$

$(3+y_{1})^{2} = 100-64 = 36$

$3+y = 6$

$y = 63 = 3$

y = 3

D =( 2,-3) & l = (10,3)

1192680_1103329_ans_a86df4f40507448b99e7261971f47003.jpg

Find the value of a , if the distance between points $P(a,-8,4)$ and $Q(-3,-5,4)$ is 5.

Given that

$P\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( a,-8,4 \right)$

$Q\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( -3,-5,4 \right)$

And distance $PQ=d=5$

Then, we know that,

Distance of

$PQ=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}}$

$PQ=\sqrt{{{\left( a+3 \right)}^{2}}+{{\left( -8+5 \right)}^{2}}+{{\left( 4-4 \right)}^{2}}}$

$5=\sqrt{{{\left( a+3 \right)}^{2}}+{{\left( -3 \right)}^{2}}}$

Squaring both side and we get,

$25={{\left( a+3 \right)}^{2}}+9$

$25-9={{\left( a+3 \right)}^{2}}$

${{\left( a+3 \right)}^{2}}=16$

$a+3=4$

$a=4-3$

$a=1$

If the equation of the locus of a point equidistant from the points $\left( {{a_1},{b_1}} \right)$ and $\left( {{a_2},{b_2}} \right)$ is $\left( {{a_1} - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y + c = 0$ , then the value of $c$ is

Let (x,y) be equidistant from

$(a_{1},b_{1})$ and

$(a_{2},b_{2})$

Then using distance formula

$\left ( a_{1}-x \right )^{2}+\left ( b_{1}-y\right )^{2}=\left ( a_{2}-x \right )^{2}+\left ( b_{2}-y\right )^{2}$

pn solving above , we will get

$2( a_{1}- a_{2})x+2( b_{1}- b_{2})y+[a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2}]$

Divide whole equation by 2

on comparing with

$(a_{1}- a_{2})x+( b_{1}- b_{2})y+c=0$

$\Rightarrow c=\dfrac{a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2}}{2}$

Find the equation of the line passing through the point $(2, 3)$ and making intercept of length $3$ unit between the lines $y + 2x = 2$ and $y + 2x = 5$ .

Dist AB

$=\dfrac{|c_1-c_2|}{\sqrt{1-m^2}}$

$=\dfrac{|5-2|}{\sqrt{1+2^2}}$

$=\dfrac{3}{\sqrt{5}}$

$AC=3$ given

$\therefore CB=\sqrt{AC^2-AB^2}$ [Pythagoras Theorem]

$=\dfrac{6}{\sqrt{5}}$

$\therefore \tan\theta =\dfrac{\dfrac{3}{\sqrt{5}}}{\dfrac{6}{\sqrt{5}}}=\dfrac{1}{2}$

$\therefore \theta =\tan^{-1}\dfrac{(m_1-m_3)}{1+m_1m_3}$

$[m_3\rightarrow s1]$

$\Rightarrow \tan\theta =\dfrac{2-m_3}{1+2m_3}$

$\Rightarrow \dfrac{1}{2}=\dfrac{2-m_3}{1+2m_3}$

$\Rightarrow m_3=1$

$\dfrac{y-3}{x-2}=m_3$

$\Rightarrow \dfrac{y-3}{x-2}=1$

$\Rightarrow y-x=1$ .

1183213_1155144_ans_5f4ec4dc8cce4c1ab16bd95e0eb01534.jpg

If the point P(x,y) is equidistant from the points A(5,1) and B(-1,5), prove that $3x=2y$ .

Let the point P(x,y) be equidistant from the points A(5,1) and B(-1,5). then,

$PA=PB$

$PA^2=PB^2$

$(x-5)^2+(y-1)^2=(x+1)^2+(y-5)^2$

$x^2+y^2-10x-2y+26=x^2+y^2+2x-10y+26$

$12x-8y=0$

$12x=8y$

$3x=2y$

Find the values of x for which the distance between the points P(4,-5) and Q(12,x) is 10 units.

$PQ=10$

$PQ^2=100$

$(12-4)^2+[x-(-5)]^2=100$

$8^2+(x+5)^2=100$

$(x+5)^2=36$

$x+5=\pm 6$

$x=1$ or

$-11$

Using slopes, find the value of $x$ so that the points $(6,\ -1),\ (5,\ 0)$ and $(x,\ 3)$ are collinear.

Let

$A \left( 6, -1 \right), B \left( 5, 0 \right)$ and

$C \left( x, 3 \right)$

Given that

$A, B$ and

$C$ are collinear.

$\therefore$ Slope of line

$AB =$ Slope of line

$BC$

$\cfrac{ 0 - \left( -1 \right)}{5 - 6} = \cfrac{3 - 0}{x - 5}$

$\cfrac{1}{-1} = \cfrac{3}{x - 5}$

$\Rightarrow x - 5 = -3$

$\Rightarrow x = 5 - 3 = 2$

Hence the value of

$x$ is

$2$ .

The co-ordinate of the mid-points of the sides of a triangle ABC are $D(2, 1), E(5, 3)$ and $F(3, 7)$ . Find the lengths and equations of its sides.

Let

$A(x_1,y_1) , B(x_2,y_2) , C(x_3,y_3)$

$4 = x_1 + x_2..........(i)$ and

$2 = y_1 + y_2.....(iv)$

$10 = x_1 + x_3..........(ii)$ and

$6 = y_1 + y_3.....(v)$

$6 = x_2+x_3............(iii)$ and

$14 = y_2 + y_3.....(vi)$

Subtracting

$(iii)$ from

$(ii)$ ,we get

$4 = x_1 - x_2$

Solving it with

$(i)$ , we get

$x_1 =4 , x_2 = 0,x_3 = 6$

Similarly,

$y_1 = -3 , y_2 = 5 , y_3 = 9$

$AB = \sqrt{16+64}$ and equation,

$(y+3) = \cfrac{5+3}{0-4}(x-4)$

$AB = \sqrt{80}$ and

$-y - 3 = 2x-8 \implies y = -2x+5$

$AC = \sqrt{2^2+12^2}$ and equation

$(y+3) = \cfrac{9+3}{6-4} (x-4)$

$AC = \sqrt{148}$ and

$y+3 = 6x-24$

$BC = \sqrt{60}$ and

$3y-15 = 2x$

Find the angle between the lines joining the points $(0,\ 0),\ (2,\ 3)$ and $(2,\ -2),\ (3,\ 5)$ .

slope of line L1 =

$\dfrac{3-0}{2-0} = \dfrac{3}{2}$

slope of line L2 =

$\dfrac{5+2}{3-2} = 7$

Angle between

$\displaystyle L_{1}, L_{2} = tan^{-1}\left | \dfrac{m_{1}-m_{2}}{1+m_{1}m_{2}} \right | = tan^{-1}\left | \dfrac{\dfrac{3}{2}-7}{1+\dfrac{3}{2}(7)} \right |$

$\displaystyle = tan^{-1}\frac{11/2}{23/2} = tan^{-1}\frac{11}{23}$

Find the area of $\triangle PQR$ whose vertices are $P(-1, 1) Q(0, 5) R(3, 2)$

$Area=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

By substituting the values we get,

$=\dfrac{1}{2}[-1(5-2)+0(2-1)+3(1-5)]$

$=\dfrac{1}{2}[-1(3)+0+3(-4)]$

$=7.5$ sq units

Find the point on y-axis which is equidistant from the points (5,-2) and (-3,2).

Let

$P(0,y)$ be any point on y-axis.

Let

$A(5,-2)$ and

$B(-3,2)$ . Then,

$PA=PB$

$PA^2=PB^2$

$(5-0)^2+(-2-y)^2=(-3-0)^2+(2-y)^2$

$25+4+y^2+4y=9+4+y^2-4y$

$8y=-16$

$y=-2$

Therefore, the required point is

$(0,-2)$ .

Prove that area if triangle $ABC$ with vertices $A(3,2), B(11,8), C(8,12)$ is $25$ square units.

Given that,

$A = (3, 2)$

$B = (11, 8)$

$C = (8, 12)$

$Area(\triangle ABC) = \dfrac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

$= \dfrac{1}{2} |3(8-12) + 11(12-2) + 8(2-8)|$

$= \dfrac{1}{2} |3(-4) + 11(10) + 8(-6)|$

$= \dfrac{1}{2} |-12 + 110 - 48|$

$= \dfrac{1}{2} (50)$

$= 25$ Sq.units

Prove that the distance between the origin and the point $(-6,\ -8)$ is twice the distance between the points $(4,\ 0)$ and $(0,\ 3)$ .

Let

${d}_{1}$ and

${d}_{2}$ be the distance between the points

$\left( 0, 0 \right) \& \left( -6, -8 \right)$ and

$\left( 4, 0 \right) \& \left( 0, 3 \right)$ respectively.

To prove:-

${d}_{1} = 2 {d}_{2}$

As we know that the distance between the points

$\left( {x}_{1}, {y}_{1} \right)$ and

$\left( {x}_{2}, {y}_{2} \right)$ is given as-

$d = \sqrt{{\left( {x}_{2} - {x}_{1} \right)}^{2} + {\left( {y}_{2} - {y}_{1} \right)}^{2}}$

Therefore,

${d}_{1} = \sqrt{{\left( -6 - 0 \right)}^{2} + {\left( -8 - 0 \right)}^{2}}$

${d}_{1} = \sqrt{{\left( -6 \right)}^{2} + {\left( -8 \right)}^{2}}$

$\Rightarrow {d}_{1} = \sqrt{36 + 64}$

$\Rightarrow {d}_{1} = \sqrt{100} = 10$

$\Rightarrow {d}_{1} = 2 \times 5 ..... \left( 1 \right)$

Now,

${d}_{2} = \sqrt{{\left( 3 - 0 \right)}^{2} + {\left( 0 - 4 \right)}^{2}}$

${d}_{2} = \sqrt{{\left( 3 \right)}^{2} + {\left( -4 \right)}^{2}}$

$\Rightarrow {d}_{2} = \sqrt{9 + 16}$

$\Rightarrow {d}_{2} = \sqrt{25} = 5 ..... \left( 1 \right)$

From

${eq}^{n} \left( 1 \right) \& \left( 2 \right)$ , we have

${d}_{1} = 2 \times {d}_{2} \; \left( \because {d}_{2} = 5 \right)$

Hence proved.

The medians AD and BE of a triangle with vertices $A(0,b),B(0,0)$ and $C(a,0)$ are perpendicular to each other, if

A median in a triangle is a line passing thought a vertex and through the midpoint of the side opposite to the vertex.

$D$ is the midpoint of side

$BC$

$D=(\dfrac {a}{2},0)$

slope of

$AD= \dfrac {b-0}{0-a/2} = \dfrac {-2b}{a}$

$E$ is the midpoint of side

$AC$

$D=(\dfrac {a}{2},\dfrac{b}{2})$

slope of

$BE= \dfrac {0-b/2}{0-a/2} = \dfrac {b}{a}$

If two lines are perpendicular, product of their slopes is -1.

$(\dfrac {-2b}{a} )(\dfrac {b}{a})=-1$

$2b^2=a^2$

$a=\pm b\sqrt {2}$

Find the area of the triangle formed by the point $(0, 0) (4, 0) (0, 3)$ .

As we can see in the diagram the triangle formed is a right-angled triangle.

$\therefore$ Area of a right-angled triangle =

$\dfrac{1}{2}\times base\times height$

Also, Area of the right-angled triangle formed by the points

$(0, 0), (x, 0), (0, y) = \dfrac{1}{2}|xy|$ sq.units

$\Rightarrow$ Area of the right-angled triangle formed by the points

$(0, 0), (4, 0), (0, 3) = \dfrac{1}{2}|4\times 3|$ sq.units

$= \dfrac{1}{2}\times(12)$

$= 6$ sq.units

Find the equation of line equally inclined to coordinate axes and passes through $(-5, 1, -2)$ .

d,r of the line making equal angle with co-ordinate axis is (1,1,1)

d,r of line

$\vec{b}$ is (1,1,1) & point on line

$\vec{a}$ is (-5,1,-2)

$\therefore$ equation of line is:

$\frac{x-(-5)}{1} = \frac{y-1}{1} = \frac{z-(-2)}{1}$

$\frac{x+5}{1} = \frac{y-1}{1} = \frac{z+2}{1}$

1193828_1295575_ans_b7ed77bf700e4c8f95b21f398750d882.jpg

Find the area (sq.units) of the triangle whose coordinates are
$\left(2,3\right),\left(-1,0\right),\left(2,-4\right)$ .

1898549_1189987_ans_801c0a1b18e74f1890489403a5eb2767.jpeg

Find a point on $y-axis$ which is equidistant from $( - 5, - 2)$ and $( 3, 2)$ .

Let the point

$A$ on y-axis be

$(0,y)$

Given that point

$A$ is equidistant from

$(-5,-2)\ and\ (3,2)$

Hence, using distance formula we have

$\sqrt{(0-(-5))^{2}+(y-(-2))^2}=\sqrt{(0-3)^{2}+(y-2)^{2}}$

$\sqrt{25+(y+2)^{2}}=\sqrt{9+(y-2)^{2}}$

squaring both sides we get

$25+(y^2+4+4y)=9+(y^{2}+4-4y)$

$4y+4y=y^{2}+4+9-y^2-4-25$

$8y=-16$

$y=\dfrac{-16}{8}=-2$

Hence, the point

$A$ on y-axis is

$(0,-2)$

Find the inclination of a line whose slope is
(i) $1$
(ii) $-1$
(iii) $\sqrt 3$
(iv) $-\sqrt 3$
(v) $\frac{1}{{\sqrt 3 }}$

ACW

$\rightarrow$ anticlockwise

(i)

$\tan\theta =1$

$\Rightarrow \theta =45^o$ ACW

(ii)

$\tan\theta =-1$

$\Rightarrow \theta =135^o$ ACW

(iii)

$\tan\theta =\sqrt{3}$

$\Rightarrow \theta =60^o$ ACW

(iv)

$\tan\theta =-\sqrt{3}$

$\Rightarrow \theta =120^o$ ACW

(v)

$\tan\theta =\dfrac{1}{\sqrt{3}}$

$\Rightarrow \theta =30^o$ ACW.

1213298_1301815_ans_3b573ecea093401aa14edda3f3e3e8f3.JPG

Find the coordinates of a point on $x+y+3=0$ , whose distance from $x+2y+2=0$ is $\sqrt{5}$ .

1339996_1332929_ans_240b5337bd9e4b07b32e43a71680b957.jpg

The line joining two points $A ( 2,0 ) , B ( 3,1 )$ is rotated about $A$ in anticlockwise direction through an angle of $15 ^ { \circ } .$ Find the equation of the line in the new position. If $B$ goes to $C$ in the new position, what will be the coordinates of $C$ ?

A(2, 0) B (3, 1)

$m_{AB}=\frac{1-0}{3-2}=1\Rightarrow m=1\Rightarrow 45^{\circ}$

angle of inclination of AC =

$60^{\circ}$

r = AB = AC

$r=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$

$r=\sqrt{2}$

coordinate of c

$(r cos 60^{\circ}, r sin 60^{\circ})$

$(\sqrt{2}\times \frac{1}{2},\sqrt{2}\times \frac{\sqrt{3}}{2})$

$(\frac{1}{\sqrt{2}},\frac{\sqrt{3}}{\sqrt{2}})$

1210671_1325416_ans_881b8d397a534d87972b27896cc1a197.jpg

Find the slope of the line $\dfrac{x}{3}+\dfrac{y}{2}=1$ .

now we have,

$\begin{array}{l} =2x+3y=6 \\ \Rightarrow 3y=-2x+6 \\ \Rightarrow y=\frac { { -2 } }{ 3 } x+2 \\ \Rightarrow y=\frac { { -2 } }{ 3 } \\ hence,\, \, the\, \, slope\, \, is\frac { { -2 } }{ 3 } \end{array}$

Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points $P ( 0 , - 4 )$ and $B ( 8,0 )$

$P(0 -4)$

$B(8, 0)$

Mid point of PB

$=\left(\dfrac{0+8}{2}, \dfrac{-4+0}{2}\right)$

$=\left(\dfrac{8}{2}, \dfrac{-4}{2}\right)=(4, -2)$

Slope of the line joining the origin and

$(4, -2)$ is

$\dfrac{-2-0}{4-0}=\dfrac{-2}{4}=-\dfrac{1}{2}$ .

Find the slope of the line whose inclination is
$5\pi/6$

b'Slope of line

$= tan\theta = tan 5\pi/ 6 = tan 150^o = -1 /\sqrt3$ '

Find the area of a triangle with sides $3\ cm,4\ cm,5\ cm.$

This triangle is a right

$-$ angled triangle

$,$ because

${3^2} + {4^2} = {5^2}$

Area of the triangle

$= \dfrac{1}{2} \times 3 \times 4 = 6\ cm^2$

Hence, this is the answer.

Check whether $(5, -2)$ , $(6, 4)$ and $(7, -2)$ are the vertices of an isosceles triangles.

Let ABC be the triangle

$A(5, -2), B(6, 4), C(7, -2)$

By distance formula

$AB=\sqrt{(5-6)^2+(-2-4)^2}=\sqrt{(-1)^2+(-6)^2}\sqrt{37}$

$BC=\sqrt{(6-7)^2+(4-(-2)^2}=\sqrt{(-1)^2+(6)^2}=\sqrt{37}$

$CA=\sqrt{(7-5)^2+(-2-(-2))^2}=\sqrt{(2)^2+0}=2$

Since

$AB=BC=\sqrt{37}$

Hence

$(5, -2), (6, 4)$ &

$(7, -2)$ are vertices of isosceles triangle.

1173501_1299864_ans_6074e49f10d54250bcf882167d060015.jpg

Find the slope of the line passing the two given points
$( a , 0 )$ and $(0, b )$

We have the given points

$\left( {a,0} \right)\,\,and\,\,\left( {0,\,b} \right)$

$Now, \\ Slope\, \left( m \right) =\dfrac { { { y_{ 2 } }-{ y_{ 1 } } } }{ { { x_{ 2 } }-{ x_{ 1 } } } }$

$\\ =\dfrac { { b-0 } }{ { 0-a } } \\ =\dfrac { { -b } }{ a }$

Hence, this is the answer.

A point moves such that its distance from the point (4,0) is half that of its distance from the line $x=16$ , find its locus.

1339998_1332932_ans_3ef977f31eff415b9c46c2398572c4e1.jpg

The distance of a point $P$ on x-axis from $A(11, 12)$ is $13$ units. Find the co-ordinates of $P.$

1350854_1346633_ans_78af4dc691d94291b170aeadbcf768f2.jpg

Find the distance between the following pairs of points:-
$(-5,7),(-1,3)$

Let

$A\left(-5,7\right)$ and

$B\left(-1,3\right)$ be the points of a straight line.

$AB=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$

$AB=\sqrt{{\left(-1+5\right)}^{2}+{\left(3-7\right)}^{2}}$

$=\sqrt{{4}^{2}+{\left(-4\right)}^{2}}$

$=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$ units.

The distance between the points $\left(0,0\right)$ and $\left(-4,3\right)$ is

Let

$A\left({x}_{1},{y}_{1}\right)=\left(0,0\right)$ and

$B\left({x}_{2},{y}_{2}\right)=\left(-4,3\right)$

$\therefore$ Distance

$AB=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$

$\therefore \,AB=\sqrt{{\left(-4-0\right)}^{2}+{\left(3-0\right)}^{2}}$

$=\sqrt{16+9}=\sqrt{25}=5$ units

Let A(6,4) and B(2,12) be two given points. Find the slope of a line perpendicular to AB.

1340011_1344494_ans_d897a6b153f94d24b1bb428ba131a6ac.jpg

Find the area of the triangle formed by the mid-points of the sides of $\triangle$ ABC, where A(3,2), B(-5,6) and C(8,3).

Area of

$\triangle$ ABC =

$\dfrac{1}{2} \begin{vmatrix} 3-(-5) & 2-6 \\ -5-8 & 6-3 \end{vmatrix}$

$=\dfrac{1}{2} \begin{vmatrix} 8 & -4 \\ -13 & 3 \end{vmatrix} =\dfrac{1}{2} |-28|=14$ sq. units

Hence, area of triangle formed by the mid-points of the sides of the

$\\triangle ABC$ =

$\\dfrac{1}{4}(14)=3.5$ sq. units

Find the slope of a non-vertical line $ax+by+c=0$

Given line is

$ax+by+c=0$

$\Rightarrow by=-ax-c$

$\Rightarrow y=\dfrac{-a}{b}x+\dfrac{-c}{b}$ is in the form of

$y=mx+c$

where slope

$=m=\dfrac{-a}{b}$

Find the distance between the points (3,-5) and (5,-1).

Let the given points be A(3,-5) and B(5,-1).

$AB=\sqrt{(5-3)^{2}+(-1+5)^{2}}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$ units

Find the area of triangle formed by the points (8,-5) , (-2,-7) and (5,1) .

We know,

Area of triangle =

$\dfrac{1}{2}|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|$

Therefore,

$A=\dfrac{1}{2}|8(-7-1)-2(1+5)+5(-5+7)|$

$A=\dfrac{1}{2}|-64-12+10|$

$A=\dfrac{1}{2}|-66|$

$A=33$

$sq.$

$units$

Show that the path of a moving point which remains at equal distance from the points $(2, 1)$ and $(-3, -2)$ is a straight line.

It is very well known that locus of point which is equidistant from the two given point is the perpendicular bisector of line joining those points .

It can be easily shown by making distance of point (say (h,k)) equal from the two given points i.e

$(h-2)^2+(k-1)^2=(h+3)^2+(k+2)^2$

which gives

$5h+3k+4=0$

Find the slope of the line joining the points (3, -2)and (-1,4)

Slope of the line joining

$(x_1 \, y_1)$ and

$(x_2\, y_2) =\frac {y_1-y_2}{x_1-x_2}$

$\therefore$ Slope of the line joining

$(3, -2)$ and

$(-1, 4)=\frac {-2-4}{3-(-1)}$

$=\frac {-6}{4}$

$\frac {-3}{2}$

1199641_1346046_ans_fed5b107434940e7bfa3ac0e5773bf08.JPG

The equation of line with slope $\dfrac 54$ nd passing through $(8,4)$

The slope of line is

$\dfrac 54$

The point is

$(8,4)$

The equation is

$y-4=\dfrac 54(x-8)\\4y-16=5x-40\\5x-4y-24=0$

Find the value k. for which the point (-1, 3) lies on the graph of the equation 2x - y + k = 0.

Given equation is

$2x-y+k=0$

Given point

$(-1,3)$

Accoding to question the point lies on line

$2(-1)-3+k=0\\-2-3+k=0\\k=5$

Find the distance between $\left(-2,-4\right)$ from the origin.

Distance between

$\left(-2,-4\right)$ from the origin

$=$ Distance between

$\left(-2,-4\right)$ from

$\left(0,0\right)$

$=\sqrt{ { \left( {x_2}-{ { x_1 } } \right) }^{ 2}+{ \left( {y_2 }-{ y_1 } \right) }^{ 2 } }$

$=\sqrt{{\left(-2-0\right)}^{2}+{\left(-4-0\right)}^{2}}$

$=\sqrt{4+16}$

$=\sqrt{20}=2\sqrt{5}$ units

Find the distance of the points P (-6,8) from the origin.

Distance between

$\left(-6,8\right)$ from the origin

$=$ Distance between

$\left(-6,8\right)$ from

$\left(0,0\right)$

$=\sqrt{{\left(-6-0\right)}^{2}+{\left(8-0\right)}^{2}}$ using distance formula

$=\sqrt{36+64}$

$=\sqrt{100}=10$ units

Find the distance between $\left(8,-8\right)$ from the origin.

Distance between

$\left(8,-8\right)$ from the origin

$=$ Distance between

$\left(8,-8\right)$ from

$\left(0,0\right)$

$=\sqrt{ { \left( {x_2}-{ { x_1 } } \right) }^{ 2}+{ \left( {y_2 }-{ y_1 } \right) }^{ 2 } }$

$=\sqrt{{\left(8-0\right)}^{2}+{\left(-8-0\right)}^{2}}$

$=\sqrt{64+64}$

$=\sqrt{2\times 64}=8\sqrt{2}$ units

Find the distance between $\left(-9,-1\right)$ from the origin.

Distance between

$\left(-9,-1\right)$ from the origin

$=$ Distance between

$\left(-9,-1\right)$ from

$\left(0,0\right)$

$=\sqrt{{\left(-9-0\right)}^{2}+{\left(-1-0\right)}^{2}}$ using distance formula

$=\sqrt{81+1}$

$=\sqrt{82}$ units.

Find the distance between the points $A(a,b)$ and $B(-a,-b)$

Given

$A(a,b),B(-a,-b)$

$AB=\sqrt{(a+a)^2+(b+b)^2}=\sqrt{4 a^2+4 b^2}=2\sqrt{a^2+b^2}$

Distance between

$A$ and

$B$ is

$2\sqrt{a^2+b^2}$ units

Fill in the blacks.
If the area of a triangle is 0 square units then the vertices of a triangle are _____

If the area of triangle is

$0$ sq. units, then the vertices of the triangle are collinear, i.e.,

$\begin{vmatrix} {x}_{1} & {y}_{1} & 1 \\ {x}_{2} & {y}_{2} & 1 \\ {x}_{3} & {y}_{3} & 1 \end{vmatrix} = 0$

Hence the required answer is collinear.

Prove that the point $\left( { - \dfrac{1}{{14}},\dfrac{{39}}{{14}}} \right)$ is the centre of the circle circumscribing the triangle whose vertices are $(1,1),(2,3),$ and $(-2,2)$

1563796_1466298_ans_3e719650960f43c4988b9ea57f149e13.png

If the vertices of a triangle are $(1,-1) , (-4,6), (-3,-5)$ then the area of triangle is :

$\begin{array}{l} Let\, the\, two\, po{ { ints } }\, be\, A\left( { 1,-1 } \right) ,\, \, B\left( { -4,6 } \right) \, and\, c\left( { -3,-5 } \right) \\ Now, \\ Area\, \, of\, triangle\, \, ABC=\frac { 1 }{ 2 } \left[ { { x_{ 1 } }\left( { { y_{ 2 } }-{ y_{ 3 } } } \right) +{ x_{ 2 } }\left( { { y_{ 3 } }-{ y_{ 1 } } } \right) +{ x_{ 3 } }\left( { { y_{ 1 } }-{ y_{ 2 } } } \right) } \right] \\ =\frac { 1 }{ 2 } \left[ { 1\left( { 6+5 } \right) -4\left( { -5+1 } \right) -3\left( { -1-6 } \right) } \right] \\ =\frac { 1 }{ 2 } \left[ { 11+16+21 } \right] \\ =\frac { 1 }{ 2 } \left[ { 58 } \right] \\ =\frac { { 58 } }{ 2 } \\ =29\, sq.units \end{array}$

Find the slope of the line $AB$ joining $A(a,0),B(0,b)$ ?

As we know that

Slope of line joining

$(x_1,y_1),(x_2,y_2)$ is

$\dfrac{y_2-y_1}{x_2-x_1}$

Slope of line

$AB$ joining

$A(a,0),B(0,b)$ is

$\dfrac{b-0}{0-a}=-\dfrac{b}{a}$

Find the shortest distance of (3,4) from origin.

As we know that

Distance between two point

$(x_1,y_1),(x_2,y_2)$ is

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance between

$(3,4)$ from origin is

$\sqrt{(3-0)^2+(4-0)^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

Check which at the following are solutions of the equation 7 x - 5 y = - 3i) (-1, -2) ii) (-4, -5)

1319091_1504085_ans_d152d90936754ce8a195c8a8799caee8.jpeg

Write the formula for area of triangle $ABC$ where A $\left( { x }_{ 1 },{ y }_{ 1 } \right) ,B\left( { x }_{ 2 },{ y }_{ 2 } \right)$ and $C\left( { x }_{ 3 },{ y }_{ 3 } \right)$

$A(x_{1}, y_{1}) B(x_{2},y_{2}) C(x_{3},y_{3})$

Area of

$\Delta = \frac{1}{2} { x_{1} (y_{2}-y_{3}) + x_{2} (y_{3}-y_{1}) + (y_{1}-y_{2})}$

Find the locus of a point, so that the join of (-5,1) and (3,2) subtends a right angle at the moving point.

Let

$P(h,k)$ be a moving point and let

$A(-5,1)$ and

$B(3,2)$ be given points. By the given condition

$\angle APB=90^o$

Therefore,

$\triangle$ APB is a righta angle triangle.

$AB^2=AP^2+PB^2$

$\Rightarrow (3+5)^2+(2-1)^2=(h+5)^2+(k-1)^2+(h-3)^2+(k-2)^2$

$\Rightarrow 65=2(h^2+k^2+2h-3k)+39$

$\Rightarrow h^2+k^2+2h-3k-13=0$

Hence, locus of (h,k) is

$x^2+y^2+2x-3y-13=0$ .

Find the area of the triangle with vertices as given below:
$(-1,-8),(-2,-3)$ and $(3,2)$

Given Points are

$(-1,-8),(-2,-3),(3,2)$

comparing with

$(x_1,y_1),(x_2,y_2),(x_3,y_3)$

Area of triangle is given as,

$A=\dfrac 12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) |$

$=\dfrac 12|-1(-3-2)-2(2+8)+3(-8+3) |$

$=\dfrac 12|5-20-15|$

$=15$

Show that the product of perpendiculars on the line $\dfrac {x}{a}\cos \theta +\dfrac {y}{b}\sin \theta =1$ from the points $(\pm \sqrt {a^2 -b^2}, 0)$ is $b^2$ .

Let

${p}_{1}$ and

${p}_{1}$ be the lengths of perpendicular draw from point

$P\left(\sqrt{{a}^{2}-{b}^{2}},0\right)$

$Q\left(-\sqrt{{a}^{2}-{b}^{2}},0\right)$ on the line

$\dfrac{x}{a}\cos \theta +\dfrac{y}{b} \sin \theta =1$ .

Then,

${ P }_{ 1 }=\left| \dfrac { \dfrac { \sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a } \cos { \theta } +\dfrac { 0 }{ b } \sin { \theta } -1 }{ \sqrt { \dfrac { \cos ^{ 2 }{ \theta } }{ { a }^{ 2 } } +\dfrac { \sin ^{ 2 }{ \theta } }{ { b }^{ 2 } } } } \right| =\left| \dfrac { \dfrac { \sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a } \cos { \theta \quad -1 } }{ \sqrt { \dfrac { \cos ^{ 2 }{ \theta } }{ { a }^{ 2 } } +\dfrac { \sin ^{ 2 }{ \theta } }{ { b }^{ 2 } } } } \right|$

${ P }_{ 2 }=$

$\left| \dfrac {- \dfrac { \sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a } \cos { \theta } +\dfrac { 0 }{ b } \sin { \theta } -1 }{ \sqrt { \dfrac { \cos ^{ 2 }{ \theta } }{ { a }^{ 2 } } +\dfrac { \sin ^{ 2 }{ \theta } }{ { b }^{ 2 } } } } \right| = \left| \dfrac { \dfrac { \sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a } \cos { \theta \quad +1 } }{ \sqrt { \dfrac { \cos ^{ 2 }{ \theta } }{ { a }^{ 2 } } +\dfrac { \sin ^{ 2 }{ \theta } }{ { b }^{ 2 } } } } \right|$

$\therefore \quad { P }_{ 1 }{ P }_{ 2 }=\left| \dfrac { \dfrac { \sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a } \cos { \theta \quad -1 } }{ \sqrt { \dfrac { \cos ^{ 2 }{ \theta } }{ { a }^{ 2 } } +\dfrac { \sin ^{ 2 }{ \theta } }{ { b }^{ 2 } } } } \right| \times \left| \dfrac { \dfrac { \sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a } \cos { \theta \quad +1 } }{ \sqrt { \dfrac { \cos ^{ 2 }{ \theta } }{ { a }^{ 2 } } +\dfrac { \sin ^{ 2 }{ \theta } }{ { b }^{ 2 } } } } \right|$

$\Rightarrow \ { P }_{ 1 }{ P }_{ 2 }=\left| \dfrac { \left( \dfrac { \sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a } \cos { \theta -1 } \right) \left( \dfrac { \sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a } \cos { \theta +1 } \right) }{ \dfrac { \cos ^{ 2 }{ \theta } }{ { a }^{ 2 } } +\dfrac { \sin ^{ 2 }{ \theta } }{ { b }^{ 2 } } } \right|$

$\Rightarrow \ { P }_{ 1 }{ P }_{ 2 }= \ \left| \dfrac { \dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ a } \cos ^{ 2 }{ \theta } -1 }{ \dfrac { \cos ^{ 2 }{ \theta } }{ { a }^{ 2 } } +\dfrac { \sin ^{ 2 }{ \theta } }{ { b }^{ 2 } } } \right| =\left| \dfrac { \left( { a }^{ 2 }-{ b }^{ 2 } \right) \cos ^{ 2 }{ \theta - } { a }^{ 2 } }{ { b }^{ 2 }\cos ^{ 2 }{ \theta +{ a }^{ 2 }\sin ^{ 2 }{ \theta } } } \right| { b }^{ 2 }$

$\Rightarrow \ { P }_{ 1 }{ P }_{ 2 }= \ \left| \dfrac { -{ b }^{ 2 }\cos ^{ 2 }{ \theta } -{ a }^{ 2 }\sin ^{ 2 }{ \theta } }{ { b }^{ 2 }\cos ^{ 2 }{ \theta } +{ a }^{ 2 }\sin ^{ 2 }{ \theta } } \right| { b }^{ 2 }={ b }^{ 2 }$

Find k, if R(1, -1), S (-2, k) and slope of line RS is -2.

Slope of line RS is -2
Slope of RS will be

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{k-(-1)}{-2-1}=\dfrac{k+1}{-3}=-2$

$\Rightarrow k+1=6$

$\Rightarrow k=5$

Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (-1, -1) are vertices of a right angles triangle.

The vertices of the given triangle are A(4, 4), B(3, 5) and C(-1, -1).
It is known that the slope (m) of a non-vertical line passing through the points

$\left (x_1, y_1 \right )$ and

$\left (x_2, y_2 \right )$ is given by

$m = \frac{y_2 - y_1} {x_2 - x_1}, x_2 \neq x_1$

$\therefore$ Slope of

$AB \left (m_2 \right ) = \frac{5 - 4} {3 - 4} = -1$

Slope of

$BC \left (m_2 \right ) = \frac{-1 - 5} {-3 - 3} = \frac{-6} {-4} = \frac{3} {2}$

Slope of

$CA \left (m_3 \right ) = \frac{4 + 1} {4 + 1} = \frac{5} {5} = +1$

This shows that line segments AB and CA are perpendicular to each other i.e. the given triangle is right-angled ar A(4, 4).
Thus, the points (4, 4), (3, 5) and (-1, -1) are the vertices of a right-angled triangle.

Find the angle between the x-axis and the line joining the points (3, -1) and (4, -2)

The slope of the line joining the points (3, -1) and (4, -2) is

$m = \frac{-2-(-1)} {4-3} = -2+1 = -1$
Now, the inclination

$(\theta)$ of the line joining the points (3, -1) and (4, -2) is given by

$\tan \theta = -1 \Rightarrow \theta = \left( 90^{0} + 45^{0} \right) = 135^{0}$
Thus, the angle between the x-axis and the line joining the points (3, -1) and (4, -2) is

$135^{0}$

Write the formula to find the area of triangle if the vertices is given

$Let A(x_{1},y_{1}) \space , B(x_{2},y_{2}) \space C(x_{3},y_{3})$

$Area \space of \space triangle \space ABC=| \dfrac{1}{2} [x_{1} (y_2 - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] |$

Fill in the blanks.
If point $A(-3,12),B(7,6)$ and $C(x,9)$ are collinear, then the value of $x$ is_______.

If three points are collinear there area must be zero

$Area = 0.5 [ x_1 ( y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2)]$ = 0

$0 = 0.5 [ -3 (6-9) + 7 (9-12) + x(12-6)]$

$0 = 0.5 [ -3 (-3) + 7 (-3) + x(6)]$

$0 = 0.5[9 - 21 + 6x]$

$0 = -12+6x$

$12=6x$

$x = 2$

Find the slope of a line, which passes through the origin, and the mid-point of the segment joining the points P(0, -4) and B(8, 0)

The coordinates of the mid-point of the line segment joining the points
P(0, -4) and B(8, 0) are

$\left (\frac{0 + 8} {2}, \frac{-4 + 0} {2} \right ) = \left (4, -2 \right )$
It is known that the slope (m) of a non-vertical line passing through the point

$\left (x_1, y_1\right)$ and

$\left (x_2, y_2\right)$ is given by

$m = \frac{y_2 - y_1} {x_2 - x_1}, x_2 \neq x_1$
Therefore, the slope of the line passing through (0, 0) and (4, -2) is

$\frac{-1 - 0 } {4 - 0} = \frac{-2} {4} = - \frac{1} {2}$
Hence, the required slope of the line is

$- \frac{1} {2}$

A point which is equidistant from the points A (5, 4) and B (1, 6) is (3,5)
If true then enter $1$ and if false then enter $0$

$Let\quad us\quad take\quad the\quad point\quad as\quad P(x,y)\quad which\quad should\quad be\quad equidistant\quad \\ from\quad the\quad given\quad points\quad A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 5,4 \right) \quad \& \quad B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 1,6 \right) .\\ Using\quad the\quad distance\quad formula\quad d=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } ,\quad we\quad get\quad \\ AP=\sqrt { { \left( { x }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }-{ y }_{ 1 } \right) }^{ 2 } } =\sqrt { { \left( x-5 \right) }^{ 2 }+{ \left( y-4 \right) }^{ 2 } } \quad and\\ BP=\sqrt { { \left( { x }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }-{ y }_{ 2 } \right) }^{ 2 } } =\sqrt { { \left( x-1 \right) }^{ 2 }+{ \left( y-6 \right) }^{ 2 } } .\\ Since\quad P\quad is\quad equidistant\quad from\quad the\quad points\quad A\quad \& \quad B,\\ AP=BP.\\ \Longrightarrow \sqrt { { \left( x-5 \right) }^{ 2 }+{ \left( y-4 \right) }^{ 2 } } =\sqrt { { \left( x-1 \right) }^{ 2 }+{ \left( y-6 \right) }^{ 2 } } \\ \Longrightarrow 2x-y-1=0........(i)\\ This\quad is\quad the\quad locus\quad (L)\quad of\quad P.\\ It\quad will\quad pass\quad through\quad the\quad mid\quad point\quad between\quad A\quad \& \quad B\\ the\quad co-ordinates\quad of\quad which\quad are\quad \\ \left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) =\left( \frac { 5+1 }{ 2 } ,\frac { 4+6 }{ 2 } \right) =\left( 3,5 \right) .\\ Putting\quad \left( x,y \right) =\left( 3,5 \right) \quad in\quad (i)\quad we\quad have\\ L.H.S.=2\times 3-5-1=0=R.H.S.\\ So\quad L\quad passes\quad through\quad the\quad mid\quad point\quad of\quad AB.\\ \therefore \quad L\quad is\quad the\quad perpendicular\quad bisector\quad of\quad AB.\\ There\quad are\quad infinite\quad number\quad of\quad points\quad on\quad L\quad one\quad of\quad which\\ is\quad (3,5).\\ Ans-There\quad are\quad infinite\quad number\quad of\quad points\quad on\quad L\quad one\quad of\quad which\\ is\quad (3,5).\\$

State whether the following statement is true or false.
Enter $1$ for true and $0$ for false
$\triangle$ ABC with vertices A (-2, 0), B (2, 0) and C (0, 2) is similar to $\triangle DEF$ with vertices D (-4, 0) E (4, 0) and F (0, 4).

$If\quad \Delta ABC\quad \& \quad \quad \Delta DEF\quad are\quad similar\quad then$

$\dfrac { AB }{ DE } =\dfrac { AC }{ DF } =\dfrac { BC }{ EF } .$

$We\quad can\quad obtain\quad the\quad above\quad lengths\quad by\quad the\quad distance\quad formula\quad$

$d=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } }$

$and\quad investigate\quad if\quad the\quad ratios\quad are\quad equal.$

$Statement-$

$\quad \Delta ABC\quad \& \quad \Delta DEF\quad are\quad similar\quad since\quad all\quad the\quad 3\quad sides\quad of\quad the\quad triangles$

$are\quad in\quad proportion.\quad$

$Justification-$

$Let\quad us\quad obtain\quad the\quad lengths\quad of\quad the\quad sides\quad of\quad \Delta ABC\quad \& \quad \Delta DEF\quad \quad$

$by\quad the\quad distance\quad formula\quad d=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } .$

$In\quad \Delta ABC\quad the\quad vertices\quad are\quad A(2,0),\quad B(-2,0),\quad C(0,2).$

$\therefore \quad AB=\sqrt { { \left( -2-2 \right) }^{ 2 }+{ \left( 0 \right) }^{ 2 } } =4\quad units.$

$Similarly\quad BC=\sqrt { { \left( 0+2 \right) }^{ 2 }+{ \left( 2-0 \right) }^{ 2 } } =2\sqrt { 2 } \quad units$

$and\quad AC=\sqrt { { \left( 0-2 \right) }^{ 2 }+{ \left( 2-0 \right) }^{ 2 } } =2\sqrt { 2 } \quad units.$

$Again\quad in\quad \Delta DEF\quad the\quad vertices\quad are\quad A(4,0),\quad B(-4,0),\quad C(0,4).$

$\therefore \quad DE=\sqrt { { \left( -4-4 \right) }^{ 2 }+{ \left( 0 \right) }^{ 2 } } =8\quad units.$

$Similarly\quad EF=\sqrt { { \left( 0+4 \right) }^{ 2 }+{ \left( 4-0 \right) }^{ 2 } } =4\sqrt { 2 } \quad units$

$and\quad DF=\sqrt { { \left( 0-4 \right) }^{ 2 }+{ \left( 4-0 \right) }^{ 2 } } =4\sqrt { 2 } \quad units.$

$So\quad \dfrac { AB }{ DE } =\dfrac { 4 }{ 8 } =\dfrac { 1 }{ 2 } .$

$\dfrac { AC }{ DF } =\dfrac { 2\sqrt { 2 } }{ 4\sqrt { 2 } } =\dfrac { 1 }{ 2 } \quad and$

$\dfrac { BC }{ EF } =\dfrac { 2\sqrt { 2 } }{ 4\sqrt { 2 } } =\dfrac { 1 }{ 2 }$

$So\quad \dfrac { AB }{ DE } =\dfrac { AC }{ DF } =\dfrac { BC }{ EF } .$

$\therefore \quad The\quad given\quad statement\quad is\quad true$

$\\ Ans-\quad TRUE.\\ \\ \\ \\ \\$

Find the area of the triangle whose vertices are $(8, 4), (6, 6)$ and $(3, 9)$ .

Let the

$\Delta ABC$ have vertices

$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 8,4 \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 6,6 \right)$ and

$C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 3,9 \right)$ .

Now

$ar\Delta ABC$ with vertices

$A\left( { x }_{ 1 },{ y }_{ 1 } \right) , B\left( { x }_{ 2 },{ y }_{ 2 } \right)$ and

$C\left( { x }_{ 3 },{ y }_{ 3 } \right)$ is

$ar\Delta ABC=\dfrac { 1 }{ 2 } \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right]$

$\therefore ar\Delta ABC=\dfrac { 1 }{ 2 } \left[ 8\left( 6-9 \right) +6\left( 9-4 \right) +3\left( 4-6 \right) \right]$ units

$\Rightarrow ar\Delta ABC=0$

Therefore,

$ar\Delta ABC=0$

The points

$A, B$ and

$C$ are collinear, since

$ar\Delta ABC=0$ .

If D $\left ( \frac{-1}{2},\frac{5}{2} \right )$ , E (7,3) and F $\left ( \frac{7}{2},\frac{7}{2} \right )$ are the midpoints of sides of $\Delta$ ABC, find the area of the $\Delta$ ABC.

$Let\quad the\quad mid\quad points\quad of\quad the\quad given\quad \Delta ABC\quad are\quad D,\quad E\quad \& \quad F\quad and\\ let\quad the\quad \Delta DEF\quad have\quad vertices\quad \\ D\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( -\frac { 1 }{ 2 } ,\frac { 5 }{ 2 } \right) ,\quad E\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 7,3 \right) \quad \& \quad F\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( \frac { 7 }{ 2 } ,\frac { 7 }{ 2 } \right) .\\ \therefore \quad Applying\quad the\quad area\quad formula\quad of\quad a\quad triangle\quad we\quad get\\ ar\Delta DEF=\frac { 1 }{ 2 } \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right] .\\ \therefore \quad ar\Delta DEF=\frac { 1 }{ 2 } \left[ -\frac { 1 }{ 2 } \left( 3-\frac { 7 }{ 2 } \right) +7\left( \frac { 7 }{ 2 } -\frac { 5 }{ 2 } \right) +\frac { 7 }{ 2 } \left( \frac { 5 }{ 2 } -3 \right) \right] \quad sq.\quad units.\\ =\frac { 11 }{ 4 } \quad sq.\quad units.\\ Now\quad the\quad area\quad of\quad the\quad \Delta DEF\quad joining\quad the\quad mid-points\quad of\quad a\quad \Delta ABC\\ =\frac { 1 }{ 4 } ar\Delta ABC.\\ \Longrightarrow ar\Delta ABC=4ar\Delta DEF=4\times \frac { 11 }{ 4 } \quad sq.\quad units.=11\quad sq.\quad units.\\ Ans-\quad ar\Delta ABC=11\quad sq.\quad units.\\ \\$

Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughters school and then reaches the office. What is the extra distance travelled by Ayush in reaching his office? (Assume that all distances covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at
(13, 26) and coordinates are in km
write answer in meter

$Let\quad the\quad co-ordinates\quad of\quad the\quad house\quad be\quad H\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 2,4 \right) ,\quad Bank\quad be\quad B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 5,8 \right) ,\\ the\quad school\quad be\quad S\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 13,14 \right) \quad \& \quad the\quad office\quad be\quad O\left( { x }_{ 4 },{ y }_{ 4 } \right) =\left( 13,26 \right) .\\ Applying\quad the\quad distanceformula...........d=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } } \quad we\quad get\\ HB=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 2 } \right) }^{ 2 } } =\sqrt { { \left( 2-5 \right) }^{ 2 }+{ \left( 4-8 \right) }^{ 2 } } km=5km,\\ BS=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 3 } \right) }^{ 2 }+{ \left( { y }_{ 2 }-y_{ 3 } \right) }^{ 2 } } =\sqrt { { \left( 5-13 \right) }^{ 2 }+{ \left( 8-14 \right) }^{ 2 } } km=10km,\\ SO=\sqrt { { \left( { x }_{ 3 }-{ x }_{ 4 } \right) }^{ 2 }+{ \left( { y }_{ 3 }-y_{ 4 } \right) }^{ 2 } } =\sqrt { { \left( 13-13 \right) }^{ 2 }+{ \left( 14-26 \right) }^{ 2 } } km=12km.\\ \therefore \quad The\quad total\quad distance\quad covered\quad D=HB+BS+SO=(5+10+12)km=27km.\\ The\quad straight\quad distance(displacement)\quad d=HO=\sqrt { { \left( { x }_{ 1 }-{ x }_{ 4 } \right) }^{ 2 }+{ \left( { y }_{ 1 }-y_{ 4 } \right) }^{ 2 } } \\ =\sqrt { { \left( 2-13 \right) }^{ 2 }+{ \left( 4-26 \right) }^{ 2 } } km=24.6km.\\ So\quad the\quad extra\quad distance\quad travelled\quad by\quad Ayush\quad is\\ D-d=(27-24.6)km=2.4km.\\ Ans-\quad The\quad extra\quad distance\quad travelled\quad by\quad Ayush\quad is\quad 2.4km.\\$

Column II gives the area of triangles whose vertices are given in column I, match them correctly.

$We\quad shall\quad apply\quad ar\Delta \quad formula\quad which\quad is\\ ar\Delta ABC=\frac { 1 }{ 2 } \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right] .\\ when\quad \Delta ABC\quad has\quad vertices\quad A\left( { x }_{ 1 },{ y }_{ 1 } \right) ,\quad B\left( { x }_{ 2 },{ y }_{ 2 } \right) \quad \& \quad C\left( { x }_{ 3 },{ y }_{ 3 } \right) \\ A)\quad Let\quad the\quad \Delta ABC\quad have\quad vertices\quad \\ A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 2,3 \right) ,\quad B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( -1,0 \right) \quad \& \quad C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 2,-4 \right) .\\ \therefore \quad ar\Delta ABC=\frac { 1 }{ 2 } \left[ 2\left( 0+4 \right) +(-1)\left( -4-3 \right) +2\left( 3-0 \right) \right] \quad sq.\quad units=10.5sq.\quad units.\\ \quad \Longrightarrow \quad ar\Delta ABC=10.5\quad sq.units\quad which\quad matches\quad with\quad 4.\\ B)\quad Let\quad the\quad \Delta ABC\quad have\quad vertices\quad \\ A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( -5,-1 \right) ,\quad B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 3,-5 \right) \quad \& \quad C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 5,2 \right) .\\ \therefore \quad ar\Delta ABC=\frac { 1 }{ 2 } \left[ -5\left( -5-2 \right) +3\left( 2+1 \right) +5\left( -1+5 \right) \right] \quad sq.\quad units=32sq.\quad units.\\ \quad \Longrightarrow \quad ar\Delta ABC=32sq.units\quad which\quad matches\quad with\quad 3.\\ C)\quad Let\quad the\quad \Delta ABC\quad have\quad vertices\quad \\ A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 1,-1 \right) ,\quad B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( -4,6 \right) \quad \& \quad C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( -3,-5 \right) .\\ \therefore \quad ar\Delta ABC=\frac { 1 }{ 2 } \left[ 1\left( 6+5 \right) +(-4)\left( -5+1 \right) +(-3)\left( -1-6 \right) \right] \quad sq.\quad units=24sq.\quad units.\\ \quad \Longrightarrow \quad ar\Delta ABC=24\quad sq.units\quad which\quad matches\quad with\quad 2.\\ D)\quad Let\quad the\quad \Delta ABC\quad have\quad vertices\quad \\ A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 0,0 \right) ,\quad B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 8,0 \right) \quad \& \quad C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 0,10 \right) .\\ \therefore \quad ar\Delta ABC=\frac { 1 }{ 2 } \left[ 0\left( 0-0 \right) +8\left( 10-0 \right) +0\left( 0-0 \right) \right] \quad sq.\quad units=40sq.\quad units.\\ \quad \Longrightarrow \quad ar\Delta ABC=40sq.units\quad which\quad matches\quad with\quad 1.\\ Ans-\quad A\longrightarrow 4\\ \quad \quad \quad \quad \quad B\longrightarrow 3\\ \quad \quad \quad \quad \quad C\longrightarrow 2\\ \quad \quad \quad \quad \quad D\longrightarrow 1\\ \\ \\$

Find the perimeter of the triangle formed by $(0, 0), (1, 0)$ and $(0, 1)$ to nearest integer.

Perimeter

$=$ Sum of length of all sides

$=1+1+\sqrt{2}=2+\sqrt{2} = 2+1.4 = 3.4$ ...using distance formula

The points $P(6, -1), Q(1, 3),$ $R = (x,8)$ are connected such that $PQ=QR$ then positive value of $x=$

A point moves so that its distance from y-axis is half of its distance from the origin. Find the locus of the point.

$x=\frac{1}{2}\sqrt{x^{2}+y^{2}}$

$\Rightarrow 2x=\sqrt{x^{2}+y^{2}}$

$\Rightarrow 4x^{2}=x^{2}+y^{2}$

$\Rightarrow 3x^{2}-y^{2}=0$

In given figure, find the inclination of line $AB$ :

The line AB forms a triangle with the x - axis and the y - axis, by intersecting them at two points.
In the triangle formed, the angle between the x - axis and the y- axis is

${ 90 }^{ 0 }$ .
Since the sum of all angles of a triangle is

${ 180 }^{ 0 }$ , the sum of the other two angles will be

${ 90 }^{ 0 }$ .
So,

$x + 2x = { 90 }^{ 0 }$

$3x = { 90 }^{ 0 }$

$x = { 30 }^{ 0 }$
Inclination of a line is the angle it makes with the positive x- axis.
Hence, inclination of line AB is

$x = { 30 }^{ 0 }$ .

Plot the points A (1, -1), B (-1, 4) and C (-3, -1) on a graph paper to obtain the triangle ABC. Give a special name to the triangle ABC and, if possible, find its area.

ABC is an isosceles triangle and lengths of sides

$AB = BC = \sqrt {29} units$

Now, area of the triangle

$= \frac {1}{2} \times base \times height = \frac {1}{2} \times 4 \times 5 = 20 sq units$ (Since, base is distance between A and C and height is the distance between B and the base AC

If $\Delta$ denotes the area of the triangle with vertices $(0, 0), (5, 0)$ and $\left(\dfrac56, \dfrac{25}6\right),$ then $\Delta$ is equal to

Area of a triangle by formula,

$\Delta = \left| \dfrac {{ a }_{ x }({ b }_{ y }-{ c }_{ y })+{ b }_{ x }({ c }_{ y }-{ a }_{ y })+{ c }_{ x }({ a }_{ y }-{ b }_{ y }) }{ 2 } \right|$
where

${a}_{x}$ and

${a}_{y}$ are the

$x$ and

$y$ coordinates of point

$a$ .

now using determinant formula of area of triangle we can say that

$12 \Delta =12 \left| \dfrac { 0+\dfrac { 125 }{ 6 }+0 }{ 2 } \right|$

$\Rightarrow 12 \Delta = 12\times \dfrac { 125 }{ 12} =125$

If O is the origin and $A_{n}$ is the point with coordinates $(n, n+1)$ then $(OA_{1})^{2}+(OA_{2})^{2}+ ... +(OA_{7})^{2}$ is equal to

The value of

$(OA_{1})^{2}+(OA_{2})^{2}+ ... +(OA_{7})^{2}$

$=1^{2}+2^{2}+2^{2}+3^{2}+ ... +7^{2}+8^{2}$

$=2(1^{2}+2^{2}+ ... +7^{2})-1+8^{2}$

$=\displaystyle \frac{2\times 7\times 8\times 15}{6}-1+8^{2}=343$

$\left (using \displaystyle 1^{2}+2^{2}+ ... +n^{2}=\frac{n(n+1)(2n+1)}{6} \right )$ .

In the following, the coordinates of the three vertices of a rectangle ABCD are given. By plotting the given points; find, the coordinates of the fourth vertex:
A (2, 0), B (8, 0) and C (8, 4)
Claim: The co-ordinates of the fourth vertex are (2,4)
If true then enter $1$ and if false then enter $0$

Plot the points A, B and C on the graph paper. Join the points to complete the rectangle ABCD.

Hence, the fourth point D

$= (2,4)$

Through the point P(3, 5), a line is drawn inclined at $45^{\circ}$ with the positive direction of x-axis. It meets the line $x+y-6=0$ at the point Q, O being the origin, then
$9\left ( PQ \right )^{2}+14\left ( OP \right )^{2}+10\left ( OQ \right )^{2}$ is equal to

Equation of line passing through the point and having slope

$m=1$

$y=mx + C$

In this question,

$m=tan 45^{0} = 1$

So the equation becomes

$y=x + C$

New line passes through (3,5)

Putting the values in the equation,

$5=3 + c$

$C=2$

New equation of line becomes

$y = x + 2$

Another line

$x+y-6=0$

These two lines intersect at Q

$y=x+2$

$y=-x+6$

$2y=8$

$y=4$

Put the value of y=4 in line equation,

$y=x+2$

$x=2$

Q(2,4)

P(3,5)

O(0,0)

Distance between two points

$\sqrt { { (x2-x1) }^{ 2 }+{ (y2-y1) }^{ 2 } }$

$(PQ)^{2} = \sqrt { { (2-3) }^{ 2 }+{ (4-5) }^{ 2 } }$

$\quad\quad=2$

$(OP)^{2} = \sqrt { { (3-0) }^{ 2 }+{ (5-0) }^{ 2 } }$

$\quad\quad=34$

$(OQ)^{2} = \sqrt { { (2-0) }^{ 2 }+{ (4-0) }^{ 2 } }$

$\quad\quad=20$

$=9\times {PQ}^{2} + 14\times{OP}^{2} + 10\times{OQ}^{2}$

$= 9\times 2 + 14\times 34 + 10\times 20$

$=18+476+200$

$=694$

If the point $P(k - 1, 2)$ is equidistant from the points $A(3, k)$ and $B(k, 5)$ , then how many values of $k$ are obtained?
Write $0$ , if the value cannot be determined.

Point

$P (k - 1, 2)$ is equidistant from

$A (3, k)$ and

$B (k, 5)$
Distance between

$P$ and

$A$ :
Distance

$=$

$D_A$ =

$\sqrt{(k - 1 - 3)^2 + (2 - k)^2}$
Distance between P and B:
Distance =

$D_B$ =

$\sqrt{(k - 1 - k)^2 + (2 - 5)^2}$

Now,

$D_A = D_B$

$\Rightarrow \sqrt{(k - 1 - 3)^2 + (2 - k)^2}$ =

$\sqrt{(k - 1 - k)^2 + (2 - 5)^2}$

$\Rightarrow k^2 + 16 - 8k + 4 + k^2 - 4k = 1 + 9$

$\Rightarrow 2k^2 - 12k + 10 = 0$

$\Rightarrow k^2 - 6k + 5 = 0$

$\Rightarrow (k - 5) (k - 1) = 0$

$\Rightarrow k = 1,5$

Find the value of $x$ so that the points $(x, -1), (2, 1)$ and $(4, 5)$ are collinear.

Let

$A(x,-1) B(2,1)$ and

$C(4,5)$

They are collinear if slope of

$AB=$ slope

$BC$

Slope of

$AB=\dfrac{1-(-1)}{2-x}=\dfrac{2}{2-x}$

Slope of

$BC=\dfrac{5-1}{4-2}=\dfrac{4}2$

$\dfrac{2}{2-x}=2$

$2-x=1$

$\therefore x=1$

$\displaystyle 120^{\circ}$ ?

Here

$\displaystyle \theta =120^{\circ}$

$\displaystyle \therefore Slope=\tan\theta =\tan 120^{\circ}=\tan\left ( 180^{\circ}-60^{\circ} \right )=-\tan 60^{\circ}=-\sqrt{3}$

The vertices of a triangle are $(-2, 0), (2, 3)$ and $(1, -3)$ .
Is the triangle equilateral, isosceles or scalene?
Write answer as $'1'$ for equilateral, $'2'$ for isosceles and $'3'$ for scalene.

Let $A (2,0), B (2,3)$ and $C (1,-3)$
Distance between two points $\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and $\left( { x }_{ 2 },{ y }_{ 2 } \right)$ can be calculated using the formula $\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y}_{ 1 } \right) ^{ 2 } }$

Distance between the points BC $= \sqrt { \left(1-2 \right) ^{ 2 }+\left( -3-3 \right) ^{ 2 } } = \sqrt { 1 + 37 } = \sqrt { 37 }$

Distance between the points AC $= \sqrt { \left(1 +2 \right) ^{ 2 }+\left( -3 0 \right) ^{ 2 } } = \sqrt { 9 + 9 } = \sqrt { 18 }$

And similarly distance between points AB = 5

Since all the sides are of different length, the triangle is a scalene triangle.

The vertices of $\displaystyle \Delta ABC$ are (-2, 1), (5, 4) and (2, -3) respectively. Find the area of triangle.

A(-2, 1), B(5, 4) and C(2, -3) be the vertices of triangle
So

$\displaystyle x_{1}=-2,y_{1}=1;x_{2}=5,y_{2}=4;x_{3}=2,y_{3}=-3$
Area of

$\displaystyle \Delta ABC$
=

$\displaystyle \frac{1}{2}\left | \left [ x_{1}\left ( y_{2}-y_{3} \right )+x_{2}\left ( y_{3}-y_{1} \right )+x_{3}\left ( y_{1}-y_{2} \right ) \right ] \right |$
=

$\displaystyle \frac{1}{2}\left | \left [ \left ( -2 \right )\left ( 4+3 \right )+\left ( 5 \right )\left ( -3-1 \right )+2\left ( 1-4 \right ) \right ] \right |$
=

$\displaystyle \frac{1}{2}\left | \left [ -14+\left ( -20 \right )+\left ( -6 \right ) \right ] \right |$
=

$\displaystyle \frac{1}{2}\left | -40 \right |$
= 20 Sq unit

The length of a line-segment isIf one end is at (2, -3) and the abscissa of the second end is 10, then find the number of values of ordinate the second end can take.

Let

$(2, -3)$ be the point A Let the ordinate of the second end B be y Then its coordinates will be

$(10, y)$
Given,

$AB = \sqrt { \left(10 - 2 \right) ^{ 2 }+\left( y + 3 \right) ^{ 2 } } = 10$

$\sqrt { 64 + {y}^{2} + 6y + 9} = 10$

${y}^{2} + 6y + 73 = 100$

${y}^{2} + 6y -27 = 0$

$(y + 9) (y - 3) = 0$
Therefore

$y = -9\ , y = 3$
i.e. the ordinate is

$3$ or

$-9$

The distance of $P(x, y)$ from $A(5, 1)$ and $V (-1, 5)$ are equal then $\dfrac xy = \dfrac pq$ . Find the value of $p+q$

Distance between two points $\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and $\left( { x }_{ 2 },{ y }_{ 2 } \right)$ can be calculated using the formula $\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } }$

Given,

Distance between the points $(x,y) ; (5,1) =$ Distance between $(x,y) ; (-1,5)$

$\sqrt { \left( 5-x \right) ^{ 2 }+\left( 1- y \right) ^{ 2 } } = \sqrt { \left( -1-x \right) ^{ 2 }+\left( 5 - y \right) ^{ 2 } }$

$\Rightarrow \left( 5-x \right) ^{ 2 }+\left( 1- y \right) ^{ 2 } = \left( -1-x \right) ^{ 2 }+\left( 5 - y \right) ^{ 2 }$

$\Rightarrow {5}^{2} + {x}^{2} - 10x + {1}^{2} + {y}^{2} - 2y ={1}^{2} + {x}^{2} + 2x + {5}^{2} + {y}^{2} - 10y$

$\Rightarrow -10x -2y = +2x - 10y$

$\Rightarrow 12x = 8y$

$\Rightarrow \dfrac xy = \dfrac 23$

Hence, $p+q = 5$ .

Enter 1 if it is true else enter 0.
The points $(-3, 2), (1, -2) and (9, -10)$ are collinear.

1897502_328881_ans_ffb46d7655414373bb00ee5947933b13.jpg

If the coordinates of a point on the y-axis which is equidistant from the points $(13, 2)$ and $(12, -3)$ is $(p,q)$ , then find the value of $p+q$ .

Let P(0, y) be the required point and the given points be A(12 , -3) and B (13 , 2) Then PA = PB (given)

$\displaystyle \sqrt{\left ( 12-0 \right )^{2}+\left ( -3-y \right )^{2}}= \sqrt{\left ( 13-0 \right )^{2}+\left ( 2-y \right )^{2}}$

$\displaystyle \Rightarrow \sqrt{144+\left ( y+3 \right )^{2}}= \sqrt{169+\left ( 2-y \right )^{2}}$

$\displaystyle \Rightarrow y= 2$

$\displaystyle \therefore$ The required point on Y-axis is (0, 2). ( hint: since x cordinates are equal to '0' on y-aixs)

Find the area of the triangle formed by the midpoints of the sides of $\Delta ABC$ , where $A = (3, 2), B = (-5, 6)$ and $C = (8, 3)$ .

Solution: Area of

$\Delta ABC\, =\, \displaystyle \frac{1}{2}\begin{vmatrix}3-(-5) & 2-6 & \\ -5-8 & 6-3 & \end{vmatrix}$

$=\, \displaystyle \frac{1}{2}\, \begin{vmatrix}8 & -4 & \\ -13 & 3 & \end{vmatrix}$
$$=\, \displaystyle \frac{1}{2}

$\begin{vmatrix}8(3)-4(13)\end{vmatrix}$ \, =\, \displaystyle \frac{1}{2}\begin{vmatrix}24-52\end{vmatrix}\, =\, 14 sq. units.$$
Hence the area of the midpoints of the sides

$\Delta ABC$ = 1/4 (Area of

$\Delta ABC$ ) = 1/4 (14) = 3.5 sq. units

Find sum of values of $a$ for which the distance between the points $P(11, -2)$ and $Q(a, 1)$ is $5$ units.

Distance between two points

$\displaystyle \left ( x_{1},y_{1} \right ),\left ( x_{2},y_{2} \right )$ is

$\displaystyle \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
Given PQ = 5

$\displaystyle \Rightarrow$

$\displaystyle \sqrt{\left ( a-11 \right )^{2}+\left ( 1-\left ( -2 \right ) \right )^{2}}= 5$
Taking square on both sides we get

$\displaystyle \left ( a-11 \right )^{2}= 25-9= 16\Rightarrow a-11$ =

$\displaystyle \pm \sqrt{16}=a-11= \pm 4$

$\displaystyle \Rightarrow a= 15$ or 7

Find the area of the triangle formed by the midpoints of the sides of $\displaystyle \triangle ABC$ where $A=(3,2), B=(-5,6)$ and $C=(8,3)$

Area of

$\displaystyle \triangle ABC$

$\displaystyle =\frac{1}{2} \begin{vmatrix} 3-(-5) & 2-6 \\ -5-8 & 6-3 \end{vmatrix}$
=

$\displaystyle \frac{1}{2} \begin{vmatrix} 8 & -4 \\ -13 & 3 \end{vmatrix}$
=

$\displaystyle \frac{1}{2}\left | 8\left ( 3 \right )-4\left ( 13 \right ) \right |$ =

$\displaystyle \frac{1}{2}$

$\displaystyle \left | 24-52 \right |= 14$ sq.units
Hence the area of triangle formed by the midpoints of the sides of

$\displaystyle \triangle ABC= \dfrac 14$ (Area of

$\displaystyle \triangle ABC$ )=

$\dfrac 14(14)=3.5$ sq.units

Find the angle between the $x$ -axis and the line joining the points $(3,-1)$ and $(4,-2)$ .

The slope of the line joining the points

$(3,-1)$ and

$(4,-2)$ is

$\displaystyle m= \frac{-2-(-1)}{4-3}= -2+1= -1$

Now the inclination

$\displaystyle \left ( \theta \right )$ of the line joining the points

$(3,1)$ and

$(4,2)$ is given by

$\displaystyle \tan \theta = -1$

$\displaystyle\Rightarrow \theta = \left ( 90^{\circ}+45^{\circ} \right )= 135^{\circ}$

A slope of the line is an inclination with the x-axis.

Thus, the angle between

$x$ -axis and the line joining the points

$(3,1)$ and

$(4,2)$ is

$\displaystyle 135^{\circ}.$

If $2p$ is the length of perpendicular from the origin to the line whose intercepts on the axes are $2a$ and $b$ , then show that $\displaystyle {\frac{1}{p^{2}}= \frac{1}{a^{2}}+\frac{1}{b^{2}}}$ .

Equation of line with intercepts

$a$ and

$b$ is,

$\displaystyle \frac{x}{a}+\frac{y}{b}=1\Rightarrow bx+ay-ab=0$ ....(1)

Now if

$p$ is the length of the perpendicular from point

$(0,0)$ to line (1),then,

$\displaystyle p= \frac{\left | b(0)+a(0)-ab \right |}{\sqrt{b^{2}+a^{2}}}$

$\displaystyle \Rightarrow p= \frac{\left | -ab \right |}{\sqrt{a^{2}+b^{2}}}$

On squaring both sides, we obtain

$\displaystyle p^{2}= \frac{\left | -ab \right |^2}{(\sqrt{a^{2}+b^{2})}^2}$

$\displaystyle \Rightarrow p^{2}\left ( a^{2}+b^{2} \right )= a^{2}b^{2}$

$\displaystyle \Rightarrow \frac{a^{2}+b^{2}}{a^{2}b^{2}}= \frac{1}{p^{2}}$

$\displaystyle \Rightarrow \frac{1}{p^{2}}= \frac{1}{a^{2}}+\frac{1}{b^{2}}$

What are the points on the $y$ -axis whose distance from the line $\displaystyle \frac{x}{3}+\frac{y}{4}= 1$ is $4$ units.

Let

$(0,b)$ be the point on the

$y$ -axis whose distance from line

$\displaystyle \frac{x}{3}+\frac{y}{4}= 1$ is

$4$ units.
The given line can be written as

$\displaystyle 4x+3y-12= 0$ ....(1)
Thus using distance formula,

$\displaystyle 4= \frac{\left | 4(0)+3(b)-12 \right |}{\sqrt{4^{2}+3^{2}}}$

$\displaystyle \Rightarrow 4= \frac{\left | 3b-12 \right |}{5}$

$\displaystyle \Rightarrow 20= \left | 3b-12 \right |$

$\displaystyle \Rightarrow 20= \pm (3b-12)$

$\displaystyle \Rightarrow 20= (3b-12)$ or

$20= -\left \{ 3b-12 \right \}$

$\displaystyle \Rightarrow 3b= 20+12$ or

$3b= -20+12$

$\displaystyle \Rightarrow b= \frac{32}{3}$ or

$b= -\cfrac{8}{3}$
Thus, the required points are

$\displaystyle \left( 0,\frac{32}{3} \right) \text{and} \left ( 0,-\frac{8}{3} \right)$ .

Find the area of square whose one pair of the opposite vertices are (3, 4) and (5, 6)

Let the given vertices be A(3, 4) and C(5, 6)
Length of AC=

$\displaystyle \sqrt{\left ( 3-5 \right )^{2}+\left ( 4-6 \right )^{2}}= \sqrt{8}$ units
Area of the square

$\displaystyle= \frac{AC^{2}}{2}= \frac{\left ( \sqrt{8} \right )^{2}}{2}= 4$ sq.units

Find a point on the $x$ -axis, which is equidistant from the points $(7, 6)$ and $(3, 4)$ .

Let

$(a,0)$ be the point on

$x$ -axis that is equidistant from the points

$(7,6), (3,4)$ .

$\Rightarrow \displaystyle \sqrt{(7-a)^{2}+(6-0)^{2}}$

$\displaystyle= \sqrt{(3-a)^{2}+(4-0)^{2}}$

$\displaystyle\Rightarrow\sqrt{49+a^{2}-14a+36}$

$\displaystyle= \sqrt{9+a^{2}-6a+16}$

$\displaystyle\Rightarrow \sqrt{a^{2}-14a+85}$

$\displaystyle= \sqrt{a^{2}-6a+25}$
On squaring both sides, we obtain

$\displaystyle a^{2}-14a+85= a^{2}-6a+25$

$\displaystyle\Rightarrow -14a+6a= 25-85$

$\displaystyle\Rightarrow 8a=60$

$\displaystyle\Rightarrow a=\frac{60}{8}= \frac{15}{2}$
Thus, the required point on

$x$ -axis is

$\displaystyle \left ( \frac{15}{2},0 \right )$

Find the area of the triangle formed by the lines $\displaystyle y-x= 0,x+y= 0$ and $x-k= 0$

The equation of the given lines are

$y-x\displaystyle = 0$ ....(1)

$x+y\displaystyle =$ ...(2)

$x-k\displaystyle = 0$ ...(3)
The point of intersection of lines (1) and (2) is given by

$x\displaystyle = 0$ and

$y\displaystyle = 0$
The point of intersection of lines (2) and (3) is given by

$x\displaystyle = k$ and

$y\displaystyle = -k$
The point of intersection of lines(3) and (1) is given by

$x\displaystyle = k$ and

$y\displaystyle = k$
Thus, the vertices of the triangle formed by the three given lines are

$(0,0),(k,-k),$ and

$(k,k)$

We know that the area of the triangle whose vertices are

$\displaystyle (x_{1},y_{1}),(x_{2},y_{2}),$ and

$(x_{3},y_{3})$ is

$\cfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1}) +x_{3}(y_{1}-y_{2})\right |$

Therefore, area of the triangle formed by the three given lines,

$\displaystyle = \frac{1}{2}\left | 0(-k-k)+k(k-0)+k(0+k) \right |$ square unts

$\displaystyle = \frac{1}{2}\left | k^{2}+k^{2} \right |$ square units

$\displaystyle = \frac{1}{2}\left | 2k^{2} \right | = k^{2}$ square units

Give the equations of two lines passing through $\left(2, 14\right)$ . How many more such lines are there, and why?

Since the given solution is

$(2, 14)$

$\therefore x=2$ and

$y=14$

Then, one equation is

$x+y=2+14=16$

$x+y=16$

And, second equation is

$x-y=2-14=-12$

$x-y=-12$

And, third equation is

$y=7x$

$7x-y=0$

So, we can find infinite equations because through one point infinite lines can pass.

If $Q(0, 1)$ is equidistant from $P(5, -3)$ and $R(x, 6)$ , find the values of $x$ . Also find the distances $QR$ and $PR$ .

As given

$Q(0,1)$ is equidistant from

$P(5,-3)$ and

$R(x,6)$

$\Rightarrow PQ=QR$

$\Rightarrow \sqrt{(0-5)^2+(1-(-3))^2}$

$=\sqrt{(x-0)^2+(1-6)^2}$ [ By using disatnce formula

$=\sqrt{ { \left( {x_2}-{ { x_1 } } \right) }^{ 2}+{ \left( {y_2 }-{ y_1 } \right) }^{ 2 } }$ ]

$\Rightarrow 25+16=x^2+25$

$\Rightarrow 41=x^2+25$

$\Rightarrow x^2=16$

$\Rightarrow x=4$

Distance between

$Q(0,1)$ and

$R(4,6)$

$QR=\sqrt{(4-0)^2+(6-1)^2}=\sqrt{4^2+5^2}=\sqrt{16+25}=\sqrt{41}$

Distance between

$P(5,-3)$ and

$R(4,6)$

$PR=\sqrt{(4-5)^2+(6+3)^2}=\sqrt{1+81}=\sqrt{82}$

Find the distance between the following pairs of points :
(i) $(2, 3), (4, 1)$
(ii) $( -5, 7), ( -1, 3)$
(iii) $(a, b), ( -a, -b)$

Distance between two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

(i) Distance between

$(2,3)$ and

$(4,1)$ is

$= \sqrt{(2-4)^2+(3-1)^2}$

$= \sqrt{(-2)^2+(2)^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$

(ii) Distance between

$(-5,7)$ and

$(-1,3)$ is

$= \sqrt{(-5-(-1))^2+(7-3)^2}$

$= \sqrt{(-4)^2+(4)^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$

(iii )Distance between

$(a,b)$ and

$(-a,-b)$ is

$= \sqrt{(a-(-a))^2+(b-(-b))^2}$

$= \sqrt{(2a)^2+(2b)^2}=\sqrt{4a^2+4b^2}=2\sqrt{a^2+b^2}$

If $(1, 2), (4, y), (x, 6)$ and $(3, 5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$ .

$\textbf{Step 1: Find x, y by equating coordinates of mid points of both diagonals.}$

$\text{Let the vertices be } A= (1,2), B = (4,y), C = (x,6)$

$\text{and }D = (3,5)$

$\text{We know that the diagonals of a parallelogram bisect each other.}$

$\text{So, the midpoint of AC is same as the mid point of BD.}$

$\text{So, midpoint of } AC =$

$\text{Mid point of } BD$

$\Rightarrow \left( \dfrac { 1+x }{ 2 } ,\dfrac {2+6 }{ 2 } \right) \quad = \left( \dfrac { 4+3}{ 2 } ,\dfrac { y + 5 }{ 2 } \right) \quad$

$\left[\because\textbf{Mid point}=\left( \boldsymbol{\dfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+y_{ 2 } }{ 2 } }\right) \right]$

$\Rightarrow \left( \dfrac { 1+x }{ 2 } ,\dfrac { 8 }{ 2 } \right) \quad = \left( \dfrac { 7 }{ 2 } ,\dfrac { y+5 }{ 2 } \right) \quad$

$\Rightarrow 1+ x = 7 \ ;\ y + 5 = 8$

$\Rightarrow x = 6 \ ;\ y = 3$

$\textbf{Hence, }\boldsymbol{x=6\ \ \&\ \ y=3.}$

Find the distance between the points $(0, 0)$ and $(36, 15)$ .

Distance Between two given point=

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Here

$x_1=0,x_2=26$

and

$y_1=0,y_2=15$

$\therefore$ Distance between the points

$(0,0)$ and

$(36,15)$ =

$\sqrt{(36-0)^2+(15-0)^2}$

$= \sqrt{36^2+15^2}$

$= \sqrt{1296+225}$

$= \sqrt{1521}=39$

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) $( -1, -2), (1, 0), ( -1, 2), ( -3, 0)$
(ii) $(-3, 5), (3, 1), (0, 3), (-1, -4)$
(iii) $(4, 5), (7, 6), (4, 3), (1, 2)$

(i) Let the given points are

$A(-1,-2)$ ,

$B(1,0)$ ,

$C(-1,2)$ and

$D(-3,0)$ Then,

$AB=\sqrt{(1+1)^2+(0+2)^2}=\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}$

$BC=\sqrt{(-1-1)^2+(2-0)^2}=\sqrt{(2^2+2^2)}=\sqrt{4+4}=\sqrt{8}$

$CD=\sqrt{((-3)-(-1))^2+(0-2)^2}=\sqrt{2^2+(-2)^2}=\sqrt{4+4}=\sqrt{8}$

$DA=\sqrt{(-3)-(-1))^2+(0-(-2))^2}=\sqrt{(-2)^2+2^2}=\sqrt{4+4}=\sqrt{8}$

$AC\sqrt{((-1)-(-1))^2+(2-(-2))^2}=\sqrt{0+4^2}=\sqrt{16}=4$

$BD=\sqrt{(-3-1)^2+(0-0)^2}=\sqrt{-4^2}=\sqrt{16}=4$

Since the four sides

$AB,BC,CD$ and

$DA$ are equal and the diagonals

$AC$ and

$BD$ are equal .

$\therefore$ Quadrilateral

$ABCD$ is a square.

(ii)Let the given points are

$A(-3,5)$ ,

$B(3,1)$ ,

$C(0,3)$ and

$D(-1,-4)$ Then

$AB=\sqrt{(-3-3)^2+(5-1)^2}=\sqrt{(-6)^2+4^2}=\sqrt{36+16}=\sqrt{52}$

$BC=\sqrt{(3-0)^2+(1-3)^2}=\sqrt{(3^2+(-2)2^2)}=\sqrt{9+4}=\sqrt{11}$

$CD=\sqrt{(0-(-1))^2+(3-(-4))^2}=\sqrt{1^2+(7)^2}=\sqrt{1+49}=\sqrt{50}$

$DA=\sqrt{(-1)-(-3))^2+((-4)-5))^2}=\sqrt{(2)^2+(-9)^2}=\sqrt{4+81}=\sqrt{85}$

Here

$AB\neq BC \neq CD\neq DA$

$\therefore$ it is a quadrilateral.

(iii)Let the given points are

$A(4,5)$ ,

$B(7,6)$ ,

$C(4,3)$ and

$D(1,2)$ Then

$AB=\sqrt{(7-4)^2+(6-5)^2}=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}$

$BC=\sqrt{(4-7)^2+(3-6)^2}=\sqrt{((-3)^2+(-3)^2)}=\sqrt{9+9}=\sqrt{18}$

$CD=\sqrt{(1-4)^2+(2-3)^2}=\sqrt{(-3)^2+(-1)^2}=\sqrt{9+1}=\sqrt{10}$

$DA=\sqrt{(1-4)^2+(2-5)^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{9+9}=\sqrt{18}$

$AC\sqrt{(4-4)^2+(3-5)^2}=\sqrt{0+(-2)^2}=\sqrt{4}=2$

$BD=\sqrt{(1-7)^2+(2-6)^2}=\sqrt{(-6)^2+(-4)^2}=\sqrt{36+16}=\sqrt{52}$

Here

$AB=CD,BC=DA$ . But

$AC \neq BD$

Hence the pairs of opposite sides are equal but diagonal are not equal so it is a parallelogram.

In each of the following find the value of $k$ , for which the points are collinear.
(i) $(7, -2), (5, 1), (3, k)$
(ii) $(8, 1), (k, -4), (2, -5)$

(i) let the given points are

$A(7,-2),B(5,1)$ and

$C(3,k)$

These point are collinear if area (

$\triangle ABC$ )=0

$\Rightarrow x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0$

Here

$x_1,y_1)=(7.-2)$

$(x_2,y_2)=(5,1)$

$(x_3,y_3)=(3,k)$

$\Rightarrow 7(1-k)+5(k+2)+3(-2-1)=0$

$\Rightarrow 7-7k+5k+10-9=0$

$\Rightarrow 8-2k=0$

$\Rightarrow 2k=8$

$\Rightarrow k=4$

Hence the given points are collinear for

$k=4$

(i) let the given points are

$A(8,1$ ),

$B(k,-4)$ and

$C(2,-5)$

These point are collinear if area (

$\triangle ABC$ )=0

$\Rightarrow x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0$

Here

$x_1,y_1)=(8,1)$

$(x_2,y_2)=(k,-4)$

$(x_3,y_3)=(2,-5)$

$\Rightarrow 8(-4+5)+k(-5-1)+2(1+4)=0$

$\Rightarrow 8-6k+10=0$

$\Rightarrow -6k=-18$

$\Rightarrow k=3$

$\Rightarrow k=4$

Hence the given points are collinear for

$k=3$

In a classroom, $4$ friends are seated at the points $A$ , $B$ , $C$ and $D$ as shown in Fig. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, Dont you think $ABCD$ is a square? Chameli disagrees. Using distance formula, find which of them is correct. why?

According to the fig.

$A(3,4$ ),

$B(6,7)$ ,

$C(9,4)$ and

$D(6,1)$ are the position of

$4$ friends .

If we join all these points then

By using distance formula

$=\sqrt{ { \left( {x_2}-{ { x_1 } } \right) }^{ 2}+{ \left( {y_2 }-{ y_1 } \right) }^{ 2 } }$

$AB=\sqrt{(3-6)^2+(4-7)^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

$BC=\sqrt{(6-9)^2+(7-4)^2}=\sqrt{(-3)^2+(3)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

$CB=\sqrt{(9-6)^2+(4-1)^2}=\sqrt{(3)^2+(3^2)}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

$AD=\sqrt{(3-6)^3+(4-1)^2}=\sqrt{-3)^2+(3)^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$

if we join the point

$AC$ and

$BD$ than

Diagonal

$AC=\sqrt{(3-9)^2+(4-4)^2}=\sqrt{(-6)^2+(0)^2}=\sqrt{36}=6$

Diagonal

$BD=\sqrt{(6-6)^2+(7-1)^2}=\sqrt{(0)^2+(6)^2}=\sqrt{36}=6$

Hence all the sides of a quadrilateral are the same and the diagonals are also the same so

$ABCD$ is a square.

Hence Champa is correct.

Find a relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the point $(3, 6)$ and $( -3, 4)$ .

Let the point

$P(x,y)$ is equidistant from the points

$A(3,6)$ and

$B(-3,4)$

$\therefore PA=PB$

By using the distance formula

$=\sqrt{ { \left( {x_2}-{ { x_1 } } \right) }^{ 2}+{ \left( {y_2 }-{ y_1 } \right) }^{ 2 } }$ we have

$\Rightarrow \sqrt{(x-3)^2+(y-6)^2}=\sqrt{(x+3)^2+(y-4)^2}$

$\Rightarrow (x-3)^2+(y-6)^2=(x+3)^2+(y-4)^2$

$\Rightarrow x^2-6x+9+y^2-12y+36=x^2+6x+9+y^2-8y+16$

$\Rightarrow -6x-6x-12y+8y+36-16=0$

$\Rightarrow -12x-4y+20=0$

$\Rightarrow 3x+y-5=0$

Hence this is a relation between

$x$ and

$y$ .

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are $(0, +1), (2, 1)$ and $(0, 3)$ . Find the ratio of this area to the area of the given triangle.

Let

$A(0,-1)$ ,

$B(2,1)$ and

$C(0,3)$ are the vertices of

$\triangle ABC$

$\therefore$ Area of

$\triangle ABC$

$=\dfrac{1}{2}\left[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \right]$

Here

$(x_1,y_1)=(0,-1)$

$(x_2,y_2)=(2,1)$

$(x_3,y_3)=(0,3)$

$= \dfrac{1}{2}\left[0(1-3)+2(3+1)+0(-1-1) \right]$

$= \dfrac{8}{2}=4$ sq.unit

Let

$P,Q,R$ are the mid-point of

$AB,AC$ and

$BC.$

then coordinates of

$P$

$=\left[\dfrac{0+2}{2},\dfrac{-1+1}{2}\right]=\left[1,0\right]$

Coordinates of

$Q$

$=\left[\dfrac{0+0}{2},\dfrac{-3+1}{2}\right]=\left[0,1\right]$

Coordinates of

$R$

$=\left[\dfrac{2+0}{2},\dfrac{1+3}{2}\right]=\left[1,2\right]$

$\therefore$ Area of

$\triangle PQR$

$= \dfrac{1}{2}\left[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \right]$

$=\dfrac{1}{2}\left[1(1-2)+0(2-0)+1(0-1)\right]$

$= \dfrac{-1-1}{2}=\dfrac{-2}{2}=-1=1$ sq.unit.

$\therefore$ Ratio of

$\triangle ABC$ and

$\triangle PQR$

$=4:1$

Find the area of the triangle whose vertices are :
(i) $(2, 3), (-1, 0), (2, -4)$
(ii) $(-5, -1), (3, -5), (5, 2)$

(i) Let

$A(2,3)$ ,

$B(-1,0)$ and

$C(2,-4)$ are the vertices of

$\triangle ABC$

Area of triangle=

$\dfrac{1}{2}\left[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \right]$

Here

$(x_1,y_1)=(2,3)$

$(x_2,y_2)=(-1,0)$

$(x_3,y_3)=(2,-4)$

$= \dfrac{1}{2}\left[2(0+4)-1(-4-3)+2(3-0)\right]$

$= \dfrac{1}{2}\left[8+7+6\right]$

$= \dfrac{21}{2}$ sq.unit

(i) Let

$A(-5,1)$ ,

$B(3,-5)$ and

$C(5,2)$ are the vertices of

$\triangle ABC$

Area of triangle=

$\dfrac{1}{2}\left[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \right]$

Here

$(x_1,y_1)=(-5,-1)$

$=(x_2,y_2)=(3,-5)$

$=(x_3,y_3)=(5,2)$

$= \dfrac{1}{2}\left[-5(-5-2)+3(2+1)+5(-1+5)\right]$

$= \dfrac{1}{2}\left[35+9+20\right]$

$= \dfrac{64}{2}=32$ sq.unit

Find the centre of a circle passing through the points $(6, -6), (3, -7)$ and $(3, 3)$ .

Let

$O(x,y)$ is the center of the circle and

$A(6.-6), B(3,-7)$ and

$C(3,3)$ are the points on the circumference of the circle.

$\therefore OA=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$\Rightarrow OA=\sqrt{(x-6)^2+(y+6)^2}$

$\Rightarrow OB=\sqrt{(x-3)^2+(y+7)^2}$

$\Rightarrow OC=\sqrt{(x-3)^3+(y-3)^2}$

$\because$ Radii of the circle are equal

$\therefore OA=OB$

$\Rightarrow \sqrt{(x-6)^2+(y+6)^2}=\sqrt{(x-3)^2+(y+7)^2}$

$\Rightarrow x^2+36-12x+y^2+36+12y=x^2+9-6x+y^2+49+14y$

$\Rightarrow -6x-2y+14=0$

$\Rightarrow 3x+y=7$ ....................................................................(i)

Similarly,

$OA=OC$

$\Rightarrow \sqrt{(x-6)^2+(y+6)^2}=\sqrt{(x-3)^2+(y-3)^2}$

$\Rightarrow x^2+36-y+y^2+36+12y=x^2+9-6x+y^2+9-6y$

$\Rightarrow -6x+18y+54=0$

$\Rightarrow -3x+9y=-27$ ............................................................(ii)

Adding (i) and (ii)

$\Rightarrow 10y=-20$

$\Rightarrow y=-2$

Substitute the value of

$y$ in (i)

$\Rightarrow 3x+-2=7$

$\Rightarrow 3x=9$

$\Rightarrow x=3$

$\therefore$ The center of the circle is

$(3,-2)$ .

The Class $X$ students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmoharare planted on the boundary at a distance of $1$ m from each other. There is a triangular grassy lawn in the plot as shown in the Fig. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking $A$ as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of $\displaystyle \Delta PQR$ if $C$ is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?

(i) If we take

$A$ as origin then

$AD$ is

$X$ -axis and

$AB$ is the

$Y$ -axis Then the coordinates of the vertices of

$\triangle PQR$ is

$P=(4,6),Q=(3,2) ,R=(6,5)$

(ii) If we take

$C$ as origin then

$CB$ is

$X$ -axis and

$CD$ is the

$Y$ -axis Then the coordinates of the vertices of

$\triangle PQR$ is

$P=(12,2),Q=(13,6) ,R=(10,3)$

Area of the triangle =

$\dfrac{1}{2}\left[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right]$

First case-

Area of

$\triangle PQR=\dfrac{1}{2}\left[4(2-5)+3(5-6)+6(6-2)\right]$

$= \dfrac{1}{2}\left(4\times -3+3\times -1+6\times 4\right)$

$= \dfrac{1}{2}\left(-12-3+24\right)=\dfrac{9}{2}$ Sq.unit

Second case-

Area of

$\triangle PQR=\dfrac{1}{2}\left[12(6-3)+13(3-2)+10(2-6)\right]$

$= \dfrac{1}{2}\left(12\times 3+13\times 1+10\times- 4\right)$

$= \dfrac{1}{2}\left(36+13-40\right)=\dfrac{9}{2}$ Sq.unit

Hence we observed that the area of the triangle in both case is equal.

$ABCD$ is a rectangle formed by the points $A(-1, -1)$ , $B( -1, 4)$ , $C(5, 4)$ and $D(5, -1)$ . $P, Q,R$ and $S$ are the mid-points of $AB$ , $BC$ , $CD$ and $DA$ respectively. Is the quadrilateral $PQRS$ a square? a rectangle? or a rhombus? Justify your answer.

Let

$P,Q,R$ and

$S$ are the midpoint of

$AB,BC,CD$ and

$DA$ .

$\therefore$ Co-ordinates of

$P=\left[\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right]=\left[\dfrac{-1-1}{2},\dfrac{-1+4}{2}\right]=\left[-1,\dfrac{3}{2}\right]$

$\therefore$ Co-ordinates of

$Q=\left[\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right]=\left[\dfrac{-1+5}{2},\dfrac{4+4}{2}\right]=\left[2,4\right]$

$\therefore$ Co-ordinates of

$R=\left[\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right]=\left[\dfrac{5+5}{2},\dfrac{4-1}{2}\right]=\left[5,\dfrac{3}{2}\right]$

$\therefore$ Co-ordinates of

$S=\left[\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right]=\left[\dfrac{5-1}{2},\dfrac{-1-1}{2}\right]=\left[4,-1\right]$

Now, length of

$PQ=\sqrt{(-1-2)^2+(\dfrac{3}{2}-4)^2}=\sqrt{(-3)^2+(-\dfrac{5}{2})^2}=\sqrt{9+\dfrac{25}{4}}=\sqrt{\dfrac{61}{4}}$

Length of

$QR=\sqrt{(2-5)^2+(4-\dfrac{3}{2})^2}=\sqrt{(-3)^2+(\dfrac{5}{2})^2}=\sqrt{9+\dfrac{25}{4}}=\sqrt{\dfrac{61}{4}}$

Length of

$RS=\sqrt{(5-2)^2+(\dfrac{3}{2}+1)^2}=\sqrt{(3)^2+(\dfrac{5}{2})^2}=\sqrt{9+\dfrac{25}{4}}=\sqrt{\dfrac{61}{4}}$

Length of

$S=\sqrt{(2+1)^2+(-1-\dfrac{3}{2})^2}=\sqrt{(3)^2+(-\dfrac{5}{2})^2}=\sqrt{9+\dfrac{25}{4}}=\sqrt{\dfrac{61}{4}}$

Length of diagonal

$PR=\sqrt{(-1-5)^2+(\dfrac{3}{2}-\dfrac{3}{2})^2}=\sqrt{(6^2)}=\sqrt{36}=6$

Length of diagonal QS=

$\sqrt{(2-2)^2+(4+1)^2}=\sqrt{(5)^2}=\sqrt{25}=5$

Hence the all the sides of the quadrilateral

$PQRS$ are equal but the diagonals are not equal then

$PQRS$ is a rhombus.

Find, if possible, the slope of the line through the points $(-9,4)$ and $(-9,-2)$

Since slope of line passing through two points

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and

$\left( { x }_{ 2 },{ y }_{ 2 } \right)$ is

${ m }=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$

We now find the slope of the line passing through the points

$(-9,4)$ and

$(-9,-2)$ as shown below:

${ m }=\dfrac { -2-4 }{ -9-(-9) } =\dfrac { -2-4 }{ -9+9 } =\dfrac { -6 }{ 0 } =$ undefined

Hence, the slope of the line is undefined.

The vertices of a $\displaystyle \Delta ABC$ are $A(4, 6), B(1, 5)$ and $C(7, 2)$ . A line is drawn to intersect sides $AB$ and $AC$ at $D$ and $E$ respectively, such that $\displaystyle \frac { AD }{ AB } =\frac { AE }{ AC } =\frac { 1 }{ 4 }$ . Calculate the area of the $\displaystyle \Delta ADE$ and compare it with the area of $\displaystyle \Delta ABC$ .

$\triangle ABC$

$A(4,6),B(1,5)$ and

$C(7,2)$

Area of

$\triangle ABC=\dfrac{1}{2}\left[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right]$

$= \dfrac{1}{2}\left[4(5-2)+1(2-6)+7(4-5)\right]$

$= \dfrac{1}{2}\left[12-4+7\right]=\dfrac{15}{2}$ sq. unit

A line is drawn to intersect sides

$AB$ and

$AC$ at

$D$ and

$E$

Then

$\dfrac{AD}{AB}=\dfrac{1}{4}$

The point

$D$ divide the line

$AB$ in

$AD$ and

$DB$ .The ratio of

$AD$ and

$DB = m_1:m_2=1:3$

$\therefore$ Co-ordinates of

$D$ =

$\left(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2}\right)$

Here,

$(x_1,y_1)=(4:6)$ and

$(x_2,y_2)=(1,5)$

$= \left(\dfrac{1\times 1+4\times 3}{1+3},\dfrac{1\times 5+3\times 6}{3+1}\right)$

$= \left(\dfrac{13}{4},\dfrac{23}{4}\right)$

The point

$E$ divide the line

$A$ C in

$AE$ and

$EC$ . The ratio of

$AE$ and

$EC$ =

$m_1:m_2=1:3$

$\therefore$ Co-ordinates of

$E=\left(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2}\right)$

Here,

$(x_1,y_1)=(4:6)$ and

$(x_2,y_2)=(7,2)$

$= \left(\dfrac{1\times 7+3\times 4}{1+3},\dfrac{1\times 2+3\times 6}{3+1}\right)$

$= \left(\dfrac{19}{4},5\right)$

Now the area of

$\Delta ADE=\dfrac{1}{2}\left[4(\dfrac{23}{4}-5)+\dfrac{13}{4}(5-6)+\dfrac{19}{4}(6-\dfrac{23}{4})\right]$

$= \dfrac{1}{2}\left[\dfrac{12}{4}-\dfrac{13}{4}+\dfrac{19}{16} \right]$

$=\dfrac{1}{2}\left[\dfrac{48-52+19}{16}\right]$

$= \dfrac{1}{2}\times \dfrac{15}{16}=\dfrac{15}{32}$ sq.unit

Hence, Area

$(\triangle ADE):$ Area (

$\triangle ABC$ )=

$\dfrac{15}{32}:\dfrac{15}{2}=\dfrac{1}{32}:\dfrac{1}{2}=1:16$

You have studied in Class IX, that a median of a triangle divides it into two triangles of equal areas. Verify this result for $\displaystyle \Delta ABC$ whose vertices are $A(4, -6), B(3, -2)$ and $C(5, 2)$ .

Let

$AD$ be the median of

$\triangle ABC$ .

Hence,

$D$ is the midpoint of

$BC$ .

We know that the coordinates of the midpoint of the line segment joining

$(x_1,y_1) \ and\ (x_2,y_2)$ are:

$P(x,y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

$\therefore\$ Coordinates of

$D=\left(\dfrac{3+5}{2},\dfrac{-2+2}{2}\right)=(4,0)$

Median

$AD$ divides the

$\triangle ABC$ into two triangles.

$\therefore \$ Area of

$\triangle ABD=\left|\dfrac{1}{2}\left[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right]\right|$

Here,

$(x_1,y_1)=(4,-6),\ (x_2,y_2)=(3,-2)\ and \ (x_3,y_3)=(4,0)$

Hence, area of

$\triangle ABD=\dfrac{1}{2}\left[4(-2-0)+3(0+6)+4(-6+2)\right]$

$= \dfrac{1}{2}\left[-8+18-16 \right]=\left|\dfrac{-6}{2}\right|=3$

Also, area of

$\triangle ADC=\left|\dfrac{1}{2}\left[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right]\right|$

Here,

$(x_1,y_1)=(4,-6),\ (x_2,y_2)=(4,0)\ and \ (x_3,y_3)=(5,-2)$

Hence, area of

$\triangle ADC=\dfrac{1}{2}\left[4(0+2)+4(-2+6)+5(-6-0)\right]$

$= \dfrac{1}{2}\left[8+16-30 \right]=\left|\dfrac{-6}{2}\right|=3$

$\therefore \$ Area of

$\triangle ABD=$ Area of

$\triangle ADC= 3$

Hence, it is proved that a median of a triangle divides it into two triangles of equal areas.

Find the slope of the line that passes through the points $(2,0)$ and $(2,4)$

Since slope of line passing through two points

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and

$\left( { x }_{ 2 },{ y }_{ 2 } \right)$ is

${ m }=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$ .

We now find the slope of the line passing through the points

$(2,0)$ and

$(2,4)$ as shown below:

${ m }=\dfrac { 4-0 }{ 2-2 } =\dfrac { 4 }{ 0 } =$ undefined

Therefore, the slope of the line is undefined.

Find the slope of the line passing through the points $(-1,1)$ and $(2,2)$

Since slope of line passing through two points

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and

$\left( { x }_{ 2 },{ y }_{ 2 } \right)$ is

${ m }=\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$

We now find the slope of the line passing through the points

$(-1,1)$ and

$(2,2)$ as shown below:

${ m }=\frac { 2-1 }{ 2-(-1) } =\frac { 1 }{ 2+1 } =\frac { 1 }{ 3 }$

Therefore, the slope of the line is 1/3.

If a point $A(0,2)$ is equidistant from the points $B(3,p)$ and $C(p,5)$ then find the value of $p.$

Distance between two points=

$\sqrt{(x_2-x_1)^2(y_2-y_1)^2}$

$\therefore AB=\sqrt{(0-3)^2+(2-p)^2}$

$\Rightarrow \sqrt{9+4-4p+p^2}$

$\Rightarrow \sqrt{p^2-4p+13}$

$\therefore AC=\sqrt{(p-0)^2+(5-2)^2}$

$\Rightarrow \sqrt{p^2+9}$

As given

$AB=AC$

$\Rightarrow \sqrt{p^2-4p+13}=\sqrt{p^2+9}$

Squaring both sides

$\Rightarrow p^2-4p+13=p^2+9$

$\Rightarrow -4p=9-13$

$\Rightarrow -4p=-4$

$\Rightarrow p=\dfrac{-4}{-4}=1$

A $(4, - 6$ ), B $(3,- 2$ ) and C $(5, 2)$ are the vertices of a $\Delta ABC$ and AD is its median. Prove that the median AD divides $\Delta$ ABC into two triangles of equal areas.

Given: AD be the median of $\Delta ABC$

BD = DC

Sice D is the mid-point of BC

Using the mid point theorem to find the coordinates of D,

$D = [\dfrac{x_2+x_3}{2}, \frac{y_2+y_3}{2}]$

$D = [\dfrac{3+5}{2}, \frac{-2+2}{2}]=[4, 0]$

Area of $\Delta ABD = \frac{1}{2}\left | [(x_1(y_2-y_4)+x_2(y_4-y_1)+x_4(y_1-y_2))]\right |$

$= \frac{1}{2}\left | [(4(-2-0)+3(0-(-6))+4(-6-(-2)))]\right |$

$= \frac{1}{2}\left | [-8+18-26]\right |$

$= \frac{1}{2}\left | [-6]\right |=3$ Sq.units

Area of $\Delta ADC = \frac{1}{2}\left | [(x_1(y_4-y_3)+x_4(y_3-y_1)+x_3(y_1-y_4))]\right |$

$= \frac{1}{2}\left | [(4(0-2)+4(2-(-6))+5(-6-0))]\right |$

$= \frac{1}{2}\left | [-8+32-30]\right |$

$= \frac{1}{2}\left | [-6]\right |=3$ Sq.units

Area od $\Delta ABD = \Delta ADC$

Hence proved that the median of triangle AD divides the triangles into equal areas.

If the vertices of a triangle are $(1,-3), (4,p)$ and $(-9,7)$ and its' area is $15$ sq. units, find the value (s) of $p.$

The vertices of the triangle are

$A(1, -3), B(4, p)$ and

$C(-9, 7).$

Area of triangle

$= 15$ sq.units.

Area of the triangle with vertices

$(x_1, x_2); (x_2, y_2); (x_3, y_3)$ is

$\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Therefore,

$\dfrac{1}{2}[1.(p-7)+4(7+3)-9(-3-p)]=|15|$

$p-7+40+27+9p=2\times |15|$

$10p+60=\pm30$

$p+6=\pm 3$

$p=-3$ or

$-9.$

If the area of triangle $ABC$ formed by $A(x, y), B(1, 2)$ and $C(2, 1)$ is $6$ square units, then prove that $x + y = 15$ .

Given:

$A(x, y)\equiv(x_1,y_1), B(1, 2)\equiv(x_2,y_2)$ and

$C(2, 1)\equiv(x_3,y_3)$

We know than area of triangle

$\displaystyle=\frac { 1 }{ 2 } \left[ x_{ 1 }(y_{ 2 }-y_{ 3})+x_{ 2 }((y_{ 3 }-y_{ 1 }))+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right]$

(hint: using determinant formula of the area of triangle)

Then area of triangle whoes vertex is

$A(x,y) ,B(1,2)$ and

$C(2,1)$ is

Area of

$\displaystyle \Delta ABC=\left [ x(2-1)+1(1-y)+2(y-2) \right ]=\frac{1}{2}(x+1-y+2y-4)= \frac{1}{2}(x+y-3)$

But given that that the area of triangle is

$6$ sq unit

$\therefore \dfrac{1}{2}(x+y-3)=6$

$\Rightarrow x+y-3=12$

$\Rightarrow x+y=15$

Find the values of $k$ so that the area of the triangle with vertices $(1,\ -1), (-4,\ 2k)$ and $(-k,\ -5)$ is $24$ sq. units.

Given area of triangle whose vertices are

$(1,-1),(-4,2k)$ and

$(-k,-5)$ is

$24$ sq units

We know that the area of triangle

$=\dfrac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ]$

$\therefore 24=\dfrac{1}{2}\left [ 1(2k-(-5))-4(-5-(-1))+(-k)(-1-2k) \right ]$

$\Rightarrow 48=\left [ 2k+5+(-4)(-4)+k(1+2k) \right ]$

$\Rightarrow 48=\left [ 2k+5+16+k+2k^{2} \right ]$

$\Rightarrow 2k^{2}+3k+5+16-48=0$

$\Rightarrow 2k^{2}+3k-27=0$

$\Rightarrow 2k^{2}+9k-6k-27=0$

$\Rightarrow k(2k+9)-3(2k+9)=0$

$\Rightarrow (k-3)(2k+9)=0$

$\Rightarrow k-3=0\; \text{or}\; 2k+9=0$

$\Rightarrow k=3\; \text{or}\; k= \frac{-9}{2}$

Mid-points of the sides $AB$ and $AC$ of $\triangle {ABC}$ are $(3,5)$ and $(-3,-3)$ respectively, then find the length of $BC$

$\triangle (ADE)$ and

$\triangle (ABC)$

$\dfrac{AD}{AB}=\dfrac{AE}{AC} i< A$ common

$DE||BC$

$\Rightarrow ADE \approx ABC\Rightarrow \dfrac{AD}{AB}=\dfrac{DE}{BC}=\dfrac{1}{2}\Rightarrow BC=2DE$

$\Rightarrow BC=2\sqrt{(3+3)^2+(5+3)^2}$

$=2\times 10=\boxed{20}$

1352925_516791_ans_f717faa8d71d4caabb3c21b6c45ac517.png

If the points $A(7,9), B(3,-7)$ and $C(-3,3)$ are vertices of a triangle, then find the measure $\angle{C}$ (in degrees)

1353045_516679_ans_dfd1aae127154f57bf1c0f56f3ae3119.jpg

Given that three points $(1,2),(2,4),(k,6)$ are collinear, find the value of $k$ .

As Points are collinear

$\Rightarrow$ area of traiangle formed by

$3$ Points is zero.

$\Rightarrow \dfrac{1}{2} \begin{vmatrix} (x_1-x_2)& (x_2-x_3)\\ (y_1-y_2) & (y_2-y_3)\end{vmatrix}=0$

$\Rightarrow \begin{vmatrix} 1-2 &2-k\\2-4 & 4-6\end{vmatrix}$

$\Rightarrow 2-(k-2)2=0$

$\Rightarrow 1+2-k=0\Rightarrow \boxed{K=3}$

Determine the equation of the line in the given graph.

We may take the equation in the form of

$y=ax+b$ .

Observe that the straight-line passes through

$(2,0)$ .

This means that

$y=0$ when

$x=2$ .

But the substitution

$y=0, x=2$ in the equation gives

$0=2a+b$ .

Thus

$b=-2a$
We also observe that the line passes through

$(0,-3)$ ; that is

$y=-3$ , whenever

$x=0$ .

This gives

$-3=0+b$ or

$b=-3\implies a=\dfrac{3}{2}$ .

Therefore, the equation of the line is

$y=\dfrac{3}{2}x-3$ i.e.

$2y=3x-6$ or

$3x-2y=6$ .

Determine the equation of the line in the given graph.

We may take the equation in the form of

$y=ax+b$ .

Observe that the straight-line passes through

$(0,0)$ .

This means that

$y=0$ when

$x=0$ .

But the substitution

$x=0$ in the equation gives

$0=y=b$ .

Thus

$b=0$ and so that

$y=ax$ .
We also observe that the line passes through

$(1,-3)$ ; that is

$y=-3$ , whenever

$x=1$ .

This gives

$-3=a\times 1$ or

$a=-3$ .

The equation of the line is therefore

$y=-3x$ .

Write the equation $4x-5y=12$ in slope intercept form.

The equation given is $4x-5y=12$
Which implies $5y = 4x-12$ , which gives $y = \dfrac{4}{5}x-\dfrac{12}{5}$
Therefore in slope intercept form , the slope is $\dfrac {4}5$ and $Y$ -intercept is $\dfrac{-12}5$

If $a\ast b=ab+a+b$ , draw the graph of $y=3\ast x+1\ast 2$ .

Given :

$a\ast b=ab+a+b$ .......

$(i)$

Consider,

$y=3\ast x+1\ast 2$

$=3x+3+x+1\times 2+1+2$ ......... From

$(i)$

$=4x+8$

So, the equation we get is

$y=4x+8$

Let's assign some values to

$x$ and get the values for

$y$

$x$

$-2$

$-1$

$0$

$1$

$y$

$0$

$4$

$8$

$12$

Plot these points on the graph and get the same image as shown above.

Find the slope of the line joining the points $(2a, 3b)$ and $(a, -b)$ .

We know that the slope of the line joining two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is:

$m=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$

Here, the given points are

$(2a,3b)$ and

$(a,-b)$ , therefore, the slope of the line is:

$m=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\dfrac { -b-3b }{ a-2a } =\dfrac { -4b }{ -a } =\dfrac { 4b }{ a }$

Hence, the slope of the line is

$\dfrac { 4b }{ a }$ .

Find the slope of the line joining the points $(0, 0)$ and $(\sqrt 3, 3)$ .

We know that the slope of the line joining two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is:

$m=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$

Here, the given points are

$(0,0)$ and

$(\sqrt { 3 } ,3)$ , therefore, the slope of the line is:

$m=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\dfrac { 3-0 }{ \sqrt { 3 } -0 } =\dfrac { 3 }{ \sqrt { 3 } } =\dfrac { 3\times \sqrt { 3 } }{ \sqrt { 3 } \times \sqrt { 3 } } =\dfrac { 3\sqrt { 3 } }{ 3 } =\sqrt { 3 }$

Hence, the slope of the line is

$\sqrt { 3 }$ .

Find the slope of the line joining the points $(-8)$ and $(5, -2)$

We know that the slope of the line joining two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is:

$m=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$

Here, the given points are

$(4,-8)$ and

$(5,-2)$ , therefore, the slope of the line is:

$m=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\dfrac { -2-(-8) }{ 5-4 } =\dfrac { -2+8 }{ 1 } =6$

Hence, the slope of the line is

$6$ .

Find the angle of inclination of straight line whose slope is $\displaystyle\frac{1}{\sqrt 3}$ .

We know that if

$\theta$ is the inclination of a straight line, then

$\tan { \theta }$ is called the slope (or gradient) of a line. The slope of a line is usually denoted by

$m$ .

Here, it is given that the slope of the line is

$m=\dfrac { 1 }{ \sqrt { 3 } }$ , therefore, the angle of inclination is:

$m=\tan { \theta } \\ \Rightarrow \dfrac { 1 }{ \sqrt { 3 } } =\tan { \theta } \\ \Rightarrow \tan { { 30 }^{ 0 } } =\tan { \theta } \quad \quad \left( \because \quad \tan { { 30 }^{ 0 } } =\dfrac { 1 }{ \sqrt { 3 } } \right) \\ \Rightarrow \theta ={ 30 }^{ 0 }$

Hence, the angle of inclination is

$30^0$ .

Find the slope of the line whose inclination is $60^o$ .

We can see that:

$\theta =60^o$
We Know that, Slope

$=m=\tan\theta$

$m=\tan 60^o$

$\therefore =\sqrt 3$ .

Find the angle of inclination of straight line whose slope is $0$ .

We know that slope

$=\tan\theta$

$0=\tan\theta$
We know that,

$\tan 0^o=0$

$\therefore\theta =0^o$

Find the angle of inclination of straight line whose slope is $1$ .

We know that if

$\theta$ is the inclination of a straight line, then

$\tan { \theta }$ is called the slope (or gradient) of a line. The slope of a line is usually denoted by

$m$ .

Here, it is given that the slope of the line is

$m=1$ , therefore, the angle of inclination is:

$m=\tan { \theta } \\ \Rightarrow 1=\tan { \theta } \\ \Rightarrow \tan { { 45 }^{ 0 } } =\tan { \theta } \quad \quad (\because \quad \tan { { 45 }^{ 0 } } =1)\\ \Rightarrow \theta ={ 45 }^{ 0 }$

Hence, the angle of inclination is

$45^0$ .

Find the angles of inclination of straight lines whose slopes are $\sqrt 3$ .

We know that if

$\theta$ is the inclination of a straight line, then

$\tan { \theta }$ is called the slope (or gradient) of a line. The slope of a line is usually denoted by

$m$ .

Here, it is given that the slope of the line is

$m=\sqrt { 3 }$ , therefore, the angle of inclination is:

$m=\tan { \theta } \\ \Rightarrow \sqrt { 3 } =\tan { \theta } \\ \Rightarrow \tan { { 60 }^{ 0 } } =\tan { \theta } \quad \quad (\because \quad \tan { { 60 }^{ 0 } } =\sqrt { 3 } )\\ \Rightarrow \theta ={ 60 }^{ 0 }$

Hence, the angle of inclination is

$60^0$ .

Find the slope of the line joining the points $(-5, 0)$ and $(0, -7)$ .

We know that the slope of the line joining two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is:

$m=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$

Here, the given points are

$(-5,0)$ and

$(0,-7)$ , therefore, the slope of the line is:

$m=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\dfrac { -7-0 }{ 0-(-5) } =-\dfrac { 7 }{ 5 }$

Hence, the slope of the line is

$-\dfrac { 7 }{ 5 }$ .

Find the slope of the line joining the points $(-4, 1)$ and $(-5, 2)$

We know that the slope of the line joining two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is:

$m=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$

Here, the given points are

$(-4,1)$ and

$(-5,2)$ , therefore, the slope of the line is:

$m=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\dfrac { 2-1 }{ -5-(-4) } =\dfrac { 1 }{ -5+4 } =-1$

Hence, the slope of the line is

$-1$ .

Prove that $(-5,-3),(1,-11),(7,-6),(1,2)$ coordinates are the vertices of parallelograms

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Let the given vertices be

$A=(-5,-3)$ ,

$B=(1,-11)$ ,

$C=(7,-6)$ and

$D=(1,2)$

We first find the distance between

$A=(-5,-3)$ and

$B=(1,-11)$ as follows:

$AB=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (1-(-5)) }^{ 2 }+{ (-11-(-3)) }^{ 2 } } =\sqrt { { (1+5) }^{ 2 }+{ (-11+3) }^{ 2 } } =\sqrt { { 6 }^{ 2 }+{ (-8) }^{ 2 } }$

$=\sqrt { 36+64 } =\sqrt { 100 } =10$

Similarly, the distance between

$B=(1,-11)$ and

$C=(7,-6)$ is:

$BC=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (7-1) }^{ 2 }+{ (-6-(-11)) }^{ 2 } } =\sqrt { 6^{ 2 }+{ (-6+11) }^{ 2 } } =\sqrt { { 6 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 36+25 }$

$=\sqrt { 61 }$

Now, the distance between

$C=(7,-6)$ and

$D=(1,2)$ is:

$CD=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (1-7) }^{ 2 }+{ (2-(-6)) }^{ 2 } } =\sqrt { (-6)^{ 2 }+{ (2+6) }^{ 2 } } =\sqrt { (-6)^{ 2 }+{ 8 }^{ 2 } } =\sqrt { 36+64 }$

$=\sqrt { 100 } =10$

Now, the distance between

$D=(1,2)$ and

$A=(-5,-3)$ is:

$DA=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (1-(-5)) }^{ 2 }+{ (2-(-3)) }^{ 2 } } =\sqrt { { (1+5) }^{ 2 }+{ (2+3) }^{ 2 } } =\sqrt { 6^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 36+25 }$

$=\sqrt { 61 }$

We also know that if the opposite sides have equal side lengths, then

$ABCD$ is a parallelogram.

Here, since the lengths of the opposite sides are equal that is:

$AB=CD=10$ and

$BC=DA=\sqrt {61}$

Hence, the given vertices are the vertices of parallelogram.

Find the distance between the origin and the point $(5, 12)$ .

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Since, the origin has the coordinates

$(0,0)$ , thus

The given points are

$(0,0 )$ and

$(5,12 )$ and therefore,

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (5-0) }^{ 2 }+{ (12-0) }^{ 2 } } =\sqrt { { 5 }^{ 2 }+{ 12 }^{ 2 } } =\sqrt { 25+144 } =\sqrt { 169 } =\sqrt { 13^{ 2 } } =13$

Hence, the distance is

$13$ units.

Find the perimeter of the triangles whose vertices have the following coordinates $(-2, 1), (4, 6), (6, -3)$ .

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Let the given vertices be

$A=(-2,1)$ ,

$B=(4,6)$ and

$C=(6,-3)$

We first find the distance between

$A=(-2,1)$ and

$B=(4,6)$ as follows:

$AB=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (4-(-2)) }^{ 2 }+{ (6-1) }^{ 2 } }$

$=\sqrt { { (4+2) }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { { 6 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 36+25 } =\sqrt { 61 }$

Similarly, the distance between

$B=(4,6)$ and

$C=(6,-3)$ is:

$BC=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (6-4) }^{ 2 }+{ (-3-6) }^{ 2 } }$

$=\sqrt { { 2 }^{ 2 }+{ (-9) }^{ 2 } } =\sqrt { 4+81 } =\sqrt { 85 }$

Now, the distance between

$C=(6,-3)$ and

$A=(-2,1)$ is:

$CA=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (6-(-2)) }^{ 2 }+{ (-3-1) }^{ 2 } }$

$=\sqrt { { (6+2) }^{ 2 }+{ (-4) }^{ 2 } } =\sqrt { { 8 }^{ 2 }+{ (-4) }^{ 2 } } =\sqrt { 64+16 }$

$=\sqrt { 80 }$

Since the perimeter

$P$ of a triangle

$ABC$ is

$AB+BC+CA$ , therefore,

$P=\sqrt { 61 } +\sqrt { 85 } +\sqrt { 80 }$

Hence, the perimeter of the triangle is

$(\sqrt { 61 } +\sqrt { 85 } +\sqrt { 80 })$ units.

Find the perimeter of the triangles whose vertices have the following coordinates $(3, 10), (5, 2), (14, 12)$ .

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Let the given vertices be

$A=(3,10)$ ,

$B=(5,2)$ and

$C=(14,12)$

We first find the distance between

$A=(3,10)$ and

$B=(5,2)$ as follows:

$AB=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (5-3) }^{ 2 }+{ (2-10) }^{ 2 } } =\sqrt { { 2 }^{ 2 }+{ (-8) }^{ 2 } } =\sqrt { 4+64 } =\sqrt { 68 }$

Similarly, the distance between

$B=(5,2)$ and

$C=(14,12)$ is:

$BC=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (14-5) }^{ 2 }+{ (12-2) }^{ 2 } } =\sqrt { { 9 }^{ 2 }+{ 10 }^{ 2 } } =\sqrt { 81+100 } =\sqrt { 181 }$

Now, the distance between

$C=(14,12)$ and

$A=(3,10)$ is:

$CA=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (14-3) }^{ 2 }+{ (12-10) }^{ 2 } } =\sqrt { 11^{ 2 }+{ 2 }^{ 2 } } =\sqrt { 121+4 } =\sqrt { 125 }$

Since the perimeter

$P$ of a triangle

$ABC$ is

$AB+BC+CA$ , therefore,

$P=\sqrt { 68 } +\sqrt { 181 } +\sqrt { 125 }$

Hence, the perimeter of the triangle is

$\sqrt { 68 } +\sqrt { 181 } +\sqrt { 125 }$ units.

A point $P(2, -1)$ is equidistant from the points $(a, 7)$ and $(-3, a)$ . Find 'a'.

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Let the given points be

$A=(a,7)$ and

$B=(-3,a)$ and the third point given is

$P(2,-1)$ .

We first find the distance between

$P(2,-1)$ and

$A=(a,7)$ as follows:

$PA=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (a-2) }^{ 2 }+{ (7-(-1)) }^{ 2 } } =\sqrt { { (a-2) }^{ 2 }+{ (7+1) }^{ 2 } } =\sqrt { { (a-2) }^{ 2 }+8^{ 2 } }$

$=\sqrt { { (a-2) }^{ 2 }+64 }$

Similarly, the distance between

$P(2,-1)$ and

$B=(-3,a)$ is:

$PB=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (-3-2) }^{ 2 }+{ (a-(-1)) }^{ 2 } } =\sqrt { { (-5) }^{ 2 }+{ (a+1) }^{ 2 } } =\sqrt { 25+{ (a+1) }^{ 2 } }$

Since the point

$P(2,-1)$ is equidistant from the points

$A(a,7)$ and

$B=(-3,a)$ , therefore,

$PA=PB$ that is:

$\sqrt { { (a-2) }^{ 2 }+64 } =\sqrt { 25+{ (a+1) }^{ 2 } } \\ \Rightarrow (\sqrt { { (a-2) }^{ 2 }+64 } )^{ 2 }=(\sqrt { 25+{ (a+1) }^{ 2 } } )^{ 2 }\\ \Rightarrow { (a-2) }^{ 2 }+64=25+{ (a+1) }^{ 2 }\\ \Rightarrow { (a-2) }^{ 2 }-{ (a+1) }^{ 2 }=25-64\\ \Rightarrow { (a }^{ 2 }+4-4a)-{ (a }^{ 2 }+1+2a)=-39\quad \quad \quad \quad \quad \quad (\because \quad { (a-b) }^{ 2 }=a^{ 2 }+b^{ 2 }-2ab,\quad { (a+b) }^{ 2 }=a^{ 2 }+b^{ 2 }+2ab)$

$\Rightarrow { a }^{ 2 }+4-4a-{ a }^{ 2 }-1-2a=-39\\ \Rightarrow -6a+3=-39\\ \Rightarrow -6a=-39-3\\ \Rightarrow -6a=-42\\ \Rightarrow a=\dfrac { -42 }{ -6 } =7$

Hence,

$a=7$ .

Prove that the points $A(1, -3), B(-3, 0)$ and $C(4, 1)$ are the vertices of right isosceles triangle.

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Let the given vertices be

$A=(1,-3)$ ,

$B=(-3,0)$ and

$C=(4,1)$

We first find the distance between

$A=(1,-3)$ and

$B=(-3,0)$ as follows:

$AB=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (-3-1) }^{ 2 }+{ (0-(-3)) }^{ 2 } } =\sqrt { { (-4) }^{ 2 }+{ (0+3) }^{ 2 } } =\sqrt { { (-4) }^{ 2 }+{ 3 }^{ 2 } } =\sqrt { 16+9 }$

$=\sqrt { 25 } =5$

Similarly, the distance between

$B=(-3,0)$ and

$C=(4,1)$ is:

$BC=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (4-(-3)) }^{ 2 }+{ (1-0) }^{ 2 } } =\sqrt { { (4+3) }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { { 7 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 49+1 } =\sqrt { 50 }$

$=\sqrt { 5^{ 2 }\times 2 } =5\sqrt { 2 }$

Now, the distance between

$C=(4,1)$ and

$A=(1,-3)$ is:

$CA=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (1-4) }^{ 2 }+{ (-3-1) }^{ 2 } } =\sqrt { { (-3) }^{ 2 }+{ (-4) }^{ 2 } } =\sqrt { 9+16 } =\sqrt { 25 } =\sqrt { 5^{ 2 } } =5$

We also know that If any two sides have equal side lengths, then the triangle is isosceles.

Here, since the lengths of the two sides are equal that is

$AB=CA=5$ , therefore,

$ABC$ is an isosceles triangle.

Now consider,

$AB^{ 2 }+CA^{ 2 }=5^{ 2 }+5^{ 2 }=25+25=50=(5\sqrt { 2 } )^{ 2 }=BC^{ 2 }$

Since

$AB^2+CA^2=BC^2$ , therefore,

$ABC$ is a right angled triangle.

Hence, the given vertices are the vertices of isosceles right angled triangle.

The distance between the points $(3, 1)$ and $(0, x)$ is $5$ units. Find x.

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Here, the given points are

$(3,1 )$ and

$(0,x)$ and the distance is

$5$ units, therefore,

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } \\ \Rightarrow 5=\sqrt { { (0-3) }^{ 2 }+{ (x-1) }^{ 2 } } \\ \Rightarrow 5=\sqrt { { (-3) }^{ 2 }+{ (x-1) }^{ 2 } } \\ \Rightarrow 5=\sqrt { 9+{ (x-1) }^{ 2 } } \\ \Rightarrow 5^{ 2 }=(\sqrt { 9+{ (x-1) }^{ 2 } } )^{ 2 }\\ \Rightarrow 25=9+{ (x-1) }^{ 2 }$

$\Rightarrow { (x-1) }^{ 2 }=25-9\\ \Rightarrow { (x-1) }^{ 2 }=16\\ \Rightarrow x-1=\pm \sqrt { 16 } \\ \Rightarrow x-1=\pm 4\\ \Rightarrow x-1=4,\quad x-1=-4\\ \Rightarrow x=4+1,\quad x=-4+1\\ \Rightarrow x=5,\quad x=-3$

Hence,

$x=-3,5$ .

Find the radius of a circle whose centre is $(-5, 4)$ and which passes through the point $(-7, 1)$ .

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Since, the radius of the circle is the distance between the centre and the point.

Let the centre of the circle is

$)=(-5,4)$ and the point be

$A=(-7,1 )$ , then the radius

$OA$ is:

$OA=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (-7+5) }^{ 2 }+{ (1-4) }^{ 2 } } =\sqrt { { (-2) }^{ 2 }+{ (-3) }^{ 2 } } =\sqrt { 4+9 } =\sqrt { 13 }$

Hence, the radius of the circle is

$\sqrt {13}$ units.

Find the distance between the following pairs of points $(2, 0)$ and $(0, 3)$ .

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Here, the given points are

$(2,0)$ and

$(0,3)$ , therefore,

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (0-2) }^{ 2 }+{ (3-0) }^{ 2 } } =\sqrt { { (-2) }^{ 2 }+3^{ 2 } } =\sqrt { 4+9 } =\sqrt { 13 }$

Hence, the distance is

$\sqrt {13}$ units.

Find a point on y-axis which is equidistant from the points $(5, 2)$ and $(-4, 3)$ .

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Let the given points be

$A=(5,2)$ and

$B=(-4,3)$ and let the point on y-axis be

$P(0,y)$ .

We first find the distance between

$P(0,y)$ and

$A=(5,2)$ as follows:

$PA=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (5-0) }^{ 2 }+{ (2-y) }^{ 2 } } =\sqrt { { 5 }^{ 2 }+{ (2-y) }^{ 2 } } =\sqrt { 25+{ (2-y) }^{ 2 } }$

Similarly, the distance between

$P(0,y)$ and

$B=(-4,3)$ is:

$PB=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (-4-0) }^{ 2 }+{ (3-y) }^{ 2 } } =\sqrt { { (-4) }^{ 2 }+{ (3-y) }^{ 2 } } =\sqrt { 16+{ (3-y) }^{ 2 } }$

Since the point

$P(0,y)$ is equidistant from the points

$A=(5,2)$ and

$B=(-4,3)$ , therefore,

$PA=PB$ that is:

$\sqrt { 25+{ (2-y) }^{ 2 } } =\sqrt { 16+{ (3-y) }^{ 2 } }$

$\\ \Rightarrow (\sqrt { 25+{ (2-y) }^{ 2 } } )^{ 2 }=(\sqrt { 16+{ (3-y) }^{ 2 } } )^{ 2 }$

$\\ \Rightarrow 25+{ (2-y) }^{ 2 }=16+{ (3-y) }^{ 2 }$

$\\ \Rightarrow { (2-y) }^{ 2 }-{ (3-y) }^{ 2 }=16-25$

$\\ \Rightarrow { (4+y }^{ 2 }-4y)-{ (9+y }^{ 2 }-6y)=-9\quad \quad \quad \quad \quad \quad \quad \quad \quad (\because \quad { (a-b) }^{ 2 }=a^{ 2 }+b^{ 2 }-2ab)$

$\Rightarrow { 4+y }^{ 2 }-4y-{ 9-y }^{ 2 }+6y=-9$

$\\ \Rightarrow 2y-5=-9$

$\\ \Rightarrow 2y=-9+5$

$\\ \Rightarrow 2y=-4$

$\\ \Rightarrow y=\dfrac { -4 }{ 2 } =-2$

Hence, the point on the y-axis is

$(0,-2)$ .

The coordinate of vertices of triangles are given. Identify the types of triangles $(3, 5)(-1, 1)(6, 2)$ .

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Let the given vertices be

$A=(3,5)$ ,

$B=(-1,1)$ and

$C=(6,2)$

We first find the distance between

$A=(3,5)$ and

$B=(-1,1)$ as follows:

$AB=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (-1-3) }^{ 2 }+{ (1-5) }^{ 2 } } =\sqrt { { (-4) }^{ 2 }+{ (-4) }^{ 2 } } =\sqrt { 16+16 } =\sqrt { 32 } =\sqrt { 4^{ 2 }\times 2 }$

$=4\sqrt { 2 }$

Similarly, the distance between

$B=(-1,1)$ and

$C=(6,2)$ is:

$BC=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (6-(-1)) }^{ 2 }+{ (2-1) }^{ 2 } } =\sqrt { { (6+1) }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { { 7 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 49+1 } =\sqrt { 50 }$

$=\sqrt { 5^{ 2 }\times 2 } =5\sqrt { 2 }$

Now, the distance between

$C=(6,2)$ and

$A=(3,5)$ is:

$CA=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (3-6) }^{ 2 }+{ (5-2) }^{ 2 } } =\sqrt { { (-3) }^{ 2 }+{ 3 }^{ 2 } } =\sqrt { 9+9 } =\sqrt { 18 } =\sqrt { 3^{ 2 }\times 2 } =3\sqrt {2}$

We also know that if

$AB^2+CA^2=BC^2$ , then the triangle

$ABC$ is a right angled triangle.

Thus consider,

$AB^{ 2 }+CA^{ 2 }=(4\sqrt { 2 } )^{ 2 }+(3\sqrt { 2 } )^{ 2 }=(16\times 2)+(9\times 2)=32+18=50=(5\sqrt { 2 } )^{ 2 }=BC^{ 2 }$

Since

$AB^2+CA^2=BC^2$ , therefore,

$ABC$ is a right angled triangle.

Hence, the given vertices are the vertices of right angled triangle.

The coordinate of vertices of triangles are given. Identify the types of triangles $(1, 6)(3, 2)(10, 8)$ .

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Let the given vertices be

$A=(1,6)$ ,

$B=(3,2)$ and

$C=(10,8)$

We first find the distance between

$A=(1,6)$ and

$B=(3,2)$ as follows:

$AB=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (3-1) }^{ 2 }+{ (2-6) }^{ 2 } } =\sqrt { { 2 }^{ 2 }+{ (-4) }^{ 2 } } =\sqrt { 4+16 }=\sqrt {20} =\sqrt { { 2 }^{ 2 }\times 5 } =2\sqrt {5}$

Similarly, the distance between

$B=(3,2)$ and

$C=(10,8)$ is:

$BC=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (10-3) }^{ 2 }+{ (8-2) }^{ 2 } } =\sqrt { { 7 }^{ 2 }+{ 6 }^{ 2 } } =\sqrt { 49+36 } =\sqrt { 85 }$

Now, the distance between

$C=(10,8)$ and

$A=(1,6)$ is:

$CA=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (10-1) }^{ 2 }+{ (8-6) }^{ 2 } } =\sqrt { { 9 }^{ 2 }+{ 2 }^{ 2 } } =\sqrt { 81+4 } =\sqrt { 85 }$

We also know that If any two sides have equal side lengths, then the triangle is isosceles.

Here, since the lengths of the two sides are equal that is

$BC=CA=\sqrt {85}$

Hence, the given vertices are the vertices of an isosceles triangle.

Find the slope of the line passing through $A (-2, 1)$ and $B(0, 3)$

Points are

$A(-2, 1)$ and

$B(0, 3)$

$A = (x_{1}, y_{1})B = (x_{2}, y_{2})$

$\therefore$ Slope of the line

$AB = \dfrac {y_{2} - y_{1}}{x_{2} - x_{1}}$

$= \dfrac {3 - 1}{0 - (-2)}$

$= \dfrac {2}{2} = 1$

Prove that the set of coordinates are the vertices of parallelogram $(4, 0), (-2, -3), (3, 2), (-3, -1)$ .

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Let the given vertices be

$A=(4,0)$ ,

$B=(-2,-3)$ ,

$C=(3,2)$ and

$D=(-3,-1)$

We first find the distance between

$A=(4,0)$ and

$B=(-2,-3)$ as follows:

$AB=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (-2-4) }^{ 2 }+{ (-3-0) }^{ 2 } } =\sqrt { { (-6) }^{ 2 }+{ (-3) }^{ 2 } } =\sqrt { 36+9 } =\sqrt { 45 } =\sqrt { 3^{ 2 }\times 5 }$

$=3\sqrt { 5 }$

Similarly, the distance between

$B=(-2,-3)$ and

$C=(3,2)$ is:

$BC=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (3-(-2)) }^{ 2 }+{ (2-(-3)) }^{ 2 } } =\sqrt { { (3+2) }^{ 2 }+{ (2+3) }^{ 2 } } =\sqrt { { 5 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 25+25 }$

$=\sqrt { 50 } =\sqrt { 5^{ 2 }\times 2 } =5\sqrt { 2 }$

Now, the distance between

$C=(3,2)$ and

$D=(-3,-1)$ is:

$CD=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (-3-3) }^{ 2 }+{ (-1-2) }^{ 2 } } =\sqrt { { (-6) }^{ 2 }+{ (-3) }^{ 2 } } =\sqrt { 36+9 } =\sqrt { 45 } =\sqrt { 3^{ 2 }\times 5 }$

$=3\sqrt { 5 }$

Now, the distance between

$D=(-3,-1)$ and

$A=(4,0)$ is:

$DA=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (-3-4) }^{ 2 }+{ (-1-0) }^{ 2 } } =\sqrt { { (-7) }^{ 2 }+{ (-1) }^{ 2 } } =\sqrt { 49+1 } =\sqrt { 50 } =\sqrt { 5^{ 2 }\times 2 }$

$=5\sqrt { 2 }$

We also know that if the opposite sides have equal side lengths, then

$ABCD$ is a parallelogram.

Here, since the lengths of the opposite sides are equal that is:

$AB=CD=3\sqrt {5}$ and

$BC=DA=5\sqrt {2}$

Hence, the given vertices are the vertices of parallelogram.

Find the distance between the following pairs of points $(1, -3)$ and $(-4, 7)$ .

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Here, the given points are

$(1,-3)$ and

$(-4,7)$ , therefore,

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (-4-1) }^{ 2 }+{ (7-(-3)) }^{ 2 } } =\sqrt { { (-5) }^{ 2 }+{ (7+3) }^{ 2 } } =\sqrt { { (-5) }^{ 2 }+10^{ 2 } } =\sqrt { 25+100 }$

$=\sqrt { 125 } =\sqrt { 5^{ 2 }\times 5 } =5\sqrt { 5 }$

Hence, the distance is

$5\sqrt { 5 }$ units.

If the points $A(1, 2), B(4, 6), C(3, 5)$ are the vertices of a $\Delta ABC$ , find the equation of the line passing through the midpoints of $AB$ and $BC$ .

The points

$A (1, 2), B (4, 6), C (3, 5)$ are the vertices of

$\Delta ABC$ .
Find: Equation of the line passing through the midpoints of

$AB$ and

$BC$ .
Let midpoint of

$AB = P$
and midpoint of

$BC = Q$
Equation of line

$PQ$ :
By midpoint formula.

$P (x, y), A (1, 2) = (x_1, y_1), B(4, 6) = (x_2, y_2)$

$x,y = \dfrac{x_1 + x_2}{2} , \dfrac{y_2 + y_1}{2}$

$= \dfrac{1 + 4}{2} , \dfrac{2 + 6}{2}$

$= \dfrac{5}{2}, \dfrac{8}{2}$

$= \dfrac{5}{2} , 4$

$\therefore P\left( \dfrac{5}{2} , 4 \right )$
By midpoint formula:

$Q(x, y), B(4, 6) = (x_1, y_1) , C(3, 5) = (x_2 , y_2)$

$x, y = \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2}$

$= \dfrac{4 + 3}{2}, \dfrac{6 + 5}{2}$

$= \dfrac{7}{2}, \dfrac{11}{2}$

$\therefore Q \left( \dfrac{7}{2} , \dfrac{11}{2} \right )$

$\therefore P\left ( \dfrac{5}{2}, 4 \right ), Q \left ( \dfrac{7}{2} , \dfrac{11}{2} \right )$
Equation of line PQ by two point formula

$\dfrac{x - x_1}{x_1 - x_2} = \dfrac{y - y_1}{y_1 - y_2}$

$\dfrac{x - \dfrac{5}{2}}{\dfrac{5}{2} - \dfrac{7}{2}} = \dfrac{y - 4}{4 - \dfrac{11}{2}}$ .....

$\left [ P \left ( \dfrac{5}{2}, 4 \right ) , Q \left ( \dfrac{7}{2} , \dfrac{11}{2} \right ) \right ]$

$\left [ x - \dfrac{5}{2} \div \dfrac{5}{2} - \dfrac{7}{2} \right ] = y - 4 \div 4 - \dfrac{11}{2}$

$\left [ x - \dfrac{5}{2} \div \dfrac{-2}{2} \right ] = y - 4 \div \dfrac{8 - 11}{2}$

$\left [ \dfrac{2x - 5}{2} \div - 1 \right ] = y - 4 \div \left ( \dfrac{-3}{2} \right )$

$\left [ \dfrac{2x - 5}{2} \times - 1 \right ] = (y - 4) \times - \left ( \dfrac{2}{3} \right )$

$\therefore - \dfrac{2x - 5}{2} = - \dfrac{2y - 8}{3}$

$6x - 15 = 4y - 16$

$6x - 4y + 1 = 0$

$\therefore$ The equation of line

$PQ$ is

$6x - 4y + 1 = 0$
Hence, equation of the line passing though the midpoints of

$AB$ and

$BC$ is

$6x - 4y + 1 = 0.$

Prove that $(4, 0), (-2, -3), (3, 2), (-3, -1)$ coordinates are not the vertices of parallelogram.

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Let the given vertices be

$A=(4,0)$ ,

$B=(-2,-3)$ ,

$C=(3,2)$ and

$D=(-3,-1)$

We first find the distance between

$A=(4,0)$ and

$B=(-2,-3)$ as follows:

$AB=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (-2-4) }^{ 2 }+{ (-3-0) }^{ 2 } } =\sqrt { { (-6) }^{ 2 }+{ (-3) }^{ 2 } } =\sqrt { 36+9 } =\sqrt { 45 }$

Similarly, the distance between

$B=(-2,-3)$ and

$C=(3,2)$ is:

$BC=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (3-(-2)) }^{ 2 }+{ (2-(-3)) }^{ 2 } } =\sqrt { { (3+2) }^{ 2 }+{ (2+3) }^{ 2 } } =\sqrt { { 5 }^{ 2 }+5^{ 2 } } =\sqrt { 25+25 }$

$=\sqrt { 50 } =\sqrt { 5^{ 2 }\times 2 } =5\sqrt { 2 }$

Now, the distance between

$C=(3,2)$ and

$D=(-3,-1)$ is:

$CD=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (-3-3) }^{ 2 }+{ (-7-2) }^{ 2 } } =\sqrt { { (-6) }^{ 2 }+(-9)^{ 2 } } =\sqrt { 36+81 } =\sqrt { 117 }$

Now, the distance between

$D=(-3,-1)$ and

$A=(4,0)$ is:

$DA=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (-3-4) }^{ 2 }+{ (-7-0) }^{ 2 } } =\sqrt { { (-7) }^{ 2 }+(-7)^{ 2 } } =\sqrt { 49+49 } =\sqrt { 98 } =\sqrt { 7^{ 2 }\times 2 }$

$=7\sqrt { 2 }$

We also know that if the opposite sides have equal side lengths, then

$ABCD$ is a parallelogram.

Here, since the lengths of the opposite sides are not equal that is:

$AB\neq CD$ and

$BC\neq DA$

Hence, the given vertices are not the vertices of parallelogram.

The coordinate of vertices of triangles are given. Identify the types of triangles $(3, -3)(3, 5)(11, -3)$ .

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Let the given vertices be

$A=(3,-3)$ ,

$B=(3,5)$ and

$C=(11,-3)$

We first find the distance between

$A=(3,-3)$ and

$B=(3,5)$ as follows:

$AB=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (3-3) }^{ 2 }+{ (5-(-3)) }^{ 2 } } =\sqrt { { 0 }^{ 2 }+{ (5+3) }^{ 2 } } =\sqrt { { 0 }^{ 2 }+{ 8 }^{ 2 } } =\sqrt { 64 } =\sqrt { 8^{ 2 } } =8$

Similarly, the distance between

$B=(3,5)$ and

$C=(11,-3)$ is:

$BC=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (11-3) }^{ 2 }+{ (-3-5) }^{ 2 } } =\sqrt { { 8 }^{ 2 }+{ (-8) }^{ 2 } } =\sqrt { 64+64 } =\sqrt { 128 } =\sqrt { 8^{ 2 }\times 2 } =8\sqrt { 2 }$

Now, the distance between

$C=(11,-3)$ and

$A=(3,-3)$ is:

$CA=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (11-3) }^{ 2 }+{ (-3-(-3)) }^{ 2 } } =\sqrt { { 8 }^{ 2 }+{ (-3+3) }^{ 2 } } =\sqrt { { 8 }^{ 2 }+{ 0 }^{ 2 } } =\sqrt { 64 } =\sqrt { 8^{ 2 } } =8$

We also know that if

$AB^2+CA^2=BC^2$ , then the triangle

$ABC$ is a right angled triangle.

Thus consider,

$AB^{ 2 }+CA^{ 2 }=8 ^{ 2 }+8^{ 2 }=64+64=128=(8\sqrt { 2 } )^{ 2 }=BC^{ 2 }$

Since

$AB^2+CA^2=BC^2$ , therefore,

$ABC$ is a right angled triangle.

Hence, the given vertices are the vertices of right angled triangle.

The coordinate of vertices of triangles are given. Identify the types of triangles $(2, 1)(10, 1)(6, 9)$ .

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Let the given vertices be

$A=(2,1)$ ,

$B=(10,1)$ and

$C=(6,9)$

We first find the distance between

$A=(2,1)$ and

$B=(10,1)$ as follows:

$AB=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (10-2) }^{ 2 }+{ (1-1) }^{ 2 } } =\sqrt { { 8 }^{ 2 }+{ 0 }^{ 2 } } =\sqrt { 64 } =\sqrt { { 8 }^{ 2 } } =8$

Similarly, the distance between

$B=(10,1)$ and

$C=(6,9)$ is:

$BC=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (6-10) }^{ 2 }+{ (9-1) }^{ 2 } } =\sqrt { { (-4) }^{ 2 }+{ 8 }^{ 2 } } =\sqrt { 16+64 } =\sqrt { 80 } =\sqrt { 4^{ 2 }\times 5 } =4\sqrt { 5 }$

Now, the distance between

$C=(6,9)$ and

$A=(2,1)$ is:

$CA=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (6-2) }^{ 2 }+{ (9-1) }^{ 2 } } =\sqrt { { 4 }^{ 2 }+{ 8 }^{ 2 } } =\sqrt { 16+64 } =\sqrt { 80 } =\sqrt { 4^{ 2 }\times 5 } =4\sqrt { 5 }$

We also know that If any two sides have equal side lengths, then the triangle is isosceles.

Here, since the lengths of the two sides are equal that is

$BC=CA=4\sqrt {5}$

Hence, the given vertices are the vertices of an isosceles triangle.

If the middle point of two points A(-2, 5) and B(-5, y) is $\left( \dfrac{-7}{2}, 3 \right )$ , then find the distance between points A and B.

Let the middle point be

$P$

Using distance formula,

$AP = \sqrt {\left(-\cfrac{7}{2}-(-2)\right)^2 + \left(3-5 \right)^2}$

$AP = \sqrt {\cfrac{9}{4}+4}$

$AP = \cfrac{5}{2}$

$AB = 2AP$ as

$P$ is the mid point of

$AB$

$AB = 5$ .

If the area of the $\triangle {ABC}$ is $68$ sq.units and the vertices are $A(6,7), B(-4,1)$ and $C(a,-9)$ taken in order, then find the value of $a$ .

Area of

$ABC = 5$ sq. Units

Area of a triangle with vertices

$({ x }_{ 1 },{ y }_{ 1 })$ ;

$({ x }_{ 2},{ y }_{ 2 })$ and

$({ x }_{ 3 },{ y }_{ 3 })$ is given as

$\left|\dfrac { { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1})+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|$

According to the given condition

$\Rightarrow \dfrac{1}{2}({a (7-1) + 6(1+9) -4 (-9-7)}) = 68$

$\Rightarrow 136=60+64+6a$

$\Rightarrow 136-124=6a$

$\Rightarrow a=2$

Show that (4, 1) is equidistant from the points (-10, 6) and (9, -13)

A(4,1) is equidistant from

$B(-10,6)$ &

$C(3,-13)$

$\therefore AB=AC$ (To prove)

$AB=\sqrt{(-14)^{2}+5^{2}}$

$AC=\sqrt{5^{2}+14^{2}}$

$\Rightarrow$ A is equidistant from

$BC$

Show that the points $(a, a), (-a, -a)$ and $(-a \sqrt 3, a \sqrt 3)$ form an equilateral triangle.

Let the points be represented by

$A(a, a), B(-a, -a)$ and

$C(- a \sqrt 3, a \sqrt 3)$ . Using the distance formula

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ , we have

$AB = \sqrt{(a+ a)^2 +(a+ a)^2}$

$= \sqrt{(2a)^2 + (2a)^2} = \sqrt{4a^2 + 4a^2} = \sqrt{8a^2} = 2\sqrt{2}a$

$BC = \sqrt{(-a\sqrt 3 + a)^2 + (a \sqrt 3 + a)^2} = \sqrt{3a^2 + a^2 - 2a^2 \sqrt 3 + 3a^2 + a^2 + 2a^2 \sqrt 3}$

$= \sqrt{8a^2} = \sqrt{4 \times 2a^2} = 2 \sqrt 2 a$

$CA = \sqrt{(a + a\sqrt{3})^2 + (a - a \sqrt 3)^2} = \sqrt{a^2 + 2a^2 \sqrt 3 + 3a^2 + a^2 - 2a^2 \sqrt 3 + 3a^2}$

$= \sqrt{8a}^2= 2 \sqrt 2a$

$\therefore AB = BC = CA = 2 \sqrt 2 a$ .
Since all the sides are equal the points form an equilateral triangle.

Find the area of the triangle formed by joining the mid points of the sides of the triangle, whose vertices are $(0, 1); (2, 1)$ and $(0, 3)$ . Find the ratio of this area to the area of the given triangle

Let

$A\left(0,1\right), B\left(2,1\right), C\left(0,3\right)$ be the vertices of the triangle and

$D,E$ and

$F$ are mid-points of

$AB,BC$ and

$CD$ respectively.

Then, by midpoint formula

$D\equiv\left(\dfrac{0+2}{2},\dfrac{1+1}2\right), E\equiv\left(\dfrac{2+0}2,\dfrac{1+3}2\right),$ and

$F\equiv \left(\dfrac{0+0}2,\dfrac{1+3}2\right)$

$D\equiv\left(1,1,\right) E \equiv\left(1,2\right),$ and

$F \equiv\left(0,2\right)$

Area of the

$\triangle DEF$

$=\dfrac12|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)|$

$=\cfrac12\times |1\times\left(2-2\right)+1\times\left(2-1\right)+0\times\left(1-2\right)|$

$=\cfrac12\times|0+1+0|$

$=\cfrac12$ square units.

Area of the

$\triangle ABC$

$=\cfrac12|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)|$

$=\cfrac12\times |0\times\left(1-3\right)+2\times\left(3-1\right)+0\times\left(1-1\right)|$

$=\cfrac12\times|0+4+0|$

$=2$ square units.

$\dfrac{\text{Area of the}\, \triangle DEF}{\text{Area of the}\, \triangle ABC}=\dfrac1{2\times2}=\dfrac14$ .

Find the area of the triangle whose vertices are $(1,2),(-3,4)$ and $(-5,-6)$

Plot the points in a rough diagram and take them in order.
Let the vertices be

$A(1,2),B(-3,4)$ and

$C(-5,-6)$
Now the area of

$\triangle {ABC}$ is

$=\cfrac { 1 }{ 2 } \left\{ \left( { x }_{ 1 }{ y }_{ 2 }+{ x }_{ 2 }{ y }_{ 3 }+{ x }_{ 3 }{ y }_{ 1 } \right) -\left( { x }_{ 2 }{ y }_{ 1 }+{ x }_{ 3 }{ y }_{ 2 }+{ x }_{ 1 }{ y }_{ 3 } \right) \right\}$

$=\cfrac { 1 }{ 2 } \left\{ \left( 4+18-10 \right) -\left( -6-20-6 \right) \right\}$

$=\cfrac{1}{2}(12+32)=22$ sq.units.

Find the area of the triangle formed by the points $(0,0), (3,0)$ and $(0,4)$ .

1348979_622340_ans_bf59816bd07840ddb40063d28613404b.png

Find the area of the triangle whose vertices are $(-5, 7), (4, 5)$ and $(-4, -5).$

Coordinates of vertices of triangle

$ABC$ is given by:

$A(x_1,y_1)\equiv(-5, 7), B(x_2,y_2)\equiv(4, 5)$ and

$C(x_3,y_3)\equiv(-4, -5)$

Area of triangle in coordinate system is given by:

$= \cfrac{1}{2} \left[ (x_1-x_3)(y_1-y_2) - (x_1-x_2)(y_2-y_3) \right]$

$= \cfrac{(-5-(-4)) (7-5) - (-5-4)(5-(-5))}{2}$

$= 44\ sq.\ units$

If the distance of P(x,y) from A(5,1) and B(-1,5) are equal, then prove that 3x = 2y.

$A(5, 1) \equiv (x_1, y_1)$

$B(-1, 5) \equiv (x_2, y_2)$

$P\equiv (x, y)$

$PA=PB$ ....(given)

$\therefore$ By distance formula

$\sqrt{ { \left( {x}-{ { x_1 } } \right) }^{ 2}+{ \left( {y}-{ y_1 } \right) }^{ 2 } }$ we have,

$\Rightarrow (x-5)^2+(y-1)^2=(x+1)^2+(y-5)2$

$\Rightarrow x^2-10x+25+y^2-2y+1=x^2+2x+1+y^2-10y+25$

$\therefore -2y+10y=2x+10x$

$\therefore 8y=12y$

$2y=3x$ .

$[hence \, proved]$

Find the area of the $\triangle ABC$ , coordinates of whose vertices are $A(4, 1), B(6, 6)$ and $C(8, 4)$ .

Area of

$\Delta ABC = \cfrac{1}{2} \begin{vmatrix} 4 & 6 & 1 \\ 6 & 6 & 1\\8&4&1 \end{vmatrix}$

$=\cfrac12|4(6-4)-6(6-8)+1(24-48)|$

$=\cfrac12|8+2-24|$

$=\cfrac12|-14|$

$=7$ square units

Find the areas of the triangles the whose coordinates of the points are respectively.
(a, b + c), (a, b-c) and (-a, c)

Area of triangle having angular points
$(x_{1} , y_{1} )$ $(x_{2} , y_{2})$ $(x_{3} , y_{3})$ is given by the formula

$\triangle=\dfrac{1}{2}\left | x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3} \right |$

or $\triangle=\dfrac{1}{2}\left |x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right |$

Points given are $(a,b+c), (a, b-c)$ and $(-a,c)$

Then, $Area= \triangle=\dfrac{1}{2}\left |a (b-c-c)+a(c-b-c)+(-a)(b+c-(b-c)) \right |$

$\Rightarrow Area= \triangle=\dfrac{1}{2}\left | ab-2ac-ab-2ac \right |$

$\Rightarrow Area= \triangle=\dfrac{1}{2}\left |-4ac \right |$

$\Rightarrow Area= \triangle=2ac$

If X(3, 1), Y(4, 5) and Z(-1, -1) are co-ordinates of vertices of $\Delta$ XYZ, then find area of $\Delta$ XYZ.

$A(x_1, y_1), B(x_2, y_2)$ and

$C(x_3, y_3)$ are the vertices of

$\Delta$ ABC, then

$A(\Delta ABC) = \dfrac{1}{2} |x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3(y_1 - y_2)|$

$\therefore A(\Delta XYZ)$ with vertices X(3, 1), Y(4, 5) and Z(-2, -1)

$= \dfrac{1}{2} |3(5- (-1)) + 4(-1 - 1) + (-2)(1- 5)|$

$=\cfrac{1}{2}|3.6+4(-2)-2(-4)|$

$= \dfrac{1}{2} |18 - 8 + 8|$

$= \dfrac{1}{2} \times 18$

$= 9$

(a) What is the slope of the line joining the points $A(2,-3)$ and $B(6,3)$ ? Find the equation of this line.
(b) Find the co-ordinates of the point $C$ at which the line cuts the x-axis.
(c) Show that $C$ is the mid-point of the line $AB$ .

(a)Slope=

$\dfrac{Y_2-Y_1}{X_2-X_1}$

$=\dfrac{3+3}{6-2}=\dfrac{6}{4}=\dfrac{3}{2}$

(b) equation of line

$y+3=\dfrac{3}{2}(x-2)$

$3x-2y=12$

On X axis y-coordinate is

$0$

So we get

$3x=12$

$x=4$

Coordinate of C

$=(4,0)$

$=(\dfrac{2+6}{2},\dfrac{-3+3}{2})=(4,0)$

Thus C is mid point of AB

Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter's school and then reaches the office What is the extra distance travelled by Ayush in reaching the office. If the house is situated at $(2, 4)$ , bank at $(5, 8),$ school at $(13, 14)$ and office at $(13, 26)$ and the coordinates are in kilometers?(Assume that all distances covered are in straight lines).

Given

House is situated at

$H(2,4)$

Bank is situated at

$B(5,8)$

School is situated at

$S(13,14)$

Office is situated at

$O(13,26)$

By distance formula

$HO=\sqrt{(13-2)^2+(26-4)^2}$

$HO=\sqrt{121+484}=\sqrt{605}$

$HO=24.59km$

$HB=\sqrt{(5-2)^2+(8-4)^2}$

$HB=\sqrt{9+16}=\sqrt{25}$

$HB=5km$

$BS=\sqrt{(13-5)^2+(14-8)^2}$

$BS=\sqrt{64+36}=\sqrt{100}$

$BS=10km$

$SO=\sqrt{(13-13)^2+(26-14)^2}$

$SO=\sqrt{144}$

$SO=12km$

Total distance travelled by Ayush is

$=HB+BS+SO$

$Total=5+10+12=27km$

Distance between office and house is

$24.59km$

Hence Extra distance

$=27-24.59=2.41km$

$(a\, \cos\, \phi _1, b \, \sin\, \phi _1), (a\, \cos\, \phi _2, b \, \sin\, \phi _2)$ and $(a\, \cos\, \phi _3, b \, \sin\, \phi _3)$

Area of the triangle having coordinates

$(a\cos{\phi_1}, b\sin{\phi_1}), (a\cos{\phi_2},b\sin{\phi_2}), (a\cos{\phi_3}, b\sin{\phi_3})$ is given by

$\cfrac{1}{2} \begin{vmatrix} a\cos{\phi_1} & a\cos{\phi_2} & a\cos{\phi_3} & a\cos{\phi_1} \\ b\sin{\phi_1} & b\sin{\phi_2} & b\sin{\phi_3} & b\sin{\phi_1} \end{vmatrix}$

$= \cfrac{1}{2} \times (a\cos{\phi_1}b\sin{\phi_2} + a\cos{\phi_2}b\sin{\phi_3} + a\cos{\phi_3}b\sin{\phi_1} - a\cos{\phi_2}b\sin{\phi_1} - a\cos{\phi_3}b\sin{\phi_2} - a\cos{\phi_1}b\sin{\phi_3})$

$= \cfrac{ab}{2} \times (\sin(\phi_2 - \phi_1) + \sin(\phi_3 - \phi_2) + \sin(\phi_1 - \phi_3))$

$\left \{ am_1m_2, a(m_1+m_2) \right \}, \left \{ am_2m_3, a(m_2+m_3) \right \} and \left \{ am_3m_1, a(m_3+m_1) \right \}$ .

Area of triangle having angular points
$(x_{1} , y_{1} )$ $(x_{2} , y_{2})$ $(x_{3} , y_{3})$ is given by the formula

$\triangle=\dfrac{1}{2}\left | x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{1}y_{3} \right |$

$\triangle = \dfrac{1}{2}\left | (x_{1}- x_{3}) y_{2}+(x_{2}- x_{1}) y_{3}+(x_{3}- x_{2}) y_{1} \right |$

$\triangle = \dfrac{1}{2}\left | (am_{1}m_{2} -am_{3}m_{1} )a(m_{2}+m_{3}) +( am_{2}m_{3}-am_{1}m_{2} )a(m_{3}+m_{1})+(am_{3}m_{1}-am_{2}m_{3})a(m_{1}+m_{2}) \right |$

$\Rightarrow \triangle = \dfrac{1}{2} a^{2}\left | m_{1}(m_{2} -m_{3} )(m_{2}+m_{3}) +m_{2} ( m_{3}-m_{1} )(m_{3}+m_{1})+m_{3}(m_{1}-m_{2})(m_{1}+m_{2}) \right |$

$\Rightarrow \triangle = \dfrac{1}{2} a^{2}\left | m_{1}(m_{2}^{2} -m_{3}^{2} ) +m_{2} ( m_{3}^{2}-m_{1}^{2} ) + m_{3}(m_{1}^{2}-m_{2}^{2}) \right |$

Prove that the polar coordinates $(0, 0), \left ( 3, \dfrac{\pi}{2} \right )$ , and $\left ( 3, \dfrac{\pi}{6} \right )$ form an equilateral triangle.

Let us convert polar co-ordinates into cartesian form,

(i)

$(0, 0)$

(ii)

$r= 3$ and

$\theta = \dfrac {\pi}{2}$

$x= r\cos{\theta}$

$= 3\cos{\dfrac {\pi}{2}}$

$=0$

And,

$y= r\sin{\theta}$

$= 3\sin{\dfrac {\pi}{2}}$

$=3$

$(x,y)=(0, 3)$

(iii)

$r= 3$ and

$\theta = \dfrac {\pi}{6}$

$x= r\cos{\theta}$

$= 3\cos{\dfrac {\pi}{6}}$

$=\dfrac{3\sqrt{3}}{2}$

$y= r\sin{\theta}$

$= 3\sin{\dfrac {\pi}{6}}$

$=\dfrac{3}{2}$

$(x,y)=\left(\dfrac{3\sqrt{3}}{2},\dfrac{3}{2} \right)$

Now, we have points

$A(0,0), B(0,3)$ and

$C\left(\dfrac{3\sqrt{3}}{2},\dfrac{3}{2} \right)$

Now, we need to prove that

$AB=BC=AC.$

$\begin{aligned}{}AB &= \sqrt {{{\left( {0 - 0} \right)}^2} + {{\left( {0 - 3} \right)}^2}} \\ &= \sqrt {0 + 9} \\& = 3\\\\BC &= \sqrt {{{\left( {\frac{{3\sqrt 3 }}{2} - 0} \right)}^2} + {{\left( {\frac{3}{2} - 3} \right)}^2}} \\& = \sqrt {{{\left( {\frac{{3\sqrt 3 }}{2}} \right)}^2} + {{\left( { - \frac{3}{2}} \right)}^2}} \\ &= \sqrt {\frac{{27}}{4} + \frac{9}{4}} \\& = \sqrt {\frac{{36}}{4}} \\ &= \sqrt 9 \\ &= 3\\\\AC &= \sqrt {{{\left( {\frac{{3\sqrt 3 }}{2} - 0} \right)}^2} + {{\left( {\frac{3}{2} - 0} \right)}^2}} \\ &= \sqrt {{{\left( {\frac{{3\sqrt 3 }}{2}} \right)}^2} + {{\left( {\frac{3}{2}} \right)}^2}} \\ &= \sqrt {\frac{{27}}{4} + \frac{9}{4}} \\& = \sqrt {\frac{{36}}{4}} \\& = \sqrt 9 \\ &= 3\end{aligned}$

Hence proved

$AB=BC=CA$ which means that

$\triangle ABC$ is an equilateral triangle.

Prove (by showing that the area of the triangle formed by them is zero) that the following set of three points are in a straight line
$(a, b + c), (b, c + a)$ and $(c, a + b).$

For points

$(x_{1} ,y_{1})$

$(x_{2} ,y_{2})$

$(x_{3} ,y_{3})$

Area of

$\triangle$ ABC is given by

$Area= \dfrac{1}{2} \left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right |$

Now, for the given points

$(a, b + c), (b, c + a)$ and

$(c, a + b).$

$Area= \dfrac{1}{2} \left |a\times(c+a-a-b)+b\times(a+b-b-c)+c\times(b+c-c-a) \right |$

$\Rightarrow Area= \dfrac{1}{2} \left |ac-ab+ab-bc+bc-ac \right |$

$\Rightarrow Area=0$

$\Rightarrow$ All 3 points are in a straight line.

$(am^2_1, 2am_1), (am^2_2, 2am_2)$ and $(am^2_3, 2am_3)$

Area of triangle having angular points

$(x_{1} , y_{1} )$

$(x_{2} , y_{2})$

$(x_{3} , y_{3})$ is given by the formula

$\triangle=\dfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right |$

$\triangle=\dfrac{1}{2}\left | am_{1}^{2}(2am_{2}-2am_{3})+am_{2}^{2}(2am_{3}-2am_{1})+am_{3}^{2}(2am_{1}-2am_{2}) \right |$

$\Rightarrow \triangle=a^{2}\left | m_{1}^{2}(m_{2}-m_{3})+m_{2}^{2}(m_{3}-m_{1})+m_{3}^{2}(m_{1}-m_{2}) \right |$

$\Rightarrow \triangle=a^{2}\left | m_{1}m_{2}(m_{1}-m_{2})-m_{3}(m_{1}+m_{2})(m_{1}-m_{2})+m_{3}^{2}(m_{1}-m_{2}) \right |$

$\Rightarrow \triangle=a^{2}\left |(m_{1}-m_{2}) ( m_{1}m_{2}-m_{3}(m_{1}+m_{2})+m_{3}^{2} ) \right |$

$\Rightarrow \triangle=a^{2}\left |(m_{1}-m_{2}) (m_{2}-m_{3})(m_{3}-m_{1} ) \right |$

Prove (by showing that the area of the triangle formed by them is zero) that the following set of three points are in a straight line $\left(-\dfrac{1}{2}, 3\right), (-5,6)$ , and $(-8,8)$ .

Given points are

$A(-5, 6), B(-8, 8)$ and

$C\left(\dfrac{-1}{2}, 3\right)$

For points

$(x_{1} ,y_{1})$

$(x_{2} ,y_{2})$

$(x_{3} ,y_{3})$

Area of

$\triangle$ ABC is given by

$Area= \dfrac{1}{2} \left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right |$

For the given points

$\Rightarrow Area= \dfrac{1}{2} \left | \dfrac{-1}{2}\times(-2)+(-5)(5)+(-8)(-3)\right |$

$\Rightarrow Area=\dfrac{1}{2}|1-25+24|$

$\Rightarrow Area=\dfrac{1}{2}|0|$

$\Rightarrow Area=0$

$\Rightarrow$ All $3$ points are in a straight line.

$A\left[ 3,4 \right]$ and $B\left[ 5,-2 \right]$ are to given points. If $PA=PB$ and area of triangle $PAB=10$ , Then $P$ is?

$A(3,4)$ and

$B(5,-2)$

$PA=PB$

Area of triangle

$PAB=10$

Let the point

$P$ be

$(x,y)$

$PA=\sqrt{(3-x)^2+(4-y)^2}$

$PB=\sqrt{(5-x)^2+(-2-y)^2}$

$\sqrt{(3-x)^2+(4-y)^2}=\sqrt{(5-x)^2+(-2-y)^2}$

squaring on both sides

$(3-x)^2+(4-y)^2=(5-x)^2+(-2-y)^2$

$\Rightarrow 9-6x+x^2+16-8y+y^2=25-10x+x^2+4+4y+y^2$

$\Rightarrow 25-6x-8y=29-10x+4y$

$\Rightarrow 4x-12y=4$

$\Rightarrow x-3y=1$ (Equation1)

area of triangle

$=\dfrac{1}{2}[x_1(y_2-y_1)+x_2(y_3-y_3)+x_3(y_1-y_2)]$

$=\dfrac{1}{2}[3(-2-y)+5(y-4)+x(4+2)]=10$

$=-6-3y+5y-20+4x+2x=20$

$=-26+2y+6x=20$

$=3x+y=23$ (Equation2)

From equation1

$x=1+3y$

put

$x$ in equation2

$3(1+3y)+y=23$

$\Rightarrow3+9y+y=23$

$\Rightarrow3+10y=23$

$\Rightarrow10y=20$

$\Rightarrow y=2$

$\therefore 3x+y=23$

$\Rightarrow 3x+2=23$

$\Rightarrow 3x=21$

$\Rightarrow x=7$

So , the coordinate of P

$=(7,2)$

Through the point $(3, 4)$ are drawn two straight lines each inclined at $45^{\circ}$ to the straight line $x - y = 2$ . Find their equations and also find the area of triangle bounded by the three lines.

Equation of the line is

$y=x-2$

Slope of the given line is

$m_1=1$

$\Rightarrow$ Given line is inclined to the x-axis at an angle of

$45 \ degrees$

So the inclination of the other two lines with the x-axis will be 0 degrees( 45-45) and 90 degrees

$(45+45)$

Now, since the lines pass through the point

$(3,4)$

So, the equation of other two lines by one point formula is

$x=3$ and

$y=4$

Now,

Line

$y=x-2$ and

$x=3$ will intersect at

$P_1(3,1)$

Line

$y=x-2$ and

$y=4$ will intersect at

$P_2(6,4)$

Line

$x=3$ and

$y=4$ will intersect at

$P_3(3,4)$

Hence,the coordinates of the three vertices of the triangle are

$P_1, P_2, P_3$

So,area of the bounded triangle=

$\left | \cfrac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2} \right |$

$\Rightarrow \left | \cfrac{3(4-4)+6(4-1)+3(1-4)}{2} \right |$

$\Rightarrow\left | \cfrac{18-9}{2} \right |$

$\Rightarrow\cfrac{9}{2}$

Starting at the origin, a beam of light hits a mirror(in the form of a line) at the point A $(4, 8)$ and reflected line passes through the point B $(8, 12)$ . Compute the slope of the mirror.

Slope of the incident ray

$= \dfrac{(8-0)}{(4-0)} = 2 =\tan A$

Slope of the reflected ray

$= \dfrac{(12-8)}{(8-4)} = 1 =\tan B$

Now let us find the line that bisects the 2 rays

$\tan (A+B) = \dfrac{(\tan A + \tan B)}{(1- \tan A \tan B)} = \dfrac{(1+2)}{(1-(1)2)} = - 3$

Let

$C$ be the angle bisector between

$A$ and

$B$

$A + B = 2C$ say

$m$ is slope of the bisector

So we get

$\dfrac{2m}{(1-m^2)} = - 3$

$2m = -3 + 3m^2$

$\implies 3m^2 – 2m – 3 = 0$

$\implies m = (\dfrac{1}{6}) [2 + \sqrt{4 + 36}$ ... [

$m$ positive taken to the correct slope]

$\implies m = \dfrac{1}{3} [1 + \sqrt{10}]$

This is the slope of the mirror.

The reason for the slope to be positive is that from

$(4,8)$ one line goes to

$(8,12)$ that is line is pointing towards

$1st \ quadrant$ and

$(0,0)$ is to the direction of

$3rd \ quadrant$ . So the tangent should have a slope between

$1$ and

$2$ and not the normal.

Find perpendicular distance from the origin of the line joining the points $(\cos \theta, \sin \theta)$ and $(\cos \phi, \sin \phi)$ .

${\textbf{Step - 1: Plotting the two points and finding the distance between them}}{\text{.}}$

${\text{The two points A and B can be plotted as above,where}}$

$A(\cos \phi ,\sin \phi ){\text{ and }}B(\cos \theta ,\sin \theta )$

${\text{Using distance formula we can calculate the distance between the two points}}{\text{.}}$

${\text{So the distance between two points }}A({x_1},{y_1}),({x_2},{y_2}){\text{can be written as}}$

$d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$

${\text{So we can write }}AB = \sqrt {{{(\cos \theta - \cos \phi )}^2} + {{(\sin \theta - \sin \phi )}^2}}$

${\textbf{Step - 2: Finding the perpendicular distance between the centre }}O{\textbf{ and the line }}AB{\text{.}}$

${\text{We can see that }}\Delta AOD{\text{ is a right - angled triangle}}$

${\text{According to the Pythagoras theorem, }}O{D^2} + D{A^2} = A{O^2}$

${\text{Also we can write }}AO = \sqrt {{{\cos }^2}\phi + {{\sin }^2}\phi }$

$OD = \sqrt {A{O^2} - A{D^2}}$

$OA = \sqrt {A{O^2} - {{\left( {\dfrac{{AB}}{2}} \right)}^2}}$

${AB = 2AD..{\text{as any perpendicular from the centre of the circle bisects the chord}}}$

$OD = \sqrt {{{\left( {\sqrt {{{\cos }^2}\phi + {{\sin }^2}\phi } } \right)}^2} - {{\left( {\dfrac{{\sqrt {{{(\cos \theta - \cos \phi )}^2} + {{(\sin \theta - \sin \phi )}^2}} }}{2}} \right)}^2}}$

$OD = \sqrt {{{\cos }^2}\phi + {{\sin }^2}\phi - \dfrac{{{{(\cos \theta - \cos \phi )}^2} + {{(\sin \theta - \sin \phi )}^2}}}{4}}$

$OD = \sqrt {\dfrac{{4 - ({{\cos }^2}\theta + {{\cos }^2}\phi - 2\cos \theta \cos \phi + {{\sin }^2}\theta + {{\sin }^2}\phi - 2\sin \theta \sin \phi )}}{4}}$ $\boldsymbol{\mathbf{{\textbf{(As si}}{{\text{n}}^2}x + \mathbf{{\cos ^2}x = 1)}}}$

$OD = \dfrac{{\sqrt {4 - 2 + 2\sin \theta \sin \phi + 2\cos \theta \cos \phi } }}{2}$

$OD = \sqrt {\dfrac{{1 + \sin \theta \sin \phi + \cos \theta \cos \phi }}{2}}$

$OD = \sqrt {\dfrac{{1 + \cos (\theta - \phi )}}{2}}$ ${\textbf{(As cos(a - b) = cos a cos b + sin a sinb)}}$

$OD = \cos \left( {\dfrac{{\theta - \phi }}{2}} \right)$ $\mathbf{\left( {{\text{As cos2}}\theta {\text{ = 2co}}{{\text{s}}^2}\theta - 1} \right)}$

$\boldsymbol{\mathbf{{\textbf{Thus,the distance d is }}\cos \left( {\dfrac{{\theta - \phi }}{2}} \right).}}$

Let $ABCD$ be a square of side $2a$ . Find the coordinates of the vertices of this square when
(i) A coincides with the origin and $AB$ and $AD$ are along $OX$ and $OY$ respectively.
(ii) The centre of the square is at the origin and coordinate axes are parallel to the sides $AB$ and $AD$ respectivey.

$(i)\$ Side of a square is

$2\ a$

$\therefore \$ coordinates of the vertices are

$A(0,0)\ B(2a, 0)$

$C(2a, 2a)\ D(0, 2a)$

$(ii)\$ centre of a square is at origin

$\therefore \$ Each side is at a distance

$a$ units from

$x$ and

$y$ axes

$\therefore \$ coordinates of the vertices of

$\Box \ ABCD$ are:

$A(-a, -a)$

$B(a, -a)$

$C(a, a)$

$D(-a, a)$

1441210_973311_ans_f63d8ad2776a4a318d8ffbd3e05c794b.png

If the point $P(x,y)$ is equidistant from the points $A(a + b, b - a)\ \text {and}\ B(a - b, a + b)$ . Prove that $bx = ay$ .

Given points

$P(x,y)\equiv(x_2,y_2)$ ,

$A(a+b,b-a)\equiv(x_1,y_1)$

Since point p(x,y) is equidistant from

$A(a+b,b-a) , B(a-b,a+b)$

$P(x,y)\equiv(x_2,y_2) , B(a-b,a+b)\equiv(x_1,y_1)$

$\therefore$

$PA=PB$

$=\sqrt{ { \left( {x_2}-{ { x_1 } } \right) }^{ 2}+{ \left( {y_2 }-{ y_1 } \right) }^{ 2 } }$

$\sqrt { \left[ x-(a+b \right)] ^{ 2 }+\left[ y-(b-a) \right] ^{ 2 } } =\sqrt { \left[ x-(a-b) \right] ^{ 2 }+\left[ y-(a+b) \right] ^{ 2 } }$

Squaring both sides

${ x }^{ 2 }+(a+b)^{ 2 }-2x(a+b)+{ y }^{ 2 }+(b-a)^{ 2 }-2y(b-a)={ x }^{ 2 }+(a-b)^{ 2 }-2x(a-b)+{ y }^{ 2 }+(a+b)^{ 2 }-2y(a+b)$

$\Rightarrow (a-b)^{ 2 }-2x(a-b)+(a+b)^{ 2 }-2y(a+b)=(a+b)^{ 2 }-2x(a+b)$

$\Rightarrow -2x(a-b)-2y(a+b)=-2x(a+b)-2y(b-a)$

$\therefore 2x(a+b-a+b)=2y(a+b-b+a)$

$\Rightarrow bx=ay$

$[hence \, proved]$

Find the area of the triangle whose vertices are $(-2, 3), (3, 2)$ and $(-1, -8)$ by using determinant method.

Area of triangle is

$\triangle = \dfrac {1}{2}\begin{vmatrix}x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1\\ x_{3} & y_{3} & 1\end{vmatrix}$
where

$(x_{1}y_{1}) = (-2, -3); (x_{2}y_{2}) = (3, 2); (x_{3}, y_{3}) = (-1, -8)$

$\therefore \triangle = \dfrac {1}{2} \begin{vmatrix}-2 & -3 & 1\\ 3 & 2 & 1\\ -1 & -8 & 1\end{vmatrix}$
Expanding

$\triangle$ using

$R_{1}$ (first row)

$\triangle = \dfrac {1}{2} [-2(2 + 8) + 3(3 + 1) + 1 (-24 + 2)]$

$= \dfrac {1}{25} [-20 + 12 - 22] = \dfrac {1}{2}|-30|= \dfrac {30}{2}$

$\triangle = 15\ sq. units$ .

Prove that the points $(2a,4a), (2a,6a)$ and $(2a+\sqrt {3}a,5a)$ are the vertices of an equilateral triangle.

PROOF :-

$AB=\sqrt{ (2a-2a)^2 + (6a-4a)^2 } \\ =2a$

$AC=\sqrt { { \left( 2a-\left( 2a+\sqrt { 3 } a \right) \right) }^{ 2 }+{ \left( 4a-5a \right) }^{ 2 } } \\ =\sqrt { 3{ a }^{ 2 }+{ a }^{ 2 } } \\ =2a\\ BC=\sqrt { { \left( 2a+\sqrt { 3 } a-2a \right) }^{ 2 }+{ \left( 6a-5a \right) }^{ 2 } } \\ =2a$

$AB=BC=AC=2a$ All sides are equal.

Hence, It is an equilateral triangle.

Find the circumcentre and circumradius of a triangle with vertices are $(3,0),\ (-1,-6),\ (4,-1)$ .

Let the vertices of the triangle be

$A(3,0), B(-1,-6)$ and

$C(4,-1)$ .

$P(x,y)$ be the required circumcentre and r is the circumradius of the

$\bigtriangleup$ ABC,

Then, we must have,

$r^2=PA^2=(x-3)^2+(y-0)^2$ --------(1)

$r^2=PB^2=(x+1)^2+(y+6)^2$ --------(2)

$r^2=PC^2=(x-4)^2+(y+1)^2$ --------(3)

From (1) and (2), we get,

$(x-3)^2+y^2=(x+1)^2+(y+6)^2$

$\Rightarrow x_2+9-6x+y^2=x^2+1+2x+y^2+36+12y$

$\Rightarrow 9-6x=37+2x+12y$

$\Rightarrow 8x+12y+28=0$

$\Rightarrow 2x+3y+7=0$

$\therefore x=\dfrac{-3y-7}{2}$ ----------(4)

From, (2) and (3), we get

$(x+1)^2+(y+6)^2=(x-4)^2+(y+1)^2$

$\Rightarrow x^2+1+2x+y^2+36+12y=x^2+16-8x+y^2+1+2y$

$\Rightarrow 2x+12y+37=2y-8x+17$

$\Rightarrow 10x+10y+20=0$

$\Rightarrow x+y+2=0$

$\Rightarrow \dfrac{-3y-7}{2}+y+2=0$

$\Rightarrow -3y-7+2y+4=0$

$\Rightarrow y=-3$

Now,

$x+y+2=0$

$\Rightarrow x-3+2=0$

$\Rightarrow x=1$

$\therefore$ Required circumcentre=

$(1,3)$

Circumradius

$=r=\sqrt{(x-3)^2+y^2}$

$\Rightarrow r=\sqrt{4+9}$

$\Rightarrow r=\sqrt{13}$

If the points $(2,1)$ and $(1,-2)$ are equidistant from the point $(x,y)$ , show that $x+3y=0$ .

Given that, Distance of point (x,y) from (2,1)

$D_{1} = \sqrt { { (x-2) }^{ 2 }+{ (y-1) }^{ 2 } }$

Distance of point (x,y) from (1,-2)

$D_{2} = \sqrt { { (x-1) }^{ 2 }+{ (y+2) }^{ 2 } }$

Given that,

$D_{1}=D_{2}$

$D_{1}^{2}=D_{2}^{2}$

$x^{2}+4-4x+y^{2}+1-2y=x^{2}+1-2x+y^{2} + 4 +4y$

$x + 3y=0$

Hence proved

Find the values of $x,y$ if the distances of the point $(x,y)$ from $(-3,0)$ as well as from $(3,0)$ are $4$ .

Distance between two points

$=\sqrt{(x_{2}-x_{1})^{_2}+(y_{2}-y_{1})^{_2}}$

$\text{(Distance between two points)}^2$

$=(x_{2}-x_{1})^{_2}+(y_{2}-y_{1})^{_2}$

Therefore, we get

$(x+3)^{2}+y^{2}=16$

Subtracting above equations we get,

$(x+3)^{2}+y^{2}=16$

$(x+3)^{2}$ _

$+y^{2}$ _

$=16$

____________________

$(x+3)^{2}=(x-3)^{2}$

$\therefore$

$x=0$ and

$y=\pm \sqrt { 7 }$

Find the angle subtended at the origin by the line segment whose end points are $(0,100)$ and $(10,0)$

Let

$O\left( 0,0 \right)$ be the origin and

$A\left( 10,0 \right) ,B\left( 0,100 \right)$

So by distance formula we have,

Distance between two points = $\sqrt { { (x_2-x_1) }^{ 2 }+{ (y_2-y_1) }^{ 2 } }$

${ OA }^{ 2 }={ \left( 10-0 \right) }^{ 2 }+{ \left( 0-0 \right) }^{ 2 }=100,{ OB }^{ 2 }={ \left( 0-0 \right) }^{ 2 }+{ \left( 100-0 \right) }^{ 2 }=10000$

${ OA }^{ 2 }+{ OB }^{ 2 }=100+10000=10100$ ---------(1)

${ AB }^{ 2 }={ \left( 10-0 \right) }^{ 2 }+{ \left( 0-100 \right) }^{ 2 }=10100$ ----------(2)

From (1) and (2) we get

${ OA }^{ 2 }+{ OB }^{ 2 }={ AB }^{ 2 }$

$In \ \triangle OAB$ ,

${ OA }^{ 2 }+{ OB }^{ 2 }={ AB }^{ 2 }$

$\therefore \angle AOB=90°$

Therefore, angle subtended the line segment AB at origin is

$90°$

Prove that $(2,-2), (-2,1)$ and $(5,2)$ are the vertices of a right angled-triangle. Find the area of the triangle and the length of the hypotenuse.

To prove:

$A(2,-2),B(-2,1)$ and C(5,2) are the vertices of a right angled-triangle.

Proof:

By distance formula,

$AB=\sqrt{ (-2-2)^2+(1+2)^2 }=\sqrt{16+9}=\sqrt{25}=5$

$BC=\sqrt{ (-2-5)^2+(1-2)^2 }=\sqrt{(-7)^2+(-1)^2}=\sqrt{50}=5\sqrt2$ Length of the hypotenuse

$AC=\sqrt{ (5-2)^2+(2+2)^2 }=\sqrt{(3)^2+(4)^2}=\sqrt{25}$

$\therefore AB^2+AC^2=25+25=50=BC^2$

$\therefore$

$\triangle$ ABC is a right-angled triangle

Area of the triangle ABC

$\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$ sq.

Area of the

$\bigtriangleup$ ABC

$=\dfrac{1}{2}\times Base \times height$

$=\dfrac{1}{2} \times 5 \times 5$

$=\dfrac{25}{2}$ sq.units.

Show that the points $(-3,2),(-5,-5),(2,-3)$ and $(4,4)$ are the vertices of a rhombus. Find the area of this rhombus.

Let the point A(3,2), B(-5,-5), C(2,-3) and D(4,4) be the vertices of a rhombus ABCD.

We know that all the sides of a rhombus ABCD are equal.

So by distance formula,

Distance between two points = $\sqrt { { (x_2-x_1) }^{ 2 }+{ (y_2-y_1) }^{ 2 } }$

$AB=\sqrt{(-5+3)^2+(-5-2)^2}=\sqrt{4+49}=\sqrt{53}$ .

$BC=\sqrt{(2+5)^2+(-3+5)^2}=\sqrt{49+4}=\sqrt{53}$ .

$CD=\sqrt{(4-2)^2+(4+3)^2}=\sqrt{4+49}=\sqrt{53}$ .

$DA=\sqrt{(4+3)^2+(4-2)^2}=\sqrt{49+4}=\sqrt{53}$ .

$\therefore AB=BC=CD=AD$

Hence, ABCD is a rhombus (Proved)

Area of the rhombus ABCD

$=\dfrac{1}{2}\mid(x_1y_2+x_2y_3+x_3y_4+x_4y_1)-(x_2y_1+x_3y_2+x_4y_3+x_1y_4)\mid$

$=\dfrac{1}{2}\mid(15+15+8+8)-(-10-10-12-12)$

$=\dfrac{1}{2}\mid46+44\mid=\dfrac{1}{2}\mid{90}\mid=45$ sq. units

Prove that the points $A(1,7), B(4,2), C(-1,-1)$ and $D(-4,4)$ are the vertices of a square.

A square is quadrilateral whose all sides are equal and all angles are right angles.

Distance between two points $=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Slope of line $=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

AB $=\sqrt{(4-1)^{2}+(2-7)^{2}}$

$=\sqrt{(3)^{2}+(5)^{2}}$

$=\sqrt{34}$

Slope of AB $=\dfrac{2-7}{4-1}$ $=\dfrac{-5}{3}$

BC $=\sqrt{(4+1)^{2}+(2+1)^{2}}$ $=\sqrt{34}$

Slope of BC $=\dfrac{-1-2}{-1-4}$ $=\dfrac{3}{5}$

CD $=\sqrt{(-4+1)^{2}+(4+1)^{2}}$ $=\sqrt{9+25}$ $=\sqrt{34}$

Slope of CD $=\dfrac{4+1}{-4+1}=\dfrac{-5}{3}$

AD $=\sqrt{(7-4)^{2}+(-4-1)^{2}}$ $=\sqrt{34}$

Slope of AD $=\dfrac{4-7}{-4-1}=\dfrac{3}{5}$

Slope of (AB x BC) $=-1$

Slope of (BC x CD) $=-1$

Slope of (CD x AD) $=-1$

Slope of (AD x AB) $=-1$

If product of two slopes are perpendicular it means they are perpendicular.

All adjacent sides are perpendicular.

Opposite sides are parallel and All sides are equal (length $=\sqrt{34}$ units )

$\therefore$ It is a square.

The length of a line segment is of $10$ units and the coordinates of one-end-point are $(2,-3)$ . If the abscissa of the other end is $10$ , find the ordinate of the other end.

Distance between two points

$=\sqrt{(x_{2}-x_{1})^{_2}+(y_{2}-y_{1})^{_2}}$

Given, one point

$=(2,-3)$

Let, other point

$=(10,b)$

absicca is x cordinate

$\therefore$

$\sqrt{(10-2)^{2}+(b+3)^{2}}=10$

$\sqrt{8^2+(b^{2}+9+6b)}=10$

Squaring both the sides, we get

$64+b^{2}+9+6b=100$

$(b+3)^{2}=36$

$\therefore$

$b+3=6$ and

$b+3=-6$

$\therefore$

$b=3$ and

$b=-9$

Find the area of a triangle whose vertices are $(a,c+a),(a,c)$ and $(-a,c-a)$

Formula:

Area of triangle

$=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Given,

$(a,a+c),(a,c),(-a,c-a)$

Area of triangle

$=\dfrac{1}{2}[a(c-c+a))+a(c-a-a-c)+(-a)(a+c-c)]$

$=\dfrac{1}{2}[a(a)+a(-2a)-a(a)]$

$=|-a^2|=a^2$ sq.units

Show that the points $A(2,3),B(-2,2),C(-1,-2)$ and $D(3,-1)$ are the vertices of a square $ABCD$ .

Length OF SIDES-

$AB = \sqrt{[(2+2)^2 +(3-2)^2 ]}$

$= \sqrt{[16+1]}$

$= \sqrt{17}$

$BC = \sqrt{[(-2+1)^2 + (2+2)^2]}$

$= \sqrt{[1+16]}$

$= \sqrt{17}$

$CD = \sqrt{[(-1-3)^2 + (-2+1)^2]}$

$= \sqrt{17}$

$DA = \sqrt{[(3-2)^2 +(-1-3)^2]}$

$= \sqrt{17}$

here

$AB=BC=CD=DA$

LENGTH OF DIAGONALS

$AC = \sqrt{[(2+1)^2 + (3+2)^2]}$

$= \sqrt{[9+25]}$

$= \sqrt{34}$

$BD = \sqrt{[(-2-3)^2 + (2+1)^2]}$

$= \sqrt{34}$

here

$AC=BD$

hence, the given points are vertices of a square

Find the value of $x$ such that $PQ=QR$ where the coordinates of $P,Q$ and $R$ are $(6,-1), (1,3)$ and $(x,8)$ respectively.

Coordinate of point P,Q and R are P(6,-1),Q(1,3) and R(x,8)

Given, PQ=QR

So by distance formula we have,

Distance between two points = $\sqrt { { (x_2-x_1) }^{ 2 }+{ (y_2-y_1) }^{ 2 } }$

PQ=

$\sqrt{(6-1)^2+(-1-3)^2}=\sqrt{25+14}=\sqrt{41}$

$\therefore PQ^2=41=QR^2$

but,

$QR^2=(x-1)^2+25$

$41=x^2+1-2x+25$

$\Rightarrow41=x^2+1-2x+25$

$\Rightarrow x^2-2x+26=41$

$\Rightarrow x^2-2x-15=0$

$\Rightarrow x^2-5x+3x-15=0$

$\Rightarrow x(x-5)+3(x-5)=0$

$\Rightarrow (x+3)(x-5)=0$

$\Rightarrow x=-3,5$

The points $A(2,0),B(9,1),C(11,6)$ and $D(4,4)$ are the vertices of a quadrilateral $ABCD$ . Determine whether $ABCD$ is a rhombus or not.

Let the point be A(2,0), B(9,1), C(11,6) and D(4,4)

So by distance formula we have,

Distance between two points = $\sqrt { { (x_2-x_1) }^{ 2 }+{ (y_2-y_1) }^{ 2 } }$

$AB=\sqrt{(9-2)^2+(1-0)^2}=\sqrt{(7)^2+(1)^2}=\sqrt{50}$

$BC=\sqrt{(11-9)^2+(6-1)^2}=\sqrt{(2)^2+(5)^2}=\sqrt{29}$

$CD=\sqrt{(4-11)^2+(4-6)^2}=\sqrt{(-7)^2+(-2)^2}=\sqrt{53}$

$AD=\sqrt{(4-2)^2+(4-0)^2}=\sqrt{(4)^2+(16)^2}=\sqrt{20}$

$\therefore AB \ne BC \ne CD \ne AD$

In rhombus , all the sides are equal

Hence , ABCD is not a rhombus.

Three consecutive vertices of a parallelogram are $(-2,-1), (1,0)$ and $(4,3)$ . Find the fourth vertex.

Diagonals of a parallelogram bisect each other,

let the point of intersection of diagonals be

$M(m,n)$

$m=\dfrac{-2+4}{2}=1$ and

$n=\dfrac{-1+3}{2}=1$

$\therefore m=1 \ and\ n=1$

Let the Fourth vertex be

$D(x,y)$

$\Rightarrow$

$m=\dfrac{1+x}{2}=1$ and

$n=\dfrac{0+y}{2}=1$

$x+1=2$ and

$y=2$

Thus

$x=1$ &

$y=2$

The fourth vertex of the parallelogram is

$D\equiv(1,2)$

Prove that the points $(4,5), (7,6), (6,3), (3,2)$ are the vertices of a parallelogram. Is it a rectangle?

let the vertices of a parallelogram be

$A (4,5), B(7,6), C(6,3) D(3,2)$

So by distance formula we have,

Distance between two points = $\sqrt { { (x_2-x_1) }^{ 2 }+{ (y_2-y_1) }^{ 2 } }$

$AB=\sqrt{(7-4)^2+(6-5)^2}=\sqrt{9+1}=\sqrt{10}$

$BC=\sqrt{(6-7)^2+(3-6)^2}=\sqrt{1+9}=\sqrt{10}$

$CD=\sqrt{(3-6)^2+(2-3)^2}=\sqrt{9+1}=\sqrt{10}$

$AD=\sqrt{(3-4)^2+(2-5)^2}=\sqrt{1+9}=\sqrt{10}$

$\therefore$

$AB=CD$ and

$BC=AD$

$\therefore$ Opposite sides are equal

Hence,

$ABCD$ is a parallelogram (Proved)

Also,

$AB=BC=CD=AD=$

$\sqrt{10}$ i.e. all the sides are

equal.

Therefore,it is not a rectangle.

Find the area of a triangle whose vertices are
$(6,3),(-3,5)$ and $(4,-2)$

Vertices of the given triangle are

$A(6,3)B(-3,5),C(4,-2)$

$\therefore$ Area of

$\bigtriangleup ABC$

$=\dfrac{1}{2}\mid y_1(x_2-x_3)+y_2(x_3-x_1)+y_3(x_1-x_2)\mid sq.unit$

Here,

$x_1=6,y_1=3, \ x_2=-3,y_2=5, \ x_3=4, y_3=-2$

$=\dfrac{1}{2}\mid 3(-3-4)+5(4-6)+(-2)(6+3)\mid sq.unit$

$=\dfrac{1}{2}\mid -21-10-18 \mid sq.unit$

$=\dfrac{1}{2}\times49$ sq.units

$=\dfrac{49}{2}$ sq.units

If $(0,-3)$ and $(0,3)$ are the two vertices of an equilateral triangle, find the coordinates of its third vertex.

Let the third vertex of the equilateral

$\Delta ABC$ be

$A(x,y)$

So by distance formula we have,

Distance between two points = $\sqrt { { (x_2-x_1) }^{ 2 }+{ (y_2-y_1) }^{ 2 } }$

$\therefore BC= \sqrt{(0-0)^2+(3+3)^2}=\sqrt6^2=\sqrt{36}=6$

$\therefore AB=\sqrt{(x-0)^2+(y+3)^2}$

$\Rightarrow AB^2=x^2+y^2+2y=27$ -----(1)

Now,

$AC=\sqrt{(x-0)^2+(y-3)^2}$

$\Rightarrow AC^2=x^2+y^2+9-6y=36$

$\Rightarrow 27-2y-6y+9=36$

$\Rightarrow -8y=36-36$

$\Rightarrow y=0$

$\therefore x^2 =27$

$\Rightarrow x=\sqrt{27}=\sqrt{3\times3\times3}=+3\sqrt3 \, or -3\sqrt3$

$\therefore$ Required points are

$A(3\sqrt3,0)$ and

$A(-3\sqrt3,0)$

Three vertices of a parallelogram are $(a+b,a-b),(2a+b,2a-b)$ and $(a-b,a+b)$ taken in a order, the fourth vertex is $(-b,b)$
If true enter $1$ else $0$

Let D be the fourth vertex.

$D(x,y)$

Given,

$C(a-b,a+b)$

Let P be the mid point of

$A(a+b,a-b),B(2a+b,2a-b)$

coordinate of mid point of

$AC =$ coordinate of mid point of

$BD$

$\left ( \dfrac{a+b+a-b}{2},\dfrac{a-b+a+b}{2} \right )=\left ( \dfrac{2a+b+x}{2},\dfrac{2a-b+y}{2} \right )$

$(a,a)=\left ( \dfrac{2a+b+x}{2},\dfrac{2a-b+y}{2} \right )$

by solving the above equation, we get

$\dfrac{2a+b+x}{2}=a\Rightarrow x=-b$

$\dfrac{2a-b+y}{2} =a\Rightarrow y=b$

$D(x,y)=D(-b,b)$ the fourth vertex.

Find the coordinates of the vertices of an equilateral triangle of side 2a as shown in Fig 14.5

Since OAB is an equilateral triangle of side 2a. Therefore,
OA=AB=OB=2a
Let BL perpendicular from B on OA. Then,
OL=LA+a
In

$\triangle OLB,$ we have

${ OB }^{ 2 }={ OL }^{ 2 }+{ LB }^{ 2 }$

$\Rightarrow { \left( 2a \right) }^{ 2 }={ a }^{ 2 }+{ LB }^{ 2 }$

$\Rightarrow { LB }^{ 2 }={ 3a }^{ 2 }$

$\Rightarrow LB=\sqrt { 3 } a$

Clearly, coordinates of O are (0,0) and that of A are (2a,0). Since OL=a and

$LB=\sqrt { 3 } a$ . So, the coordinates of B are

$\left( a,\sqrt { 3 } a \right) .$

If $a$ and $b$ are real numbers between $0$ and $1$ such that the points $(a,1),(1,b)$ and $(0,0)$ from an equilateral triangle, then $a=b=2-\sqrt { 3 }$ .

${ \left( a-1 \right) }^{ 2 }+{ \left( 1-b \right) }^{ 2 }={ a }^{ 2 }+1={ b }^{ 2 }+1={ \left( side \right) }^{ 2 }$

The last two imply

$a=b$ or

$a=-b$

The first two imply

${b}^{2}-2b-2a+1=0$

$a=-b$ , then

${b}^{2}+1=0$ not possible

$a=b$ , then

${b}^{2}-4b+1=0$

$\therefore\quad b=2+\sqrt { 3 } ,2-\sqrt { 3 }$

Since

$b<1\quad \therefore\quad b=2-\sqrt { 3 } =a$

Find the area of the triangle $PQR$ if coordinates of $Q$ are $(3,2)$ and the coordinates of mid-points of the sides through $Q$ are $(2,-1)$ and $(1,2)$

Let the coordinate of P and R are

$P(x,y)$ and

$R(a,b)$

$(X,Y)$ is the mid-point of

$x_1$ and

$y_1$ then using mid-point formula

$P(X,Y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

According to the given condition

$\dfrac{x+3}{2}=2 \Rightarrow x=1$

$\dfrac{y+2}{2}=-1 \Rightarrow y=-4$

Thus co-ordinates of

$P$ are

$(1,-4)$

Also, for point

$R$ ,

$\dfrac{3+a}{2}=1 \Rightarrow a=-1$

$\dfrac{b+2}{2}=2 \Rightarrow b=2$

$\therefore$ Area of

$\bigtriangleup PQR$

$=\dfrac{1}{2}\mid y_1(x_2-x_3)+y_2(x_3-x_1)+y_3(x_1-x_2)\mid$

Here, $x_1=1,y_1=-4, \ x_2=3,y_2=2, \ x_3=-1, y_3=2$

$\therefore$ Area of

$\bigtriangleup PQR$

$=\dfrac{1}{2}\mid -4(3+1)+2(-1-1)+2(1-3)\mid$

$=\dfrac{1}{2}\mid-16-4-4 \mid$

$=\dfrac{1}{2}\times24$

$=12$ Sq.units

Find the value of $a$ for which the area of the triangle formed by the points $A(a,2a),B(-2,6)$ and $C(3,1)$ is $10$ square units.

Area of a triangle with three vertices:

We know that, the area of the triangle formed by points $A(x_1,y_1),\ B(x_2,y_2)\ and\ C(x_3,y_3)$ is:

$\dfrac12\left| x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right|$

Here, $x_1=a,y_1=2a, \ x_2=-2,y_2=6, \ x_3=3, y_3=1$

Therefore,

Area

$=10=\dfrac{1}{2}[a(6-1) - 2(1-2a) +3(2a-6)]$

$20=5a+4a+6a-2-18$

$15a=40$

$a=\dfrac{40}{15}=\cfrac{8}{3}$

Find the area of a parallelogram $ABCD$ if three of its vertices are $A(2,4),B(2+\sqrt {3},5)$ and $C(2,6)$

We know that the diagonal of a parallelogram divides the parallelogram into $2$ equal triangles.

So, Area of parallelogram,

$ABCD=2\times$ area of $\Delta ABC$ ......... (1)

We know that,

Area of $\Delta ABC$ with vertices $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ is

$A=\dfrac{1}{2}\times \left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$

So, Area of $\Delta ABC$ with vertices $[A\left( 2,4 \right),B\left( 2+\sqrt{3},5 \right),C\left( 2,6 \right)]$ is

$A=\dfrac{1}{2}\times \left[ 2\left( 5-6 \right)+\left( 2+\sqrt{3} \right)\left( 6-4 \right)+2\left( 4-5 \right) \right]$

$A=\dfrac{1}{2}\times \left[ -2+2\left( 2+\sqrt{3} \right)-2 \right]$

$A=\dfrac{1}{2}\times \left[ -2+4+2\sqrt{3}-2 \right]$

$A=\sqrt{3}$

Put this value in equation (1), we get,

Area of parallelogram,

$ABCD=2\times$ area of $\Delta ABC$

$ABCD=2\sqrt{3}$ sq units

Hence, area of the parallelogram is $2\sqrt{3}$ sq units.

If the vertices of a triangle are $(1,-3),(4,p)$ and $(-9,7)$ and its area is $15$ sq. units, find the value(s) of $p$ .

Area of the triangle having 3 vertices

$A=\cfrac { 1 }{ 2 } \left| \begin{matrix} x_{ 1 } & y_{ 1 } & 1 \\ x_{ 2 } & y_{ 2 } & 1 \\ x_{ 3 } & y_{ 3 } & 1 \end{matrix} \right|$

Here,

$x_1=1,y_1=-3, \ x_2=4,y_2=p, \ x_3=-9, y_3=7$

$\pm 15=\cfrac { 1 }{ 2 } \left| \begin{matrix} 1 & -3 & 1 \\ 4 & p & 1 \\ -9 & 7 & 1 \end{matrix} \right|$

$\pm 30=p-7 + 3(4+9) +1(28+9p)$

$\pm 30=p-7+12+27+28+9p$

$\pm 30=10p+60$

$10p=-90,-30$

$p=-9,-3$

The vertices of $\triangle{ABC}$ are $(-2,1),(5,4)$ and $(2,-3)$ respectively. Find the area of the triangle and the length of the altitude through $A$ .

We will use the formula to find the area of the triangle when all three coordinates are given,

$\begin{aligned}{}A &= \frac{1}{2}\left[ {{y_1}\left( {{x_2} - {x_3}} \right) + {y_2}\left( {{x_3} - {x_1}} \right) + {y_3}\left( {{x_1} - {x_2}} \right)} \right]\\& = \frac{1}{2}\left[ {1\left( {5 - 2} \right) + 4\left( {2 + 2} \right) - 3\left( { - 2 - 5} \right)} \right]\\& = \frac{1}{2}\left[ {3 + 4 \times 4 - 3 \times - 7} \right]\\& = \frac{1}{2}\left( {3 + 16 + 21} \right)\\ &= \frac{1}{2} \times 40\\& = 20\text{ sq. units}\end{aligned}$

But as we know the area of a triangle is equal to half of the product of base and height. So, if we assume base as

$BC$ then the height will be equal to the length of the altitude from the vertex

$A.$

$\begin{aligned}{}BC &= \sqrt {{{\left( {2 - 5} \right)}^2} + {{\left( { - 3 - 4} \right)}^2}} \\& = \sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 7} \right)}^2}} \\& = \sqrt {9 + 49} \\ &= \sqrt {58}\text{ units} \end{aligned}$

Now,

$\begin{aligned}{}Area &= \frac{1}{2} \times BC \times AD\\20 &= \frac{1}{2} \times \sqrt {58} \times AD\\AD& = \frac{{40}}{{\sqrt {58} }}\text{ units}\end{aligned}$

Let the opposite angular points of a square be $(3,4)$ and $(1,-1)$ . Find the coordinates of the remaining angular points.

Let ABCD be a square and let A(3,4) and C(1,-1) be the given angular points. Let B(x,y) be the unknown vertex.
Then, AB=BC

$\Rightarrow \quad \quad { AB }^{ 2 }={ BC }^{ 2 }$

$\Rightarrow \quad \quad { \left( x-3 \right) }^{ 2 }+{ \left( y-4 \right) }^{ 2 }={ \left( x-1 \right) }^{ 2 }+{ \left( y+1 \right) }^{ 2 }$

$\Rightarrow \quad \quad 4x+10y-23=0$

$\Rightarrow \quad \quad x=\dfrac { 23-10y }{ 4 }$ ....(i)

In right-angled triangle ABC, we have

${ AB }^{ 2 }+{ BC }^{ 2 }={ AC }^{ 2 }$

$\Rightarrow \quad \quad { \left( x-3 \right) }^{ 2 }+{ \left( y-4 \right) }^{ 2 }+{ \left( x-1 \right) }^{ 2 }+{ \left( y+1 \right) }^{ 2 }={ \left( 3-1 \right) }^{ 2 }+{ \left( 4+1 \right) }^{ 2 }$

$\Rightarrow \quad \quad { x }^{ 2 }+{ y }^{ 2 }-4x-3y-1=0$ ...(ii)

Substituting the value of x from (i) into (ii), we get

${ \left( \dfrac { 23-10y }{ 4 } \right) }^{ 2 }+{ y }^{ 2 }-\left( 23-10y \right) -3y-1=0$

$\Rightarrow \quad \quad { 4y }^{ 2 }-12y+5=0$

$\Rightarrow \quad \quad \left( 2y-1 \right) \left( 2y-5 \right) =0$

$\Rightarrow \quad \quad y=\dfrac { 1 }{ 2 } or,\dfrac { 5 }{ 2 }$

$Putting\quad y=\dfrac { 1 }{ 2 } \quad and\quad y=\dfrac { 5 }{ 2 } \quad respectively\quad in\quad (i),\quad we\quad get$

$x=\dfrac { 9 }{ 2 } \quad and\quad x=\dfrac { -1 }{ 2 } respectively.$

Hence, the required vertices of the square are (9/2,1/2) and (-1/2,5/2).

Find the fourth vertex D of a parallelogram ABCD whose three vertices are A ( -2,3 ) B (6 ,7) and C ( 8,3)

In a parallelogram ,mid points of diagonals coincides

$\implies$ mid point of

$AC$ is

$\bigg(\dfrac{-2+8}{2},\dfrac{3+3}{2}\bigg)=(3,3)$

Let

$D$ be

$(x,y)$

mid point of

$BD$ is

$\bigg(\dfrac{x+6}{2},\dfrac{y+7}{2}\bigg)$

$\implies x+6=6,y+7=6\implies x=0,y=-1$

The fourth vertex be

$D(0,-1)$

If P and Q are two points whose coordinates are $\left( { at }^{ 2 },2at \right)$ and $\left( \dfrac { a }{ { t }^{ 2 } } ,\dfrac { 2a }{ t } \right)$ respectively and S is the point (a,o). Show that $\dfrac { 1 }{ SP } +\dfrac { 1 }{ SQ }$ is independent of t.

We have,

$SP=\sqrt { { \left( { at }^{ 2 }-a \right) }^{ 2 }+{ \left( 2at-0 \right) }^{ 2 } } =a\sqrt { { \left( { t }^{ 2 }-1 \right) }^{ 2 }+{ 4t }^{ 2 } } =a\left( { t }^{ 2 }+1 \right)$

and,

$SQ=\sqrt { { \left( \dfrac { a }{ { t }^{ 2 } } -a \right) }^{ 2 }+{ \left( \dfrac { 2a }{ t } -0 \right) }^{ 2 } }$

$\Rightarrow \quad \quad SQ=\sqrt { \dfrac { { a }^{ 2 }{ \left( 1-{ t }^{ 2 } \right) }^{ 2 } }{ { t }^{ 4 } } +\dfrac { { 4a }^{ 2 } }{ { t }^{ 2 } } }$

$\Rightarrow \quad \quad SQ=\dfrac { a }{ { t }^{ 2 } } \sqrt { { \left( 1-{ t }^{ 2 } \right) }^{ 2 }+{ 4t }^{ 2 } } =\dfrac { a }{ { t }^{ 2 } } \sqrt { { \left( 1+{ t }^{ 2 } \right) }^{ 2 } } =\dfrac { a }{ { t }^{ 2 } } \left( 1+{ t }^{ 2 } \right)$

$\therefore \quad \quad \dfrac { 1 }{ SP } +\dfrac { 1 }{ SQ } =\dfrac { 1 }{ a\left( { t }^{ 2 }+1 \right) } +\dfrac { { t }^{ 2 } }{ a\left( { t }^{ 2 }+1 \right) }$

$\Rightarrow \quad \quad \dfrac { 1 }{ SP } +\dfrac { 1 }{ SQ } =\dfrac { 1+{ t }^{ 2 } }{ a\left( { t }^{ 2 }+1 \right) } =\dfrac { 1 }{ a } ,\quad which\quad is\quad independent\quad of\quad t$

Do the points $A(3,2), B(-2,-3)$ and $C(2,3)$ form a triangle? If so, name the type of triangle formed.

We have,

$AB=\sqrt { { \left( -2-3 \right) }^{ 2 }+{ \left( -3-2 \right) }^{ 2 } } =\sqrt { 25+25 } =\sqrt { 50 }$

$BC=\sqrt { { \left( 2+2 \right) }^{ 2 }+{ \left( 3+3 \right) }^{ 2 } } =\sqrt { 16+36 } =\sqrt { 52 }$

$and\quad \quad \quad AC=\sqrt { { \left( 2-3 \right) }^{ 2 }+{ \left( 3-2 \right) }^{ 2 } } =\sqrt { 1+1 } =\sqrt { 2 }$

$Clearly,\quad AB+BC>AC,\quad AC+BC>AB\quad and\quad AB+AC>BC.$ Therefore, points A,B and c form a triangle.

$Also,\quad { BC }^{ 2 }={ AB }^{ 2 }+{ AC }^{ 2 }$
Therefore,

$\triangle ABC$ is a right triangle, right angled at A.

if A(5,2), B(2,-2) and C (-2,t) are the vertices of right angled triangle with $\angle B={ 90 }^{ 0 },$ then find the value of t.

Using Pythagoras theorem in right triangle ABC, we obtain

${ AC }^{ 2 }={ AB }^{ 2 }+{ BC }^{ 2 }$

$\Rightarrow \quad \quad { \left( 5+2 \right) }^{ 2 }+{ \left( 2-t \right) }^{ 2 }=\left\{ { \left( 5-2 \right) }^{ 2 }+{ \left( 2+2 \right) }^{ 2 } \right\} +\left\{ { \left( 2+2 \right) }^{ 2 }+{ \left( -2-t \right) }^{ 2 } \right\}$

$\Rightarrow \quad \quad 49+(4-4t+{ t }^{ 2 })=(9+16)+(16+4+4t+{ t }^{ 2 })$

$\Rightarrow \quad \quad { t }^{ 2 }-4t+53={ t }^{ 2 }+4t\times 45$

$\Rightarrow \quad \quad -8t=-8$

$\Rightarrow \quad \quad t=1$

Prove that $(4, -1), (6, 0), (7, 2)$ and $(5, 1)$ are the vertices of a rhombus. Is it a square?

Let the given points be A, B, C and D respectively. Then,Coordinates of the mid-point of Ac are

$\left( \dfrac { 4+7 }{ 2 } ,\dfrac { -1+2 }{ 2 } \right) =\left( \dfrac { 11 }{ 2 } ,\dfrac { 1 }{ 2 } \right)$ Coordinates of the mid-point of BD are

$\left( \dfrac { 6+5 }{ 2 } ,\dfrac { 0+1 }{ 2 } \right) =\left( \dfrac { 11 }{ 2 } ,\dfrac { 1 }{ 2 } \right)$ Thus, AC and BD have the same mid-point.

Hence, ABCD is a parallelogram.

Now,

$AB=\sqrt { { \left( 6-4 \right) }^{ 2 }+{ \left( 0+1 \right) }^{ 2 } } =\sqrt { 5 } ,\quad BC=\sqrt { { \left( 7-6 \right) }^{ 2 }+{ \left( 2-0 \right) }^{ 2 } } =\sqrt { 5 }$

$\therefore \quad \quad AB=BC$ So, ABCD is a parallelogram whose adjacent sides are equal.Hence,ABCD is a rhombus.

We have,

$AC=\sqrt { { \left( 7-4 \right) }^{ 2 }+{ \left( 2+1 \right) }^{ 2 } } =3\sqrt { 2 } ,\quad and,\quad BD=\sqrt { { \left( 6-5 \right) }^{ 2 }+{ \left( 0-1 \right) }^{ 2 } } =\sqrt { 2 }$

Clearly,

$AC\neq BD$ .So, ABCD is not a square.

Find the value x, if the distance between the points $(x,-1)$ and $(3,2)$ is $5$ .

$\textbf{Step -1: Write down given information and relevant formula}$

${\text{The two points are (x, - 1) and (3,2) and the distance between them is 5.}}$

${\text{Let (x, - 1) be (}}{{\text{x}}_1}{\text{,}}{{\text{y}}_1}{\text{) and (3,2) be (}}{{\text{x}}_2}{\text{,}}{{\text{y}}_2}{\text{)}}$

${\text{As we know that distance between two points (}}{{\text{x}}_{\text{1}}}{\text{,}}{{\text{y}}_{\text{1}}}{\text{) and (}}{{\text{x}}_2}{\text{,}}{{\text{y}}_2}{\text{) is}}$

${\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}}$

${\textbf{Step -2: Substitute the given values in the formula}}$

${\text{5 = }}\sqrt {{{(x - 3)}^2} + {{( - 1 - 2)}^2}}$

$\Rightarrow \sqrt {{{(x - 3)}^2} + 9} = 5$

$\Rightarrow {(x - 3)^2} + 9 = 25$

$\Rightarrow {(x - 3)^2} = 16$

$\Rightarrow \left( {x - 3} \right) = \pm 4$

$\Rightarrow {\text{x = 7 or x = - 1}}$

${\textbf{Therefore, the value of x is 7 or - 1.}}$

Prove that the points $(-3,0), (1,-3)$ and $(4,1)$ are the vertices of an isosceles right-angled triangle. Find the area of this triangle

Let A(-3,0), B(1,-3) and C(4,1) by the given points. Then,

$AB=\sqrt { { \left\{ 1-\left( -3 \right) \right\} }^{ 2 }+{ \left( -3-0 \right) }^{ 2 } } =\sqrt { { 4 }^{ 2 }+{ \left( -3 \right) }^{ 2 } } =\sqrt { 16+9 } =5units$

$BC=\sqrt { { \left( 4-1 \right) }^{ 2 }+{ \left( 1+3 \right) }^{ 2 } } =\sqrt { 19+16 } =5\quad units$
and,

$CA=\sqrt { { \left( 4+3 \right) }^{ 2 }+{ \left( 1-0 \right) }^{ 2 } } =\sqrt { 49+1 } =5\sqrt { 2 } unit$
Clearly, AB=BC. Therefore,

$\triangle$ ABC is isosceles.
Also,

${ AB }^{ 2 }+{ BC }^{ 2 }=25+25={ \left( 5\sqrt { 2 } \right) }^{ 2 }={ CA }^{ 2 }$

$\Rightarrow$

$\triangle$ ABC is right-angled at B.
Thus,

$\triangle$ ABC is right-angled isosceles triangle.
Now, Area of

$\triangle$ ABC

$=\frac { 1 }{ 2 } \left( Base\times Height \right) =\frac { 1 }{ 2 } \left( AB\times BC \right)$

$\Rightarrow$ Area of

$\triangle$ ABC=

$\left( \frac { 1 }{ 2 } \times 5\times 5 \right)$ sq. Units=

$\frac { 25 }{ 2 }$ sq. Units

Show that the points (1,-1), (5,2) and (9,5) are collinear.

Let A(1,-1), B(5,2) and (9,5) be the given points. Then,we have

$AB=\sqrt { { \left( 5-1 \right) }^{ 2 }+{ \left( 2+1 \right) }^{ 2 } } =\sqrt { 16+9 } =5$

$BC=\sqrt { { \left( 5-9 \right) }^{ 2 }+{ \left( 2-5 \right) }^{ 2 } } =\sqrt { 16+9 } =5$
and,

$AC=\sqrt { { \left( 1-9 \right) }^{ 2 }+{ \left( -1-5 \right) }^{ 2 } } =\sqrt { 64+36 } =10$
Clearly, AC=AB+BC
Hence, A,B,C are collinear points.

If two vertices of an equilateral triangle be (0,0), $\left( 3,\sqrt { 3 } \right)$ , find the third vertex.

O(0,0) and A

$\left( 3,\sqrt { 3 } \right)$ be the given points and let B(x,y) be the third vertex of equilateral

$\triangle OAB$ . Then,

$OA=OB=AB$

$\Rightarrow \quad \quad { OA }^{ 2 }={ OB }^{ 2 }={ AB }^{ 2 }$

$We\quad have,\quad { OA }^{ 2 }={ \left( 3-0 \right) }^{ 2 }+{ \left( \sqrt { 3 } -0 \right) }^{ 2 }=12,$

${ OB }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }$

$and,\quad \quad { AB }^{ 2 }={ \left( x-3 \right) }^{ 2 }+{ \left( y-\sqrt { 3 } \right) }^{ 2 }$

$\Rightarrow \quad \quad { AB }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }-{ 6x }-2\sqrt { 3 } y+12$

$\therefore \quad \quad { OA }^{ 2 }={ OB }^{ 2 }={ AB }^{ 2 }$

$\Rightarrow \quad \quad { OA }^{ 2 }={ OB }^{ 2 }\quad and\quad { OB }^{ 2 }={ AB }^{ 2 }$

$\Rightarrow \quad \quad { x }^{ 2 }+{ y }^{ 2 }=12$

$and,\quad { x }^{ 2 }+{ y }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }-{ 6x }-2\sqrt { 3 } y+12$

$\Rightarrow \quad \quad { x }^{ 2 }+{ y }^{ 2 }={ 12\quad and\quad { 6x }+2\sqrt { 3 } y=12 }$

$\Rightarrow \quad \quad { x }^{ 2 }+{ y }^{ 2 }={ 12\quad and\quad 3x+\sqrt { 3 } y=6 }$

$\Rightarrow \quad \quad { x }^{ 2 }+{ \left( \frac { 6-3x }{ \sqrt { 3 } } \right) }^{ 2 }=12\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left[ \because 3x+\sqrt { 3 } y=6\quad \therefore y=\frac { 6-3x }{ \sqrt { 3 } } \right]$

$\Rightarrow \quad \quad { 3x }^{ 2 }+{ \left( 6-3x \right) }^{ 2 }=12$

$\Rightarrow \quad \quad { 12x }^{ 2 }-36x=0$

$\Rightarrow \quad \quad x=0,3$

$\therefore \quad \quad x=0\Rightarrow \sqrt { 3 } y=6\Rightarrow y=\frac { 6 }{ \sqrt { 3 } } =2\sqrt { 3 } \quad \quad \quad \quad \quad \quad \quad \quad \left[ Putting\quad x=0\quad in\quad 3x+\sqrt { 3 } y=6 \right]$

$and,\quad \quad x=3\Rightarrow 9+\sqrt { 3 } y=6\Rightarrow y=\frac { 6-9 }{ \sqrt { 3 } } =-\sqrt { 3 } \quad \quad \quad \quad \left[ Putting\quad x=3\quad in\quad 3x+\sqrt { 3 } y=6 \right]$
Hence, the coordinates of the third vertex B are

$\left( 0,2\sqrt { 3 } \right) \quad or,\quad \left( 3,-\sqrt { 3 } \right)$ .

A line through the origin intersects $x= 1, y=2,$ and $x+y =4$ , at $A, B$ and $C$ respectively, such that $OA. OB. OC = 8\sqrt2$ . Find the equation of the line is $y=mx$ .Find the value of $m$

Let equation of line

$y=mx$

$x=1 , y=m$

$y=2 ,$

$x=\frac{2}{m}$

$x+y=4$

$x(m+1)=4$

$x=\frac{4}{m+1}$

$y=\frac{4m}{m+1}$

$A(1,m)$

$B(\frac{2}{m},2)$

$C(\frac{4}{m+1},\frac{4m}{m+1})$

$OA=\sqrt{1^{2}+m^{2}}$

$OB=\sqrt{\frac{4}{m^{2}}+4}$

$OC=\sqrt{\left ( \frac{4}{m+1} \right )^{2}+\left ( \frac{4m}{m+1} \right )^{2}}$

Acc to question

$\sqrt{1+m^{2}}*\frac{2}{m}\sqrt{1+m^{2}}*\frac{4}{m+1}\sqrt{1+m^{2}}=8\sqrt{2}$

$\frac{8}{m(m+1)}(1+m^{2})^{\frac{3}{2}}=8\sqrt{2}$

Squaring both

$\frac{(1+m^{2})^{3}}{m^{2}(m+1)^{2}}=2$

$(1+m^{2})^{3}=2m^{2}(m+1)^{2}$

$(1+m^{6})+3m^{2}+3m^{4}=2m^{2}(m^{2}+1+2m)$

$1+m^{6}+3m^{2}+3m^{4}=2m^{4}+2m^{2}+4m^{3}$

Using hit and trial method

$m=1$ Ans

In $\Delta ABC, A \equiv (6, 3), B \equiv (-3, 5), C \equiv (4, -2)$ . $P$ is any point $(x, y)$ . Show how that $\dfrac{\Delta(PBC)}{\Delta(ABC)} = \left|\dfrac{x + y - 2}{7}\right|$

$\Delta$ ABC,

$A=(6, 3)$ ;

$B(-3, 5)$ ;

$C=(4, -2)$ ;

$P=(x, y)$

$Ar(\Delta ABC)=\dfrac{1}{2}\times (BC)\times h_1$

$Ar(\Delta PBC)=\dfrac{1}{2}\times (BC)\times h_2$

BC, equation

$\Rightarrow m=\dfrac{7}{-7}=-1$

$x+y=2$

$h_1=\dfrac{|6+3-2|}{\sqrt{1^2+1^2}}=\dfrac{7}{\sqrt{2}}$

$h_2=\dfrac{|x+y-z|}{\sqrt{2}}$

$\dfrac{\Delta (PBC)}{\Delta (ABC)}=\dfrac{\dfrac{1}{2}\times BC\times h_2}{\dfrac{1}{2}\times BC\times h_1}$

$=\dfrac{h_2}{h_1}=\dfrac{|x+y-2|}{7}$ .

Find the coordinate of the point on the y axis which at a distance of $10$ units from the point $\left( { - 8,4} \right)$

Let the required point be (0,y)

$10^2=(0+8)^2+(4-y)^2$

$\Rightarrow 100=64+16+y^2-8y$

$\Rightarrow 16+y^2-8y=36$

$\Rightarrow y^2-8y-20=0$

$\Rightarrow y^2-10y+2y-20=0$

$\Rightarrow y(y-10)+2(y-10)=0$

$\Rightarrow y=-2$ or

$y=10$

$\therefore$ Coordinate of the required point be

$(0,-2)$ or

$(0,10)$

Find the area of the triangle $ABC$ with $A(1,-4)$ and mid points of sides through $A$ being $(2,-1)$ and $(0,-1)$

A(1,-4)

Considering,

$AB(2,-1)$

$AC(0,-1)$

Let

$B(x_1,y_1)$

$\left ( \dfrac{x_1+1}{2},\dfrac{y_1-4}{2} \right )=(2,-1)$

$x_1=3, y_1=2$

$B(3,2)$

Let

$C(x_2,y_2)$

$\left ( \dfrac{x_2+1}{2},\dfrac{y_2-4}{2} \right )=(0,-1)$

$x_2=-1, y_2=2$

$C(-1,2)$

Area

$= \dfrac{1}{2}\left [ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \right ]$

$= \dfrac{1}{2}\left [ 3(2+4)+(-1)(-4-2)+1(2-2) \right ]$

$=\dfrac{1}{2}[18+6+0]$

$=\dfrac{24}{2}$

Area of triangle

$= 12$ sq. units

A parallel line is drawn from point $P(5, 3)$ to $y$ -axis, what is the distance between the line and $y$ -axis.

Equation of line parallel to y axis

$= x= 5$

If line passes through (5,3)

Distance between two lines

$L_{1} = ax+by+c_{1}=0$

$\quad \quad \quad L_{2} = ax+by+c_{2}=0$

$D=\cfrac { |c_{ 1 }-c_{ 2 }| }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } }$

Here,

$L_{1} = x = 5$

$L_{2} = x = 0$

$D=\cfrac { |5-0| }{ 1 }$

$D = 5 \quad Ans$

948334_1025775_ans_7cbc4743fc78415ab967818114ed720f.PNG

If the points $(2, 1)$ and $(1, -2)$ are equidistant from the point $(x, y)$ , show that $x + 3y = 0$ .

The distances of the point

$(2,1)$ and

$(1,-2)$ from

$(x,y)$ are

$\sqrt{(x-2)^2+(y-1)^2}$ and

$\sqrt{(x-1)^2+(y+2)^2}$ units respectively.

According to the problem,

$\sqrt{(x-2)^2+(y-1)^2}=$

$\sqrt{(x-1)^2+(y+2)^2}$

or,

$(x-2)^2+(y-1)^2=(x-1)^2+(y+2)^2$

or,

$-4x+4-2y+1=-2x+1+4y+4$

or,

$2x+6y=0$

or,

$x+3y=0$ .

Find the direction cosines of the sides of the triangle whose vertices are $(3, 5, -4)$ and $(-1, 1, 2)$ and $( -5, -5, -2)$ . What type triangle is it?

$A(3,5,-4);B(-1,1,2);C(-5,-5,-2)$ are vertices of

$\triangle ABC$

$\therefore \overrightarrow{AB}=(-1-3,1-5,2-(-4))$

$\therefore \overrightarrow {BC}=(-5-(-1),-5-1,-2-2)$

$\therefore \overrightarrow {CA}=(3-(-5),5-(-5),-4-(-2))$

$\therefore \overrightarrow {AB}=(-4,-4,6)$

$\therefore \overrightarrow {BC}=(-4,-6,-4)$

$\therefore \overrightarrow {CA}=(8,10,-2)$

$\therefore$ direction cosines of

$AB=\dfrac{1(-4,-4,6)}{\sqrt{(-4)^2+(-4)^2+6^2}}$

$=\left(\dfrac{-4}{\sqrt{68}},\dfrac{-4}{\sqrt {68}},\dfrac{6}{\sqrt {68}}\right)$

$\therefore$ direction cosines of

$BC=\dfrac{(-4,-6,-4)}{\sqrt{(-4)^2+(-6)^2+(-4)^2}}$

$=\left(\dfrac{-4}{\sqrt 68},\dfrac{-6}{\sqrt{68}},\dfrac{-4}{\sqrt {68}}\right)$

$\therefore$ direction cosines of

$CA=\left(\dfrac{8,10,-2}{\sqrt{8^2+10^2+(-2)^2}}\right)$

$=\left(\dfrac{8}{\sqrt{168}},\dfrac{10}{\sqrt{168}},\dfrac{-2}{\sqrt {168}}\right)$

If (-4,0) and (4,0) are two vertices of an equilateral triangle, find the coordinates of its third vertex.

Length of

$AB = 8$ units

In equilateral triangle all sides are equal

$AC=BC=AB$

$BC=AC$

$\sqrt{(h-4)^{2}+b^{2}}=\sqrt{(h+4)^{2}+b^{2}}$

Squaring both sides

$(h-4)^{2}+b^{2}=(h+4)^{2}+b^{2}$

$h=0$

$AC=AB$

$\sqrt{(h+4)^{2}+b^{2}}=8$

Squaring both sides

$(h+4)^{2}+b^{2}=64$

$b^{2}=48$

$b=\pm \sqrt{48}$

Third vertex

$=(0,\sqrt{48}) or (0,-\sqrt{48})$

960218_1028395_ans_31b919e235614ad59c087b8a29438eec.png

Distance between parallel lines $9x^2-6xy +y^2+18x-6y +8=0$ is $\dfrac{m}{\sqrt n}$ .Find $m+n$

$9{ x }^{ 2 }-6xy+{ y }^{ 2 }+18x-6y+8=0\\ (3x-y+2)(3x-y+4)=0\\ \therefore \quad Parallel\quad lines\quad are\\ 3x-y+2=0\\ 3x-y+4=0\\ \\ \quad$

Distance between lines =

$\cfrac { 4-2 }{ \sqrt { { 3 }^{ 2 }+{ 1 }^{ 2 } } }$

$\cfrac { 2 }{ \sqrt 10}$

$\therefore \quad \cfrac { m }{ \sqrt n } =\cfrac { 2 }{ \sqrt { 10 } } \\ m+n= 12$

Find the area of $\triangle$ le with coordinate of vertices $A(-5,-1),B(3,-5)$ and $C(5,2)$ resp. Find the length of median trough $B$ .

$A\left( { - 5, - 1} \right),B\left( {3, - 5} \right)C\left( {5,2} \right)$

Area of $\Delta ABC$ = $\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$

$= \dfrac{1}{2}\left[ { - 5\left( { - 5 - 2} \right) + 3\left( {2 + 1} \right) + 5\left( { - 1 + 5} \right)} \right]$

$= \dfrac{1}{2}\left[ { - 5 \times - 7 + 3 \times 3 + 5 \times 4} \right]$

$= \dfrac{1}{2}\left[ {35 + 9 + 20} \right]$

$= \dfrac{1}{2}\left[ {64} \right]$

$= 32sq\,units$

Mid point of AC=

$\left( {\dfrac{{ - 5 + 5}}{2},\dfrac{{ - 1 + 2}}{2}} \right) = \left( {0,\dfrac{1}{2}} \right)$

Length of Median through B= $B = \sqrt {{{\left( {3 - 0} \right)}^2} + {{\left( { - 5\dfrac{{ - 1}}{2}} \right)}^2}}$

$= \sqrt {9 + \dfrac{{121}}{4}}$

$= \dfrac{{\sqrt {157} }}{2}\,unit$

If $(4, 3)$ and $(-2, -1)$ are positive vertex of a parallelogram and third vertex $(1, 0)$ then the product of co-ordinates of the four vertex?

We know that diagonals of a parallelogram bisect each other.

Let the point of intersection of diagonals be

$P(m,n)$

Co-ordinates of three vertex of parallelogram are

$(4,3) , (1,0)$ and

$(-2,-1)$

$\therefore m = \dfrac{(-2+4)}{2} = 1$

Also,

$n = \dfrac{(-1+3)}{2} = 1$

Let the fourth vertex of a parallelogram be

$(x,y)$

$\Rightarrow m = \dfrac{(1+x)}{2}$

$\Rightarrow 1 = \dfrac{(1+x)}{2}$

$\Rightarrow x = 1$

Also,

$n = \dfrac{(0+y)}{2}$

$\Rightarrow 1 = \dfrac{(0+y)}{2}$

$\Rightarrow y = 2$

Product of co-ordinates

$= 1\times2 = 2$

$\int {\dfrac{{{e^x}dx}}{{\left( {1 + {e^{2x}}} \right)}}}$

We have,

$I=\int {\dfrac{{{e^x}dx}}{{\left( {1 + {e^{2x}}} \right)}}}$

Let

$t=e^x$

$\dfrac{dt}{dx}=e^x$

$dt=e^x\ dx$

Therefore,

$I=\int \dfrac{dt}{1+t^2}$

$I=\tan^{-1} t+C$

On putting the value of

$t$ , we get

$I=\tan^{-1} e^x+C$

Hence, this is the answer.

Find the equation of a line which is equidistant from the lines $x = -2$ and $x = 6$ .

Since the given two lines are parallel so any line which is to be equidistant to those lines must be parallel to those lines.

Let the equation will be

$x=c$ ......(1).

Now the distance between the line (1) and

$x=-2$ is

$|c+2|$ units......(2).

And the distance between the line (1) and

$x=6$ is

$|c-6|$ units.....(3)

According to the problem,

$|c+2|=|c-6|$

or,

$c+2=-c+6$ [ taking the negative sign]

or,

$2c=4$

or,

$c=2$ .

So the line is

$x=2$ .

Find the distance between the points:
$\left( {\operatorname{Cos} \theta ,\operatorname{Sin} \theta } \right),\left( {\operatorname{Sin} \theta ,\operatorname-{Cos} \theta } \right)$

1350319_1050121_ans_6facf249aac5464e98b0dbeca64b7de8.png

Find the value of $k$ for which the area of the triangle with vertices $(2,-2),(-3,3k)$ and $(-2,3)$ is $20$ sq.units.

Area of

$\bigtriangleup$

$=\frac { 1 }{ 2 } \biggr[\begin{matrix} { x }_{ 1 } & { y }_{ 1 } & 1 \\ { x }_{ 2 } & { y }_{ 2 } & 1 \\ { x }_{ 3 } & { y }_{ 3 } & 1 \end{matrix}\biggr]$

Magnitude can be +ve or -ve

$\pm 20$

$=\frac { 1 }{ 2 } \biggr [\begin{matrix} 2 & -2 & 1 \\ -3 & 3b & 1 \\ -2 & 3 & 1 \end{matrix}\biggr]$

$\pm 20=\frac{1}{2}\left ( 2(3b-3)+2(-3+2)+1(-9+6k) \right )$

$\pm 40=6b-6-2-9+6b$

$\pm 40=12b-17$

$\frac{17\pm 40}{12}=b$

$b = \dfrac{57}{12}$ or

$-\dfrac{23}{12}$

The distances of point $P(x,y)$ from the points $A(1,-3)$ and $B(-2,2)$ are in the ratio $2:3$ .
Show that :
$5{x^2} + 5{y^2} - 34x + 70y + 58 = 0$

Distance between P & A is :-

${r_1} = \sqrt {{{(x - 1)}^2} + {{(y + 3)}^2}}$

and Distance between P & B is:-

${r_2} = \sqrt {{{(x + 2)}^2} + {{(y - 2)}^2}}$

Now,

$\cfrac{{{r_1}}}{{{r_2}}} = \cfrac{2}{3}$

$3{r_1} = 2{r_2}$

$3\sqrt {{{(x - 1)}^2} + {{(y + 3)}^2}} = 2\left( {\sqrt {{{(x + 2)}^2} + {{(y - 2)}^2}} } \right)$

requiring both side,

$9\left[ {{{\left( {x - 1} \right)}^2} + {{\left( {y + 3} \right)}^2}} \right] = 4\left[ {{{\left( {x + 2} \right)}^2} + {{\left( {y - 2} \right)}^2}} \right]$

$9\left[ {{x^2} + 1 - 2x + {y^2} + 9 + 6y} \right] = 4\left[ {{x^2} + 4 + 4x + {y^2} + 4 - 4y} \right]$

$9{x^2} - 18x + 9{y^2} + 54y + 90 = 4{x^2} + 16x + 4{y^2} + 6y + 32$

$5{x^2} - 34x + 5{y^2} + 70y - 58 = 0$

Check whether $(5, -2),(6,4)$ and $(7,-2)$ are the vertices of an isosceles triangle.

The vertices of a triangle ABC is given as $A\left( {5, - 2} \right)$ , $B\left( {6,4} \right)$ and $C\left( {7, - 2} \right)$ .

The distance AB is,

$AB = \sqrt {{{\left( {6 - 5} \right)}^2} + {{\left( {4 - \left( { - 2} \right)} \right)}^2}}$

$= \sqrt {{{\left( 1 \right)}^2} + {{\left( 6 \right)}^2}}$

$= \sqrt {1 + 36}$

$= \sqrt {37}$

The distance BC is,

$BC = \sqrt {{{\left( {7 - 6} \right)}^2} + {{\left( { - 2 - 4} \right)}^2}}$

$= \sqrt {{{\left( 1 \right)}^2} + {{\left( { - 6} \right)}^2}}$

$= \sqrt {1 + 36}$

$= \sqrt {37}$

The distance CA is,

$CA = \sqrt {{{\left( {5 - 7} \right)}^2} + {{\left( { - 2 - \left( { - 2} \right)} \right)}^2}}$

$= \sqrt {{{\left( { - 2} \right)}^2} + 0}$

$= \sqrt 4$

$= 2$

Since, $AB = BC$ , that is the two sides of a triangle are equal, therefore, the triangle is an isosceles triangle.

Show that the points $A(1,2,3),B(-1,-2,-3),C(2,3,2)$ and $D(4,7,6)$ are the vertices of a parallelogram.

To prove ;

$ABCD$ is a parallelogram for this we'll show that both the diagonal meet at

$0$

i.e.,

$0$ is the midpoint of

$AC$ and

$BD$

$(BD)\ 0 \equiv \left( \dfrac{-1+4}{2}, \dfrac{-2+7}{2} , \dfrac{-1+6}{2} \right)$

$(AC)\ 0 \equiv \left( \dfrac{1+2}{2} , \dfrac{2+3}{2} , \dfrac{3+2}{2} \right)$

$0 \equiv \left( \dfrac{3}{2}, \dfrac{5}{2}, \dfrac{5}{2} \right) \quad 0 \equiv \left( \dfrac{3}{2}, \dfrac{5}{2}, \dfrac{5}{2} \right)$

Hence,

$A (1,2,3), B(-1,-2,-1), C(2,3,2)$ and

$D(4, 7 , 6)$ are the vertices of a parallelogram.

1425901_1057341_ans_0de790569bd2469b8b3bd360e5f36cb3.png

Find the distance of the point $(6, 8)$ and the origin.

$\textbf{Step 1: Use the distance formula to find the distance between origin and (6,8)}$

$\because \text{the distance between two points is given by}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

$\text{Distance between the point (6,8) and origin(0,0)}$

$=\sqrt{(6-0)^2+(8-0)^2}$

$=\sqrt{36+64}$

$=\sqrt{100}$

$=10$

$\textbf{Hence, Distance between (6,8) and origin is 10.}$

The points $A(2, 9)$ , $B(a, 5)$ , $C(5, 5)$ are the vertices of a triangle ABC right angled at B. find the value of $'a'$ and hence the area of $\Delta ABC$ .

Given the points $A\left( 2,9 \right),\,B\left( a,5 \right),C\left( 5,5 \right)$ are the vertices of right angle triangle whose angle $B$ is right angle.

Then distance

$AB=\sqrt{{{\left( 2-a \right)}^{2}}+{{\left( 9-5 \right)}^{2}}}$

$=\sqrt{4+{{a}^{2}}-4a+16}$

$=\sqrt{{{a}^{2}}-4a+20}$

$BC=\sqrt{{{\left( a-5 \right)}^{2}}+{{\left( 5-5 \right)}^{2}}}$

$=\sqrt{{{a}^{2}}+25-10a}$

$CA=\sqrt{{{\left( 5-2 \right)}^{2}}+{{\left( 5-9 \right)}^{2}}}$

$=\sqrt{9+16}$

$=\sqrt{25}=5$

We know that, In a right angle triangle

By phythagoras theorm,

$A{{B}^{2}}+B{{C}^{2}}=C{{A}^{2}}$

${{\left( \sqrt{{{a}^{2}}-4a+20} \right)}^{2}}+{{\left( \sqrt{{{a}^{2}}+25-10a} \right)}^{2}}={{5}^{2}}$

$\Rightarrow {{a}^{2}}-4a+20+{{a}^{2}}+25-10a=25$

$\Rightarrow 2{{a}^{2}}-14a+20=0$

$\Rightarrow 2\left( {{a}^{2}}-7a+10 \right)=0$

$\Rightarrow {{a}^{2}}-7a+10=0$

By factorize this equation

$\Rightarrow a=2,5$

Put first $a=2$ , In $AB,BC\,\,and\,\,CA$

Then,

$AB=\sqrt{{{2}^{2}}-4\times 2+20}$

$=\sqrt{4-8+20}$

$=\sqrt{16}=4$

$BC=\sqrt{{{a}^{2}}-10a+25}$

$=\sqrt{{{2}^{2}}-10\times 2+25}$

$=\sqrt{9}=3$

Now, put $a=5$ and we get,

$AB=\sqrt{{{5}^{2}}-4\times 5+20}$

$=\sqrt{25-20+20}$

$=\sqrt{25}=5$

$BC=\sqrt{{{a}^{2}}-10a+25}$

$=\sqrt{{{5}^{2}}-10\times 5+25}$

$=\sqrt{0}=0$

Then $a=5$ is not possible

Hence $a=2$ is possible for all value of right angle triangle.

Find the area of triangles formed by the following points :
(3,4), (2, -1), (4, -6)

Area bounded by points

$(3,4),(2,-1)$ and

$(4,-6)$

=>Area

$==\cfrac { 1 }{ 2 } \left| \begin{matrix} 3 & 4 \\ 2 & -1 \\ 4 & -6 \\ 3 & 4 \end{matrix} \right| =\cfrac { 1 }{ 2 } \left| 15 \right| =\cfrac { 15 }{ 2 } sq.units\\$

Show that straight lines $x+y=0 , 3x+y-4=0$ , $x+3y-4 =0$ form isosceles

coordinates of A by intersection of

$L_{1},L_{3}$

$A(2,-2)$

coordinates of B by intersection of

$L_{1},L_{2}$

$B(-2,2)$

coordinates of C by intersection of

$L_{2},L_{3}$

$C(1,1)$

$AC=\sqrt{(2-1)^{2}+(-2-1)^{2}}=\sqrt{10}$

$BC=\sqrt{(-2-1)^{2}+(2-1)^{2}}=\sqrt{10}$

$AB=\sqrt{(2+2)^{2}+(-2-2)^{2}}=4\sqrt{2}$

$AC=BC$

Triangle is isosceles

1068385_1065900_ans_8260cc1cdfa84b25ad862e011dcfcdc4.PNG

Four points $A(6, 3)$ , $B)-3, 5)$ , $C(4, -2)$ and $D(x, 3x)$ are given such that $\dfrac{\Delta DBC}{\Delta ABC}=\dfrac{1}{2}$ , find x.

$A(6,3),B(-3,5),C(4,-2)$ and

$D(x,3x)$

Area of

$\triangle DBC=\cfrac { 1 }{ 2 } \begin{vmatrix} -3 & 5 \\ 4 & -2 \\ x & 3x \\ -3 & 5 \end{vmatrix}=\cfrac { 1 }{ 2 } \left| -14+14x+14x \right| =7(2x-1)\\$

Area of

$\triangle ABC=\cfrac { 1 }{ 2 } \begin{vmatrix} 6 & 3 \\ -3 & 5 \\ 4 & -2 \\ 6 & 3 \end{vmatrix}=\cfrac { 1 }{ 2 } (49)\\$

Given,

$\cfrac { Area\quad of\quad \triangle DBC }{ Area\quad of\quad \triangle ABC } =\cfrac { 1 }{ 2 } \\ =>\cfrac { 7(2x-1) }{ \cfrac { 1 }{ 2 } (49) } =\cfrac { 1 }{ 2 } \\ =>56x-28=49\\ =>x=\cfrac { 11 }{ 8 }$

Draw the graph of the equation $\dfrac{x}{4}+\dfrac{y}{3}=1$ . Also, find the area of the triangle formed by the line and the coordinate axes.

We are given, the equation

$\dfrac{x}{4} + \dfrac{y}{3} = 1$

For graph, let

$x=0$ we get

$y=3\ 4$

$y=0$ we get

$x=4$

So, the prints an equation(s) are

$(0,3)$ &

$(4,0)$

$X-intercept$ will be

$OA=\sqrt {(4-0)^2+(0-0)^2}$

$OA=4$

$Y-intercept$ will be

$OB=\sqrt {(0-0)^2+(0-3)^2}$

$OA=3$

So, the Area of

$OAB$ is

$\triangle OAB = \dfrac{1}{2} \times OA \times OB$

$=\dfrac{1}{2} \times 4 \times 3$

$\boxed {\triangle OAB = 6 (unit)^2}$

1424504_1066873_ans_24369dab41ae40d19d83dbec5db9de2a.png

If the vertices of a triangle have integral coordinates, prove that the triangle cannot be equilateral.

Without loss of generality, we can choose the co-ordinates as

$A=(0, 0)$

$B=(x, y)$

$C=(a, b)$

Now, slope of

$AB=\dfrac{y}{x}$ is rational slope of

$AC=\dfrac{b}{a}$ is rational.

$\tan (BAC)=\left(\dfrac{\dfrac{y}{x}-\dfrac{a}{b}}{1+\dfrac{ay}{bx}}\right)$ which is rational.

$\tan (60)=\sqrt{3}$ , there cannot be any point with rational coefficient so angle BAC cannot be

$60^o$ .

As we cannot find any rational point on the k line at

$60^o$ so getting an equilateral triangle is not possible.

Hence proved.

Given that $A(5, 4), B(-3, -2)$ and $C(1, -8)$ are the vertices of a $\triangle ABC$
Find:
i)The slope of median $AD$ ,
ii) The slope of altitude $BM$ .

Consider the following question.

i)The slope of median AD

ii) The slope of altitude BM.

$A(5,4),B(-3,-2)\,\,\And \,\,C(1,-8)$

Hence, D & M is the mid point of BC & AC.

$D=\left( \dfrac{5-3}{2},\dfrac{4-2}{2} \right)$

$D=\left( 1,1 \right)$

$=1$

$M=\left( \dfrac{-3+1}{2},\dfrac{-2-8}{2} \right)$

$M=\left( -1,-5 \right)$

Slope of AD

$AD=\left( \dfrac{1-4}{1-5} \right)$

$=-\dfrac{3}{4}$

Slope of BM.

$BM({{m}_{1}})=\left( \dfrac{-2+5}{-3+1} \right)$

$=-\dfrac{3}{2}$

The slope of altitude BM

${{m}_{1}}\times {{m}_{2}}=-1$

${{m}_{2}}=\dfrac{1}{-{{m}_{1}}}$

$=\dfrac{1}{-\left( \dfrac{-3}{2} \right)}$

$=\dfrac{2}{3}$

Hence, this is the required answer.

1065595_1085211_ans_c37767a0b00e4408a81050a8a55512d1.png

Find the area of the triangle formed by joining the mid-points of the sides whose vertices are (0, -1), (2, 1) and (0, 3). Also find the ratio of the given triangle to the newly formed triangle.

$A(0,-1),B(2,1)$ and

$C(0,3)$

Mid-points of AB,BC&CD=>

$D(1,0),E(1,2)$ and

$F(0,1)$

Area of

$\triangle DEF=\cfrac { 1 }{ 2 } \begin{vmatrix} 1 & 0 \\ 1 & 2 \\ 0 & 1 \\ 1 & 0 \end{vmatrix}=\cfrac { 1 }{ 2 } \left| 2+1-1 \right| =1sq.units\\$

Area of

$\triangle ABC=\cfrac { 1 }{ 2 } \begin{vmatrix} 0 & -1 \\ 2 & 1 \\ 0 & 3 \\ 0 & -1 \end{vmatrix}=4sq.units\\$

$\cfrac { Area\quad of\quad \triangle DEF }{ Area\quad of\quad \triangle ABC } =\cfrac { 1 }{ 4 }$

Find the acute angle between the two lines:
$AB$ and $CD$ passing through the points $A\equiv (3,1,-2),B\equiv (4,0,-4)$ and $C\equiv (4,-3,3)$ , $D\equiv (6,-2,2)$

$\bar{AB}=\bar{B}-\bar{A}$

$=\hat{i}-\hat{j}-\hat{2k}=\vec{\alpha}$

$\bar{C \Delta}=\bar{\Delta}-\bar{c}$

$=2\hat{i}+\hat{j}-\hat{k}=\vec{\beta}$

$|\bar{AB}|=\sqrt{6};|\bar{CD}|=\sqrt{6}$

$\vec{\alpha}.\vec{\beta}=|\vec{\alpha}||\vec{\beta}|\cos\theta$

$3=sqrt{6}.\sqrt{6}.\cos \theta$

$\dfrac{1}{2}=\cos \theta$

$\theta=60^o$

If a line makes angles $90^o$ , $60^o$ and $30^o$ with the positive direction of x, y and z axis respectively find its direction cosines.

Let the line be

$x \hat{i}+y \hat{j}+z\hat{k}$

makes

$90^o$ with

$x$ direction.

$(x \hat{i}+y \hat{j}+z\hat{k}).\hat{(j})=\cos 90=0$

$\Rightarrow x=0$

makes

$60^o$ with

$y$ axsis

$(x \hat{i}+y \hat{j}+z\hat{k}).\hat{(j)}=\cos 60=\dfrac{1}{2}$

$y=\dfrac{1}{2}$

makes

$30^o$ with

$z$ axis

$(x \hat{i}+y \hat{j}+z\hat{k}).\hat{(k)}=\cos 30^o=\dfrac{\sqrt 3}{2}$

$z=\dfrac{\sqrt 3}{2}$

The vector is

$\dfrac{1}{2}\hat{j}+ \dfrac{\sqrt 3}{2} \hat{k}$

it directional cosines are

$\left(0,\dfrac{1/2}{\sqrt{\dfrac{1}{4}+\dfrac{3}{4}}}, \dfrac{\sqrt 3/2}{\sqrt{\dfrac{1}{4}+\dfrac{3}{4}}}\right)$

$\Rightarrow \left(0,\dfrac{1}{2},\dfrac{\sqrt 3}{2}\right)$

If the points $(6, 9),(10, x)$ and $(-6, -7)$ are collinear then find the value of $x$ .

Consider the following question.

Let A,B & C are collinear.

$A\left( 6,9 \right)\,\,\,\,\,B\left( 10,x \right)\,\,\,\And \,\,\,C\left( -6,-7 \right)$

Hence,

Slope of AB=Slope of BC

$\dfrac{x-9}{10-6}=\dfrac{-7-x}{-6-10}$

$\dfrac{x-9}{10-6}=\dfrac{7+x}{6+10}$

$\dfrac{x-9}{4}=\dfrac{7+x}{16}$

$x=\dfrac{43}{3}$

Hence, this is the required answer.

Find the value of x, if the distance between the points (x, -1) and (3, 2) is 5.

$A(x,-1)$ and

$B(3,2)$

$=>AB=5\\ =>\sqrt { (x-3)^{ 2 }+(2+1)^{ 2 } } =5\\ =>(x-3)^{ 2 }=16\\ =>x=7,-1$

Find the relation between $x$ and $y$ if the points $A(X,Y), B(-5,7)$ and $C(-4, 5) \$ are collinear.

Consider the given points,

$A\left( x,y \right),B\left( -5,7 \right)$ , $C\left( -4,5 \right)$ are collinear

If A,B,C are collinear then area of triangle ABC=0

Area of triangle ABC= $=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$

$=\dfrac{1}{2}\left[ x\left( 7-5 \right)-5\left( 5-y \right)-4\left( y-7 \right)=0 \right]$

$2x-5y-4y+28=0$

$\therefore \,2x+y+3=0$

The line $3x+2y=24$ meets the y- axis at A and the x -axis at B. The perpendicular bisectors of AB meets the line through $(0,-1)$ parallel to x - axis at $C$ . Find the area of the triangle ABC ?

Given the line,

$3x+2y=24\quad \quad \longrightarrow (1)$

for A putting x=0 in

$(1)$ ,

$2y=24$

$\Rightarrow y=12$

$\therefore$ coordinates of

$A=\left( 0,12 \right)$

for B putting y=0 in

$(1)$ ,

$3y=24$

$\Rightarrow x=8$

$\therefore$ coordinates of

$B=\left( 8,0 \right)$

Midpoint of

$AB=\left( \dfrac { 8+0 }{ 2 } ,\dfrac { 0+12 }{ 2 } \right) =\left( 4,6 \right)$

Now, equation of line perpendicular to line

$(1)$ ,

$2x-3y=\lambda$

It will pass through

$(4,6)$

so,

$2(4)-3(6)=\lambda$

$\Rightarrow \lambda =8-18$

$\Rightarrow \lambda =-10$

Equation of the line parallel to X-axis is,

$y=constant$

it will pass through

$(0,-1)$

$\Rightarrow -1=constant$

$\Rightarrow y=-1$

To get coordinates of C,putting

$y=-1$ in

$2x-3y=-10$

$\Rightarrow 2x+3=-10$

$\Rightarrow x=-\dfrac { 13 }{ 2 }$

Hence, we have

$C\left( -\dfrac { 13 }{ 2 } ,-1 \right) ,A\left( 0,12 \right) and\quad B\left( 8,0 \right)$

$\therefore$ Area of $$\triangle ABC

$=\left| -87-4 \right|$

$=\left| -91 \right|$

$=\quad 91\quad square\quad units$

1117473_1096398_ans_3663b131742249ed85406c387b83e279.jpg

Find the distance of the line $4x - y = 0$ find the point $P(4, 1)$ measured along the line making an angle of $135^o$ with the positive x-axis.

$P(4, 1)$

$m = tan 135 = tan(160^\circ - 45^\circ) = -tan 45^\circ = -1$

equation of line with slope

$(135) \,\& \, P(4, 1)$

$\Rightarrow \dfrac{y - 1}{x - 4} = -1$

$\Rightarrow y - 1 = -x + 4$

$\Rightarrow x + y - 5 = 0 \Rightarrow x + y = 5$ ........(i)

Now,

$4x - y = 0$

(i) + (ii)

$5x = 5 \Rightarrow x = 1$ &

$x + y = 5$

$y = 5 - 1 = 4$

So,

$P(4, 1), Q(1, 4)$

$Distance = \sqrt{(4 - 1)^2 + (1 - 4)^2} = 3\sqrt{2}\,units$

If the points $(x, y), (-5, -2)$ and $(3, -5)$ are collinear, then prove that $3x + 8y + 31 = 0$ .

$(x,y)$

$(-5,-2)$

$and$

$(3,-5)$ $ are collinear.

The area bounded by them=

$0$

$\cfrac { 1 }{ 2 } \begin{vmatrix} x & y \\ -5 & -2 \\ 3 & -5 \\ x & y \end{vmatrix}=0\\ =>-2x+5y+31+3y+5x=0\\ =>3x+8y+31=0\\$

Find the perimeter of the rectangle formed by the joining the points $A(-3,-1),B(5,-1),C(5,3)$ and $D(-3,3)$ in sides.

Using distance formula

$=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$

Perimeter of rectangle

$ABCD=AB+BC+CD+DA$

$AB=\sqrt{{\left(5-\left(-3\right)\right)}^{2}+{\left(-1-\left(-1\right)\right)}^{2}}=\sqrt{{\left(5+3\right)}^{2}+{\left(-1+1\right)}^{2}}=\sqrt{64+0}=8$ units

$BC=\sqrt{{\left(5-5\right)}^{2}+{\left(3-\left(-1\right)\right)}^{2}}=\sqrt{0+{\left(3+1\right)}^{2}}=\sqrt{0+16}=4$ units

$CD=\sqrt{{\left(-3-5\right)}^{2}+{\left(3-3\right)}^{2}}=\sqrt{{\left(-8\right)}^{2}+0}=\sqrt{64}=8$ units

$DA=\sqrt{{\left(-3-\left(-3\right)\right)}^{2}+{\left(3-\left(-1\right)\right)}^{2}}=\sqrt{{\left(-3+3\right)}^{2}+{\left(3+1\right)}^{2}}=\sqrt{0+{\left(4\right)}^{2}}=\sqrt{16}=4$ units

$\therefore$ Perimeter of

$ABCD=8+4+8+4=24$ units

If two vertices of an equilateral triangle are $(3, 0)$ and $(6, 0)$ , find the third vertex.

$A(3,0)$ and

$B(6,0)$ , let third vertex

$C(a,b)$

$=>AB^{ 2 }={ BC }^{ 2 }={ CA }^{ 2 }\\ =>{ (3) }^{ 2 }+{ (0) }^{ 2 }={ (3-a) }^{ 2 }+{ b }^{ 2 }={ (6-a) }^{ 2 }+{ b }^{ 2 }\\ =>a^{ 2 }-6a+9+{ b }^{ 2 }=36+{ a }^{ 2 }-12a+{ b }^{ 2 }=9\\ =>a=\cfrac { 27 }{ 6 } \\ =>b=\left( \cfrac { 3 }{ 2 } \right) ^{ 2 }-(3)^{ 2 }\\ =>b=-\cfrac { 27 }{ 4 } \\$

So,

$C\left( \cfrac { 9 }{ 2 } ,-\cfrac { 27 }{ 4 } \right)$

Determine if the points $(1, 5) , (2, 3)$ and $(-2, -11)$ are collinear , by distance formula.

$A(1,5),B(2,3) and C(-2,-11)$

Now,

$AB=\sqrt { ({ 1 }^{ 2 })+(2^{ 2 }) } =\sqrt { 5 } \\ BC=\sqrt { ({ 4 }^{ 2 })+(14^{ 2 }) } =\sqrt { 212 } \\ CA=\sqrt { ({ 1 }^{ 2 })+(16^{ 2 }) } =\sqrt { 257 } \\$

As,

$CA>BC>AB$

If points

$A,BandC$ are collinear then

$AB+BC=CA$

But

$\sqrt { 5 } +\sqrt { 212 } \neq \sqrt { 257 }$

So they are not collinear.

If $D(3,-1) E(2,6) F(-5,7)$ are the mid points of the sides of triangle $ABC$ . find the area of $\Delta ABC$

Given:

$D=(3, 1)$

$E=(2, 6)$

$F=(-5, 7)$

It is known that if

$(A_x, A_y), (B_x, B_y)$ and

$(C_x, C_y)$ are the co-ordinates of vertices of a triangle then its area is given as

$\left[\dfrac{A_x(B_y-C_y)+B_x(C_y-A_y)+C_x(A_y-B_y)}{2}\right]$

Let

$A$ be the area of

$\Delta DEF$ ,

$A=\dfrac{1}{2}\left\{3(6-7)+2(7+1)+(-5)(-1-6)\right\}$

$=\dfrac{1}{2}\left\{3(-1)+2(8)+5(7)\right\}$

$=\dfrac{1}{2}(-3+16+35)$

$=24$

Area of

$\Delta$ ABC

$=4\times$ Area of

$\Delta DEF$

$=4\times 24$

$=96$ sq. units

Prove that the area of triangle with vertices $(t,t-2), (t+2, t+2), (t+3, t)$ is independent of t.

Area of Triangle with vertices

$A(t,t-2)$ ,

$B(t+2,t+2),C(t+3,t)$

$Area\, of\, \triangle ABC=\begin{vmatrix} t & t-2 \\ t+2 & t+2 \\ t+3 & t \\ t & t-2 \end{vmatrix}\\ =\cfrac { 1 }{ 2 } |{ t }^{ 2 }+2t-{ t }^{ 2 }+4+{ t }^{ 2 }+2t-{ t }^{ 2 }-5t-6+{ t }^{ 2 }+t-6-{ t }^{ 2 }|\\ =\cfrac { 1 }{ 2 } |-8|\\ =4$

(is independent of t)

Equation of the base of an equilateral triangle is $3x + 4y = 9$ and its Vertex is at the point $\left( {1,2} \right)$ . Find the length of each side of the triangle .

Length of the perpendicular from

$\left( x _ { 1 } , y _ { 1 } \right) \text { to the line } a x + b y + c = 0 \text { is } p = \left| \dfrac { a x _ { 1 } + b y _ { 1 } + c } { \sqrt { a ^ { 2 } + b ^ { 2 } } } \right|$

$\begin{array} { l } { \text { Let } a \text { be the length of each side of the equilateral triangle. } } \\ { \text { The length of the perpendicular p from } ( 1 , 2 ) \text { on } 3x +4 y - 9 = 0 } \\ { \Rightarrow p = \left| \dfrac { 3 +8 - 9 } { \sqrt { 3 ^ { 2 } + 4 ^ { 2 } } } \right| } \\ { = \dfrac { 2 } { { 5 } } } \end{array}$

$\begin{array} { l } { \sin 60 ^ { \circ } = \dfrac { p } { a } } \\ { \Rightarrow \dfrac { \sqrt { 3 } } { 2 } = \dfrac { \dfrac { 2 } { { 5 } } } { a } } \\ { \Rightarrow a = \dfrac { 2 } { \sqrt { 3 } } \times \dfrac { 2 } { { 5 } } } \\ { a = \dfrac { {4 } } { 5 \sqrt 3 } } \end{array}$

1043810_1122456_ans_4795ae7786e24a228a0ce93a26278c5d.png

Prove that the points $(a, b), (a_{1}, b_{1}$ and $(a - a_{1}, b - b_{1})$ are collinear if $ab_{1} = a_{1}b$ .

$A(a,b),B({ a }_{ 1 },{ b }_{ 1 })$ and

$C(a-{ a }_{ 1 }b-,{ b }_{ 1 })$

$=>\cfrac { 1 }{ 2 } \begin{vmatrix} a & b \\ { a }_{ 1 } & { b }_{ 1 } \\ a-{ a }_{ 1 } & b-{ b }_{ 1 } \\ a & b \end{vmatrix}=\cfrac { 1 }{ 2 } \left| a{ b }_{ 1 }-b{ a }_{ 1 }+{ a }_{ 1 }b-{ a }_{ 1 }{ b }_{ 1 }-{ b }_{ 1 }a+{ b }_{ 1 }{ a }_{ 1 }+ba-{ a }_{ 1 }b-ab+a{ b }_{ 1 } \right| \\ =\cfrac { 1 }{ 2 } \left| { a }_{ 1 }b-a{ b }_{ 1 } \right| \\ =0\quad (As\quad given\quad a{ b }_{ 1 }={ a }_{ 1 }b)$

The area of the triangle formed by points $\left( 0,0 \right) \quad \left( { 3 }{ \quad ,\frac { \pi }{ 2 } } \right) \quad and\quad \left( { 3 }\quad ,{ \frac { \pi }{ 6 } } \right)$ is

Area of a triangle

$=\Delta=\dfrac{1}{2}\left| \begin{matrix} {x}_{1} & {y}_{1} & 1 \\ {x}_{2} & {y}_{2} & 1 \\{x}_{3} & {y}_{3} & 1 \end{matrix} \right|$

$=\dfrac{1}{2}\left| \begin{matrix} 0 & 0 & 1 \\ 3 & \dfrac{\pi}{2} & 1 \\3 & \dfrac{\pi}{6} & 1 \end{matrix} \right|$

$=\dfrac{1}{2}\left[0-0+1\left(\dfrac{3\pi}{6}-\dfrac{3\pi}{2}\right)\right]$

$=\dfrac{1}{2}\left(\dfrac{3\pi}{6}-\dfrac{9\pi}{6}\right)$

$=\dfrac{1}{2}\times -\dfrac{6\pi}{6}$

$=\dfrac{-\pi}{2}$

But area is always positive.

Hence area of the triangle

$=\dfrac{\pi}{2}$ sq.units

Find the area of whose vertices are $A(1,1,2)$ $B(2,3,5)$ and . $C(1,5,5)$

$=\cfrac { 1 }{ 2 } \left| \begin{matrix} 1 & 1 & 2 \\ 2 & 3 & 5 \\ 1 & 5 & 5 \\ 1 & 1 & 2 \end{matrix} \right| =\cfrac { |1(-5)-1(-8)+2(3)| }{ 2 } =\cfrac { 9 }{ 2 }$

Find the values of $k$ , if the area of triangle is $35\ sq. units$ whose vertices are $(2, -6), (5, 4)$ and $(k, 4)$ .

Area of triangle is

$\triangle =\dfrac{1}{2}\left| \begin{matrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{matrix} \right|=35 \left(given\right)$

$\triangle = \left(20-4k\right)- \left(8+6k\right)+ \left(8+30\right)=70 \left[opening\ from\ {C}_{3}\right]$

$50-10k=70$

$k=-2$
taking

$+ve$ sign

$\Rightarrow -10k+50=70$

$\Rightarrow -10k=70-50$

$\Rightarrow -10k=20$

$\Rightarrow k=-2$
taking

$-ve$ sign

$\Rightarrow -10k+50=-70$

$\Rightarrow -10k=-70-50$

$\Rightarrow -10k=-120$

$\Rightarrow k=12$

$\Rightarrow k=-2,12$

Find the equation of locus of $P$ , if the ratio of the distance from $P$ to $(5, -4)$ and $(7, 6)$ is $2 : 3$ .

Find the equation of locus of

$P$ , if the ratio of the distance from

$P\left(5,-4\right)$ and

$\left(7,6\right)$ is

$2:3$

Let

$P$ be

$\left(x,y\right)$

Distance from

$P$ to

$\left(5,4\right)=\sqrt{{\left(x-5\right)}^{2}+{\left(y+4\right)}^{2}}$

Distance from

$P$ to

$\left(7,6\right)=\sqrt{{\left(x-7\right)}^{2}+{\left(y-6\right)}^{2}}$

Given ratio of distance

$=2:3$

$\Rightarrow \dfrac{\sqrt{{\left(x-5\right)}^{2}+{\left(y+4\right)}^{2}}}{\sqrt{{\left(x-7\right)}^{2}+{\left(y-6\right)}^{2}}}=\dfrac{2}{3}$

Squaring on both sides

$\Rightarrow \dfrac{{x}^{2}+25-10x+{y}^{2}+16+8y}{{x}^{2}+49-14x+{y}^{2}+36-12y}=\dfrac{4}{9}$

$\Rightarrow 9{x}^{2}+225-90x+9{y}^{2}+144+72y=4{x}^{2}+196-56x+4{y}^{2}+144-48y$

$\Rightarrow 5{x}^{2}+5{y}^{2}-.34x+120y+31=0$

If $(x, y)$ be on the line joining the two points $(1, -3)$ and $(-4, 2)$ , prove that $x + y + 2 = 0$ .

$A(x,y)$ lies on line PB where

$P(1,-3)$ and

$B(-4,2)$

IF P,B,A are on same line ->they are collinear

$=>\cfrac { 1 }{ 2 } \begin{vmatrix} x & y \\ 1 & -3 \\ -4 & 2 \\ x & y \end{vmatrix}=0\\ =>-3x-y-10-4y-2x=0\\ =>x+y+2=0$

If $A\left( 5,\ 2 \right), B\left( 3,\ 4 \right), C\left( x,\ y \right)$ are collinear and $AB=BC$ , the find $\left( x,\ y \right)$ .

1356718_1128702_ans_1d60a5a6a5974597bb70fd246efd5a49.jpg

If the points $(4,-4),(-4,4)$ and $(x,y)$ from an equilateral triangle, find $x$ and $y$ .

$\therefore$ for ABC to be equilateral

$\Delta$

$AB=BC=CA$

$\therefore$ By distance formula

$AB=\sqrt{(4-(-4))^2+(-4-4)^2}=8\sqrt2\\BC=\sqrt{(x+4)^2+(y-4)^2}\\CA=\sqrt{(x-4)^2+(y+4)^2}\\ \therefore AB^2=BC^2=CA^2\\(x+y)^2+(y+4)^2=128\rightarrow(1)\&(x-4)^2+(y+4)^2=128\rightarrow(2)$

Solving the equations

$(1)\&(2)$

$x=y=4\sqrt3$

1132847_1125735_ans_3d455c86419245da8e594b7d11f8192d.png

If the points $(1,1)(-1,-1)$ and $(-\sqrt {3},k)$ are vertices of an equilateral triangle, Find the value of $k$

$A(1,1),B(-1,-1)$ and

$C(\sqrt { 3 } ,k)$ are vertices of equilateral triangle

AB=BC=CA=>

$(AB)^{ 2 }={ (AC })^{ 2 }$

$=>2\left( 1+1 \right) ^{ 2 }=(1+\sqrt { 3 } )^{ 2 }+(1-k)^{ 2 }\\ =>3=-2k+{ k }^{ 2 }\\ =>{ k }^{ 2 }-2k-3=0\\ k=3,-1$

Find the value of $k$ , if the points $A(7, -2), B(5, 1)$ and $C(3, 2k)$ are collinear.

If these are collinear then the area formed by them is zero

$\implies \cfrac { 1 }{ 2 } \begin{vmatrix} 7 & -2 &1\\ 5 & 1 &1 \\ 3 & 2k &1\\\end{vmatrix}=0\\ \implies 17+10k-14k-9=0\\ \implies k=2$

Find the relation between $x$ and $y$ if the point $P(x,y)$ is equidistance from the point $A(7,0)$ and $B (0,5)$

$P(x,y)$ is equidistant from

$A(7,0) and B(0,5)$

$=>AP=BP\\ =>(7-x)^{ 2 }+{ y }^{ 2 }={ x }^{ 2 }+(5-y)^{ 2 }\\ =>49-14x=25-10y\\ =>5y-7x+12=0$

In PAB,PA=PB and area of PAB= sq. units. Find the coordinates of P if coordinates of A and B are (1,2) and (3,8) respectively.

We have,

In $\Delta PAB,\,\,PA=PB$

Let, $P\left( a,b \right)$ be the point.

Suppose that

$A\left( {{x}_{1}},{{y}_{1}} \right)=\left( 1,2 \right)$

$P\left( {{x}_{2}},{{y}_{2}} \right)=\left( a,b \right)$

$C\left( {{x}_{3}},{{y}_{3}} \right)=\left( 3,8 \right)$

\end{align}$$

Then,

Given that,

$PA=PB$

$\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}=\sqrt{{{\left( {{x}_{2}}-{{x}_{3}} \right)}^{2}}-{{\left( {{y}_{2}}-{{y}_{3}} \right)}^{2}}}$

$\Rightarrow \sqrt{{{\left( a-1 \right)}^{2}}+{{\left( b-2 \right)}^{2}}}=\sqrt{{{\left( a-3 \right)}^{2}}-{{\left( b-8 \right)}^{2}}}$

Squaring both side and we get,

${{\left( a-1 \right)}^{2}}+{{\left( b-2 \right)}^{2}}={{\left( a-3 \right)}^{2}}+{{\left( b-8 \right)}^{2}}$

$\Rightarrow {{a}^{2}}+1-2a+{{b}^{2}}+4-4b={{a}^{2}}+9-6a+{{b}^{2}}+64-16b$

$\Rightarrow -2a-4b+5=-6a-16b+65$

$\Rightarrow -2a-4b+5+6a+16b-65=0$

$\Rightarrow 4a+12b-60=0$

$\Rightarrow a+3b-15=0\,\,......\,\left( 1 \right)$

But again given that,

$\text{Area}\,\text{of }\Delta \text{PAB=}\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$

$10=\dfrac{1}{2}\left[ 1\left( b-8 \right)+a\left( 8-2 \right)+3\left( 2-b \right) \right]$

$20=b-8+6a+6-3b$

$20+2=6a-2b$

$6a-2b=22$

$3a-b=11$

$3a-b-11=0\,\,......\,\left( 2 \right)$

By equation (1) and (2) to and we get,

$\left( a+3b=15 \right)\times 3$

$3a-b=11$

$3a+9b=45$

$3a-b=11$

On subtracting and we get,

$10b=34$

$b=\dfrac{34}{10}$

$b=\dfrac{17}{5}$

Put the value of $b$ in equation (1) and we get,

$a+3b-15=0$

$\Rightarrow a+3\times \dfrac{17}{5}-15=0$

$\Rightarrow a+\dfrac{51}{5}-15=0$

$\Rightarrow a+\dfrac{51-75}{5}=0$

$\Rightarrow a-\dfrac{24}{5}=0$

$\Rightarrow a=\dfrac{24}{5}$

The value of $P\left( a,b \right)=\left( \dfrac{24}{5},\dfrac{17}{5} \right)$ .

Hence, this is the answer.

If the point $(x,y)$ be equidistance from the point $(a+b,b-a)$ and $(a-b,a+b)$ , prove that $bx = ay$ .

$(x,y)$ is equidistant from

$A(a+b,b-a)$ &

$B(a-b,a+b)$

$x,y$ cuts AB in equal parts.

$(x,y)=\left( \cfrac { a+b+a-b }{ 2 } ,\cfrac { b-a+a+b }{ 2 } \right) \\ (x,y)=(a,b)\\ =>x=a\quad \quad \& \quad b=y$

so,

$xb=ay$ (multiplying them both)

$A(3,-4),B(5,-2),C(-1,8)$ are the vertices of $\triangle ABC.D,E,F$ are the mid-points of sides $\overline {BC},\overline {CA}$ and $\overline {AB}$ respectively. Find area of $\triangle ABC$ . Using coordinates of $D,E,F$ , find area of $\triangle DEF$ . Hence show that the $ABC=4(DEF)$

$D=\left(\dfrac{5-1}{2}\right), \left(\dfrac{-2+8}{2}\right)$

$\Rightarrow (2, 3)$

$E=\left(\dfrac{-1+3}{2}, \dfrac{8-4}{2}\right)$

$\Rightarrow (1,2)$

$F=\left(\dfrac{3+5}{2}, \dfrac{-4-2}{2}\right)$

$\Rightarrow (4, -3)$

$\triangle ABC\Rightarrow a=\sqrt{(-1-5)^{2}+(P-(-2))^{2}}$

$=\sqrt{6^{2}+10^{2}}$

$=\sqrt{136}=2\sqrt{34}$

$b=\sqrt{(-1-3)^{2}+(8-(-4))^{2}}$

$=\sqrt{4^{2}+12^{2}}$

$=\sqrt{160}=4\sqrt{10}$

$c=\sqrt{(5-3)^{2}+(-2+4)^{2}}$

$=\sqrt{2^{2}+2^{2}}$

$\sqrt{8}=2\sqrt{2}$

$s =\dfrac{a+b+c}{2}, A=\sqrt{s(s-a)(s-b)(s-c)}$

$A=(\sqrt{34}-2\sqrt{10}+\sqrt{2})(\sqrt{34}+2\sqrt{10}-\sqrt{2})$

$\Rightarrow 16\ sq=ar(\triangle ABC)$

$\triangle DEF, r=\sqrt{(4-1)^{2}+(-3-2)^{2}}$

$\sqrt{34}=\sqrt{34}$

$c=\sqrt{(4-2)^{2}+(-3-3)^{2}}$

$=\sqrt{40}=2\sqrt{10}$

$f=\sqrt{(1-2)^{2}+(2-3)^{2}}$

$=\sqrt{1^{2}+1^{2}}$

$=\sqrt{2}$

$=\dfrac{d+e+f}{2}, A=\sqrt{s(s-d)(s-e)(s-f)}$

$=\sqrt{\left(\dfrac{\sqrt{34}+2\sqrt{10}+\sqrt{2}}{2}\right)\left(\dfrac{\sqrt{34}+2\sqrt{10}+\sqrt{2}}{2}-\sqrt{34}\right)}\times \left(\dfrac{\sqrt{34}+2\sqrt{10}+\sqrt{2}}{2}-2\sqrt{10}\right)\times \left(\dfrac{\sqrt{34}+2\sqrt{10}+\sqrt{2}}{2}-\sqrt{2}\right)$

$A'=4\ sq\ unit=ar(\triangle DEF)$

$\Rightarrow ar(\triangle ABC)=16\ sq\ unit$

$=4\times 4$

$=4\times ar(\triangle DEF)$

Hence proved

Find area of the triangle with vertices at the point given in each of the following:
$\left (1,0\right), \left (6,0\right), \left (4,3\right)$
$\left (2,7\right), \left (1,1\right), \left (10,8\right)$
$\left (-2,-3\right), \left (3,2\right), \left (-1,-8\right)$

(i) Area

$=\cfrac{1}{2}\begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix}$

$=\cfrac{1}{2}\left| -3(1-6) \right|$

$=\cfrac{15}{2}$ sq. unit

(ii) Area

$=\cfrac{1}{2}\begin{vmatrix} 2 & 7 & 1 \\ 1 & 1 & 1 \\ 10 & 8 & 1 \end{vmatrix}$

$=\cfrac{1}{2}[2(1-8)-7(1-10)+1(8-10)]$

$=\cfrac { 1 }{ 2 } \left| -14+63-2 \right|$

$=\cfrac{1}{2}(47)=\cfrac{47}{2}$ sq. units

(iii) Area

$=\cfrac { 1 }{ 2 } \begin{vmatrix} -2 & -3 & 1 \\ 3 & 2 & 1 \\ -1 & -8 & 1 \end{vmatrix}$

$=\cfrac { 1 }{ 2 } [-2(2+8)+3(3+1)+1(-24+2)]$

$=\cfrac { 1 }{ 2 } (-20+12-22)=\cfrac { 1 }{ 2 } \left| -30 \right| =15$ sq.units

Find a point which is equidistant from the points $A\left( { - 5,4} \right)$ and $B\left( { - 1,6} \right)$ .How many such points are there?

By distance formula

${ (x+5) }^{ 2 }+{ (y-4) }^{ 2 }={ (x+1) }^{ 2 }+{ (y-6) }^{ 2 }\\ { x }^{ 2 }+{ y }^{ 2 }+{ 10x }+{ (-8y) }+{ 25 }+{ 16 }={ x }^{ 2 }+{ y }^{ 2 }+{ 2x }+{ (-12y) }+{ 36 }+{ 1 }\\ { 8x }+{ 4y }+{ 4 }={ 0 }\\ { 2x }+{ y }+{ 1 }={ 0 }\\ ({ 1 },{ -3 })$

is the one of them

There will be infinite points

$\triangle {ABC}$ lies in the plane with $A=(0,0), B(0,1)$ and $C(1,0)$ . Points $M$ and $N$ are chosen on $AB$ and $AC$ , respectively, such that $MN$ is parallel to $BC$ and $MN$ divides the area of $\triangle {ABC}$ in half. Find the coordinates of $M$ .

From the figure, we can say that,

$\Delta ABC$ is a right angled triangle at

$A$ with

$AB=AC=1$

$\Delta AMN$ is a right-angled triangle with

$AM=AN=a$ {

$\because MN\parallel BC$ }

Area of

$\Delta MAN=\dfrac{1}{2}$ (area of

$\Delta BAC$ )

$\dfrac{1}{2}\times a\times a=\dfrac{1}{2}\times \dfrac{1}{2}\times 1\times 1$

$a^{2}=\dfrac{1}{2} \Rightarrow a=\dfrac{1}{\sqrt{2}}$

$M\equiv \left (0,\dfrac{1}{\sqrt{2}}\right)$

Shown that the point $(5,-1,1),(7,-4,7),(1,-6,10)$ and $(-1,-3,4)$ are the vertices of a rhombus.

For a quadrant to be rhombus, four side of quadrant be equal.

By distance formula:

${ S }_{ 1 }=\sqrt { { \left( 7-5 \right) }^{ 2 }+{ \left( -4+1 \right) }^{ 2 }{ +\left( 7-1 \right) }^{ 2 } } =7\\ { S }_{ 2 }=\sqrt { { \left( 1-7 \right) }^{ 2 }{ +\left( -6+4 \right) }^{ 2 }{ +\left( 10-7 \right) }^{ 2 } } =7\\ { S }_{ 3 }=\sqrt { { \left( -1-1 \right) }^{ 2 }{ +\left( -6+4 \right) }^{ 2 }{ +\left( 10-7 \right) }^{ 2 } } =7\\ { S }_{ 4 }=\sqrt { { \left( -1-5 \right) }^{ 2 }{ +\left( -3+1 \right) }^{ 2 }{ +\left( 4-1 \right) }^{ 2 } } =7\\ \because { S }_{ 1 }{ =S }_{ 2 }={ S }_{ 3 }={ S }_{ 4 }\\$

Hence the given quadrant is rhombus

Find the area of the triangle formed by joining the mid point of the triangle whose vertices are (0,-1),(2,1)and (0,3). Find the ratio of area of the triangle formed to the area of the given triangle.

Vertices of

$\triangle ABC$ ,

$A(0,-1),B(2,1)$ and

$C(0,3)$

Area of

$\triangle ABC$

$=\cfrac { 1 }{ 2 } \begin{vmatrix} 0 & -1 \\ 2 & 1 \\ 0 & 3 \\ 0 & -1 \end{vmatrix}=\cfrac { 1 }{ 2 } \left| 8 \right| =4$

Mid-points AB,BC,CA be O,E,F respectively.

$D(1,0),E(1,2),F(0,1)$

Area of

$\triangle DEF=\cfrac { 1 }{ 2 } \begin{vmatrix} 1 & 0 \\ 1 & 2 \\ 0 & 1 \\ 1 & 0 \end{vmatrix}=1$

$\cfrac { Area\quad of\quad \triangle DEF }{ Area\quad of\quad \triangle ABC } =\cfrac { 1 }{ 4 }$

Find the equation of the set of all points whose distances from $(0, 4)$ are $\dfrac{2}{3}$ of their distances from the line $y=9$ .

given that the distance of the set of points from

$(0, 4)$ is

$\dfrac{2}{3}$ of the distance from the line

$y=9$

Hence

$\sqrt{(x-0)^2+(y-4)^2}=\dfrac{2}{3}\sqrt{(x-0)^2+(y-9)^2}$

Squaring on both sides we get

$x^2+(y-4)^2=\dfrac{4}{9}[(x^2)+(y-9)^2]$

$9(x^2+y^2-8y+16)=4x^2+4(y^2-18y+81)$

On simplifying we get

$5x^2+5y^2+180=0$

This is the required equation of the set of points.

The vertices of triangle ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively ,such that $\dfrac{AD}{AB}=\dfrac{AE}{AC}=\dfrac{1}{4}$ . Calculate the area of the $\triangle ADE$ and compare it with the area $\triangle ABC$ ?

Given,

$\dfrac { AD }{ AB } =\dfrac { AE }{ AC } =\dfrac { 1 }{ 4 }$

$Thus,\quad \dfrac { AD }{ DB } =\dfrac { AE }{ EC } =\dfrac { 1 }{ 3 }$

For D,

$x=\dfrac { m{ x }_{ 2 }+n{ x }_{ 1 } }{ m+n } =\dfrac { 1.1+3.4 }{ 1+3 } =\dfrac { 13 }{ 4 }$

$y=\dfrac { m{ y }_{ 2 }+n{ y }_{ 1 } }{ m+n } =\dfrac { 1.5+3.6 }{ 1+3 } =\dfrac { 23 }{ 4 }$

$\therefore D=\left( \dfrac { 13 }{ 4 } ,\dfrac { 23 }{ 4 } \right)$

for E,

$x=\dfrac { 1.7+3.4 }{ 1+3 } =\dfrac { 19 }{ 4 }$

$y=\dfrac { 1.2+3.6 }{ 1+3 } =5$

$\therefore E=\left( \dfrac { 19 }{ 4 } ,5 \right)$

$ar(\Delta ABC)=\dfrac { 1 }{ 2 } \left[ 4(5-2)+1(2-6)+7(6-5) \right] =\dfrac { 15 }{ 2 }$

$ar(\Delta ADE)=\dfrac { 1 }{ 2 } \left[ 4\left( \dfrac { 23 }{ 4 } -5 \right) +\dfrac { 13 }{ 4 } (5-6)+\dfrac { 19 }{ 4 } \left( 6-\dfrac { 23 }{ 4 } \right) \right] =\dfrac { 9 }{ 4 } \\ \therefore \dfrac { ar(\Delta ADE) }{ ar(\Delta ABC) } =\dfrac { \dfrac { 9 }{ 4 } }{ \dfrac { 15 }{ 2 } } =\dfrac { 3 }{ 10 }$

1171938_1187275_ans_876ea05b94844a8b8e8eed126629e5f6.png

Given the vertices $A\left( {10,4} \right),B\left( { - 4,9} \right),C\left( { - 2, - 1} \right)$ of $\Delta$ ABC find the area of the triangle:

1352209_1183202_ans_ffd97b1eb767496b8bd52f6419a89e8d.jpg

Solve it:-
Through the point $(3,4)$ are drawn two straight lines each inclined at ${45^ \circ }$ to the straight line $x-y=2$ find their equations and also find the area of triangle bounded by the three lines.

Foundation chapter and LO

Given : Line

$x-y=2$

Slope =

$\tan 45=1$

Given that two lines are including with

$x-y=2$ , let M, and

$M_2$ be the slopes of the lines that makes an angle of

$45^o$ with

$x-y=2.$

It can clearly be observed that are line is parallel to x-axis and another is parallel to y-axis so, the lines are.

$\boxed{x=3}$

$\boxed{y=4}$

So the equations of line are

$x=3$ and

$y=3$ and

$y=4$ Area of the triangle formed by the there lines is

$\Rightarrow\begin{vmatrix}3 &4 &1 \\ 3 & 1 &1 \\ 6 &4 &1 \end{vmatrix}$

$\Rightarrow 3(1-4)-4(3-6)+1(12-6)$

$\Rightarrow 9$ rq.units

2128724_1176886_ans_61004c9a12c34321a85697b2b0f20195.png

Find the distance of the point $P ( - 6,8 )$ from the origin.

${\textbf{Step -1: Using distance formula find distance of a point from origin.}}$

${\text{We are given, a point }}P\left( { - 6,8} \right)$ ${\text{and a origin be }}Q\left( {0,0} \right).$

${\text{Distance }}PQ = \sqrt {{{\left( {0 - \left( 6 \right)} \right)}^2} + {{\left( {0 - 8} \right)}^2}} \ldots \left( 1 \right)$ $\left(\because\mathbf {{\text{}}d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} } \right)$

${\textbf{Step -2: Further simplify equation }}\left( \mathbf1 \right)$

$\therefore {\text{Distance }}PQ = \sqrt {{{\left( 6 \right)}^2} + {{\left( { - 8} \right)}^2}}$

$\therefore {\text{Distance }}PQ = \sqrt {36 + 64}$

$\therefore {\text{Distance }}PQ = \sqrt {100}$

$\therefore {\text{Distance }}PQ = 10$

${\textbf{Hence, distance of the point from the origin is 10.}}$

Find the area of the triangle formed by joining the mid points of the sides of the triangle
whose vertices are $( 0 , - 1 ) , ( 2,1 )$ and $( 0,3 )$ .

Let

$\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right)$ and

$\left({x}_{3},{y}_{3}\right)$ be the mid-points of

$\triangle{ABC}$

$\Rightarrow\,{x}_{1}=\dfrac{0+2}{2}=1,\,{y}_{1}=\dfrac{-1+1}{2}=0$

$\Rightarrow\,{x}_{2}=\dfrac{2+0}{2}=1,\,{y}_{2}=\dfrac{1+3}{2}=2$

$\Rightarrow\,{x}_{3}=\dfrac{0+0}{2}=0,\,{y}_{3}=\dfrac{3-1}{2}=1$

$\therefore\,$ Mid-points are

$\left(1,0\right),\left(1,2\right)$ and

$\left(0,1\right)$

Area of the triangle joining the midpoints

$=\dfrac{1}{2}\left| \begin{matrix} 1 & 0 & 1 \\ 1 & 2 & 1 \\0 & 1 & 1 \end{matrix} \right|$

$=\dfrac{1}{2}\left[1\left(2-1\right)-0+1\left(1-0\right)\right]=\dfrac{1}{2}\left[1-0+1\right]=\dfrac{2}{2}=1\,sq.unit$

1511728_1179174_ans_4dd0dfdbb23c494bab113151d013e446.PNG

If $\mathrm { A } ( 4,3 ) , \mathrm { B } ( 6 , - 2 )$ and $\mathrm { C } ( a , - 3 )$ are the vertices of a triangle right angled at $\mathrm { A } ,$ find $a$ .

$\\\bigtriangleup\>ABC\>is\>a\>right\>angled\>triangle\>hence\\AB^2+AC^2=BC^2\\(4-6)^2+[3-(-2)]^2+(4-a)^2+[3-(-3)]^2\\=(6-a)^2+[-2-(-3)]^2\\=4+25+(4-a)^2+36=(6-a)^2+1\\(6-a)^2-(4-a)^2=64\\(6-a-4+a)(6-a+4-a)=64\\2(10-2a)=64\\10-2a=32\\2a=-22\\a=-11$

Find the angles between the lines
$x-\sqrt{3}y=-1$

Consider the given equations,

$x-\sqrt{3}y=-1$ ….(1) and $\sqrt{3}x-y=-1$ ….(2)

From equation 1^st,

$y=\dfrac{1}{\sqrt{3}}x+\dfrac{1}{\sqrt{3}}$

Compare with, $y={{m}_{1}}x+C$

${{m}_{1}}=\dfrac{1}{\sqrt{3}}$ $$

From equation 2^nd ,

$y=\sqrt{3}x+1$

Compare with $y={{m}_{2}}x+C$

${{m}_{2}}=\sqrt{3}$

Now,let the angle between line 1^st and 2^{nd $=\theta$},

Using the formula, $\tan \theta =\dfrac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}}$

$\tan \theta =\dfrac{\sqrt{3}-\dfrac{1}{\sqrt{3}}}{1+\sqrt{3}\times \dfrac{1}{\sqrt{3}}}$

$\tan \theta =\dfrac{\sqrt{3}\times \sqrt{3}-1}{1+1}$

$\tan \theta =\dfrac{3-1}{1+1}$

$\tan \theta =1$

$\tan \theta =\tan \dfrac{\pi }{4}$

$\theta =\dfrac{\pi }{4}$

Hence, this is the answer.

Find the angles between the lines
$x+\sqrt{3}y=1$

Consider the given equations,

$x+\sqrt{3}y=1$ ..(1), $\sqrt{3}x+y=1$ ......(2)

from the equation 1^st,

$y=-\dfrac{1}{\sqrt{3}}x+\dfrac{1}{\sqrt{3}}$

Compare with, $y={{m}_{1}}x+C$

${{m}_{1}}=-\dfrac{1}{\sqrt{3}}$

From equation 2^nd ,

$y=-\sqrt{3}x+1$

Compare with $y={{m}_{2}}x+C$

${{m}_{2}}=-\sqrt{3}$

Now,let the angle between line 1^st and 2^{nd $=\theta$},

Using the formula, $\tan \theta =\dfrac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}}$

$\tan \theta =\dfrac{-\sqrt{3}-\left( -\dfrac{1}{\sqrt{3}} \right)}{1+\left( -\sqrt{3} \right)\times \left( -\dfrac{1}{\sqrt{3}} \right)}=\dfrac{-\sqrt{3}\times \sqrt{3}+1}{1+1}$

$\tan \theta =\dfrac{-3+1}{2}=-1$

$\tan \theta =-1$

$\tan \theta =-\tan \dfrac{\pi }{4}$

$\tan \theta =\tan \left( \pi -\dfrac{\pi }{4} \right)$

$\tan \theta =\tan \dfrac{3\pi }{4}$

$\theta =\dfrac{3\pi }{4}$

Hence, this is the answer.

Find the angles between the lines
$x+\sqrt{3}y=5$

Consider the given equations,

$x+\sqrt{3}y=5$ ..(1), $\sqrt{3}x+y=5$ ….(2)

from the equation 1^st,

$y=-\dfrac{1}{\sqrt{3}}x+\dfrac{5}{\sqrt{3}}$

Compare with, $y={{m}_{1}}x+C$

${{m}_{1}}=-\dfrac{1}{\sqrt{3}}$

From equation 2^nd ,

$y=-\sqrt{3}x+5$

Compare with $y={{m}_{2}}x+C$

${{m}_{2}}=-\sqrt{3}$

Now,let the angle between line 1^st and 2^{nd $=\theta$},

Using the formula, $\tan \theta =\dfrac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}}$

$\tan \theta =\dfrac{-\sqrt{3}-\left( -\dfrac{1}{\sqrt{3}} \right)}{1+\left( -\sqrt{3} \right)\times \left( -\dfrac{1}{\sqrt{3}} \right)}=\dfrac{-\sqrt{3}\times \sqrt{3}+1}{1+1}$

$\tan \theta =\dfrac{-3+1}{2}=-1$

$\tan \theta =-1$

$\tan \theta =-\tan \dfrac{\pi }{4}$

$\tan \theta =\tan \left( \pi -\dfrac{\pi }{4} \right)$

$\tan \theta =\tan \dfrac{3\pi }{4}$

$\theta =\dfrac{3\pi }{4}$

Hence, this is the answer.

Find the area of the triangle formed by joining the mid-point of the sides of the triangle whose vertices are $(0,-1),(2,1)$ and $(0,3)$ . Find the ratio of this area to the area of the given triangle.

since

$D,E$ and

$F$ are the midpoints of

$AB$ ,

$BC$ and

$CA$ of triangle respectively

coordinate of

$D$

$=\left( \dfrac { 0+2 }{ 2 } ,\dfrac { -1+1 }{ 2 } \right) =\left( 1,0 \right)$

coordinate of

$E$

$=\left( \dfrac { 2+0 }{ 2 } ,\dfrac { 1+3 }{ 2 } \right) =\left( 1,2 \right)$

coordinate of

$F$

$=\left( \dfrac { 0+0 }{ 2 } ,\dfrac { -1+3 }{ 2 } \right) =\left( 0,1 \right)$

$\therefore ar\left( \triangle DEF \right) =\dfrac { 1 }{ 2 } \left[ 1(2-1)+1(1-0)+0(0-2) \right] =1sq.unit\\ and\quad ar\left( \triangle ABC \right) =\dfrac { 1 }{ 2 } \left[ 0(1-3)+2(3+1)+0(-1-1) \right] =4sq.unit\\ \dfrac { ar\left( \triangle DEF \right) }{ ar\left( \triangle ABC \right) } =\dfrac { 1 }{ 4 } \\ \Rightarrow ar\left( \triangle DEF \right) :ar\left( \triangle ABC \right) =1:4$

1178589_1198109_ans_7e922eb138e449709a5eda4a5cfbd6da.png

Find the linear equation between $x$ and $y$ such that $P(x,y)$ is equidistant from the points $A(1,4)$ and $B(-1,2)$ .

Let the points be

$P(x,y), A(1,4), B(-1,2)$

Point

$P$ is equidistant from

$A \& B$

Now, using the distance formula

$x_1=x,\quad y_1=y$

$x_2=1,\quad y_2=4$

$PA=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$=\sqrt{(1-x)^2+(4-y)^2}$

Similarly for

$PB$

$x_1=x,\quad y_1=y$

$x_2=-1,\quad y_2=2$

$PA=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$=\sqrt{(-1-x)^2+(2-y)^2}$

we know that

$PA=PB$

$\sqrt{(1-x)^2+(4-y)^2}=\sqrt{(-1-x)^2+(2-y)^2}$

Squaring both sides

$\Rightarrow (1-x)^2+(4-y)^2=(-1-x)^2+(2-y)^2$

$\Rightarrow 1+x^2-2x+16+y^2-8y=1+x^2+2x+4+y^2-4y$

$\Rightarrow -2x-2x-8y+4y+16-4=0$

$\Rightarrow -4x-4y+12=0$

$\Rightarrow x+y-3=0$

Find the area of the triangle whose vertices are (-5,-1)(3,-5)(5,2)

Given three vertices of a triangle

$(-5,-1)(3,-5)(5,2)$ the area is given by the formula

$=\left| \dfrac { 1 }{ 2 } \left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right] \right|$

$=\left| \dfrac { 1 }{ 2 } \left[ -5\left( -5-2 \right) +3\left( 2+1 \right) +5\left( -1+5 \right) \right] \right|$

$=\left| \dfrac { 1 }{ 2 } \left[ +35+9+20 \right] \right|$

$=\left| \dfrac { 1 }{ 2 } \left[ 29+35 \right] \right|$

$=\dfrac { 64 }{ 2 }$

$=32$ sq units.

Find a relation between $a$ and $b$ if the points $A(-2,1), B(a,b)$ and $C(4,-1)$ are collinear

Given

$A(-2,1),B(a,b),c(4,-1)$ are collinear

Slope of

$AB=$ slope of

$BC$

$\dfrac {b-1}{a+2}=\dfrac {-1-b}{4-a}$

$\Rightarrow \quad \quad (b-1)(4-a)=-(b+1)(a+2)$

$\Rightarrow \quad \quad 4b-ab-4+a=-ab-2b-a-2$

$\therefore \quad \quad 2a+6b-2=0$

$\therefore \quad \quad a+3b=1$

Find the area of the triangle whose vertices are $( - 5, - 1) ,(3, - 5)$ and $(5,2)$

Let the given points be

$A ( - 5 , - 1 ) , B ( 3 , - 5 )$ and

$C ( 5,2 ) .$

Here, we have

$x _ { 1 } = - 5 , y _ { 1 } = - 1$

$x _ { 2 } = 3 , y _ { 2 } = - 5$

and

$x _ { 3 } = 5 , y _ { 3 } = 2$

Now, Area of

$\Delta A B C$

$= \dfrac { 1 } { 2 } \left[ x _ { 1 } \left( y _ { 2 } - y _ { 3 } \right) + x _ { 2 } \left( y _ { 3 } - y _ { 1 } \right) + x _ { 3 } \left( v _ { 1 } - y _ { 2 } \right) \right]$

$= \dfrac { 1 } { 2 } [ ( - 5 ) \{ - 5 - 2 \} + ( 3 ) \{ 2 - ( - 1 ) \} + ( 5 ) \{ ( - 1 ) - ( - 5 ) \} ]$

$= \dfrac { 1 } { 2 } [ 35 + 9 + 20 ] = 32 \ sq$ units

If $A = \left( {1, - 2, - 1} \right),\,B = \left( {4,0, - 3} \right),\,C = \left( {1,2, - 1} \right)$ and $D = \left( {2, - 4, - 5} \right)$ , find the distance between $AB$ and $CD$ .

$A(1,-2,-1) B(4,0,-3)$

$C(1,2,-1) D(2,-4,-5)$

distance

$AB=\sqrt{(1-4)^{2}+(-2-0)^{2}+(-1+3)^{2}}$

$=\sqrt{9+4+4}= \sqrt{17}$

$CD=\sqrt{(1-2)^{2}+(2+4)^{2}+(-1+5)^{2}}$

$=\sqrt{1+36+16}=\sqrt{93}$

If the distance between the points $A(4, k)$ and $B(1, 0)$ is $5 units$ , then what can be the possible values of $k$ ?

Using distance formula

$AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

From the question

$5=\sqrt{(1-4)^2+(0-k)^2}$

$\Rightarrow 25=9+k^2$ on squaring

$\Rightarrow k^2=25-9=16$

$\therefore k=\pm 4$

$\therefore k$ takes

$2$ values

$\{-4, +4\}$ .

The abscissa of a point A is twice its ordinate and $B \equiv ( 10,0 ) .$ Find the co-ordinates of A if $A B = 5$ units.

$\\given\>x=2y,\>where\>(x,y)\>is\>co-ordinate\>of\>A\\AB=5\\\sqrt{(x-10)^2+(y-0)^2}=5\\(2y-10)^2+y^2=5^2\\4y^2-40y+100+y^2=25\\5y^2-40y+75=0\\y^2-8y+15=0\\y^2-5y-3y+15=0\\y(y-5)-3(y-5)=0\\(y-3)(y-5)=0\\\therefore\>y=3\>or\>y=5\\If\>y=3\>then\>x=6\\\therefore\>point(6,3)\\If\>y=5\>then\>x=10\\\therefore\>point(10,5)$

The points $A (2a,4a), B(2a,6a)$ and $C (2a + \sqrt 3 a,5a)$ , $\left( {\text{when}\ a > o} \right)$ vertices of

$\begin{array}{l} A\left( { 2a,4a } \right) ,\, \, B\left( { 2a,6a } \right) and\, \, C\left( { 2a+\sqrt { 3 } a,5a } \right) \\ \Rightarrow Dis\tan ce\, \, of\, \, AB=\sqrt { { { \left( { 2a-2a } \right) }^{ 2 } }+{ { \left( { 6a-4a } \right) }^{ 2 } } } \\ AB=\sqrt { 0+4{ a^{ 2 } } } =2a\, \, unit \\ \Rightarrow BC=\sqrt { { { \left( { \sqrt { 3a } } \right) }^{ 2 } }+{ { \left( { -a } \right) }^{ 2 } } } \\ \Rightarrow BC=\sqrt { 3{ a^{ 2 } }+{ a^{ 2 } } } \\ BC=\sqrt { 4{ a^{ 2 } } } =2a\, \, unit \\ \Rightarrow AC=\sqrt { \left( { \sqrt { 3 } { a^{ 2 } } } \right) +{ { \left( a \right) }^{ 2 } } } \\ \Rightarrow AC=\sqrt { 3{ a^{ 2 } }+{ a^{ 2 } } } \\ \Rightarrow AC=\sqrt { 4{ a^{ 2 } } } =2a\, \, unit \\ AB=BC=CA \\ Hence,\, \, A,\, \, B\, and\, \, C\, \, are\, Vertices\, \, of\, \, equilateral\, \, triangle \end{array}$

If the vertices of a triangle are (1, k), (4, -3), (-9, 7) and its area is 15 sq units. Find the values of k.

f the vertices of a triangle are

$(1,k),(4,-3)(-9,7)$

area =

$15$ sq.units. find the value of k.

Area of

$\triangle$

$\dfrac{1}{2} \left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ] = 15$

$\dfrac{1}{2} \left [ 1(-3-7)+4(7-k)+(-9)(k-(-3) \right ] = 15$

$\dfrac{1}{2} (-10+8-4k-9 (k+3)] = 15$

$\dfrac{1}{2}\left [ \left ( -10+28-4k-9k-27 \right ) \right ] = 15$

$-10+28-4k-9k-27 = 30$

$-10+28-13k-27 = 30$

$-13k = 30+10+27-28$

$-13k = 39$

$k = \dfrac{39}{13}$

$k = -3$

Find the area of the triangle formed by the line $\dfrac{x}{a}+\dfrac{y}{b}=1$ and the $x-$ axis and $y-$ axis.

Given the equation of the line

$\dfrac{x}{a}+\dfrac{y}{b}=1$ .....

$(1)$ .

From (1) it is clear that the

$x$ and

$y$ intercept made by the line are

$a$ and

$b$ respectively.

Now the triangle formed by the line

$(1)$ and

$x$ and

$y$ axis is right angled triangle whose base and height are

$a$ and

$b$ respectively.

$\text{Area}=\dfrac{1}{2}\times a\times b\\\ \ \ \ \ \ \ =\dfrac{ab}{2}\text{sq. units}$

If $P(x, y)$ is equidistant from $A(a+b, b-a)$ and $B(a-b, a+b)$ , show that $bx=ay$ .

$PA=PB\Rightarrow { PA }^{ 2 }={ PB }^{ 2 }$

$\Rightarrow { \left( a+b-x \right) }^{ 2 }+{ \left( b-a-y \right) }^{ 2 }={ \left( a-b-x \right) }^{ 2 }+{ \left( a+b-y \right) }^{ 2 }$

$\Rightarrow { \left( a+b-x \right) }^{ 2 }-{ \left( a-b-x \right) }^{ 2 }={ \left( a+b-y \right) }^{ 2 }-{ \left( b-a-y \right) }^{ 2 }$

$2\left( a-x \right) .2b=2\left( b-y \right) 2a$

$\Rightarrow ab-bx=ab-ay$

$\Rightarrow bx=ay$ Proved.

Find the area of the triangle whose vertices are $(-5,-1)(3,-5)(5,2).$

$A(-5, -1) B(3, -5) C(5, 2)$

Area of

$\triangle ABC = \dfrac {1}{2}$

$\begin{vmatrix} -5 & -1 & 1\\ 3 & -5 & 2\\ 5 & 2 & 1\end{vmatrix}$

$= \dfrac {1}{2} |-5 (-7) +1(-2) + 1(31)|$

$= \dfrac {1}{2} |35 - 2 + 31|$

$=\dfrac {1}{2}\times 64 =32\ sq\ units$ .

If the point (x , y) is equivalent from the points (a + b, b - a) and (a - b, a+ b) prove that bx $=$ ay.

1334562_1263460_ans_25e867310457422a99dab3de9563185d.PNG

Find the area of the triangle whose vertices are $\left(10,-6\right),\left(2,5\right)$ and $\left(-1,3\right)$ .

$(10,-6);(2,5);(-1,3)$

$x_1 \,\,\,\,\,y_1$

$x_2\,y_2$

$x_3\,\,\,y_3$

$=Area=\dfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

$=\dfrac{1}{2}|10(5-3)+2(3+6)-1(-6-5)|$

$=\dfrac{1}{2}|10(2)+2(9)-1(-11)|$

$=\dfrac{1}{2}|20+18+11|$

$=\dfrac{1}{2}|49|$

$=24.5$

If the coordinates of two points $A,\ B$ are $(1,\ 2)(3,\ 8)$ respectively, find a point $P$ such that $|PA|=|PB|$ and area of $\triangle PAB=10$ .

Given that :

$A$ and

$B$ are two points

Let

$A (x_{1}, y_{1}) = (1, 2)$

$B(x_{2}, y_{2}) = (3, 8)$

and

$|PA| = |PB|$

Lt the point

$P(x, y)$

Find

$P(x, y) =?$

Since, It is given that,

$|PA| = |PB|$

$\sqrt {(x - x_{1})^{2} + (y - y_{1})^{2}} - \sqrt {(x - x_{2})^{2} + (y - y_{2})^{2}}$

$\sqrt {(x - 1)^{2} + (y - 2)^{2}} = \sqrt {(x - 3)^{2} + (y - 8)^{2}}$

Squaring both side and we get,

$(x - 1)^{2} + (y - 2)^{2} = (x - 3)^{2} + (y - 8)^{2}$

$x^{2} + 1^{2} - 2x + y^{2} + 4 - 4y = x^{2} + 9 - 6x + y^{2} + 64 - 16y$

$-2x - 4y + 5 + 6x + 16y = 73$

$4x + 12y = 73 - 5$

$4x + 12y = 68$

$4(x + 3y) = 68$

$x + 3y = \dfrac {68}{4}$

$x + 3y = 17 ..... (1)$

Area of

$\triangle PAB = 10$

$\dfrac {1}{2} [x(y_{1} - y_{2}) + x_{1} (y_{2} - y) + x_{2}(y - y_{1})] = 10$

$x (2 - 8) + 1 (8 - y) + 3(y - 2) = 10$

$x (-6) + (8 - y) + 3y - 6 = 10$

$-6x + 8 - y + 3y - 6 - 10 = 0$

$-6x + 2y - 8 = 0$

$6x - 2y +8 = 0$

$3x - y + 4 = 0$

$3x - y = -4 ....(2)$

From equation (1) and (2) to

$(x + 3y = 17)\times 3$

$3x - y = -4$

$3x + 9y = 51$

$3x - y = -9$

$3x + 9y = 51$

$\underline {\underset {-}{3}x \underset {+}{-} y \underset {+}{=} -4}$

$\log = 55$

$y = \dfrac {55}{10}$

$y = 5.5$

put in equation (1)

$x + 3y = 17$

$x + 3\times 5.5 = 17$

$x + 16.5 = 17$

$x = 17 - 16.5$

$x = 0.5$

Hence, the point

$P(x, y) = (0.5,5.5)$

Hence, this is the answer.

The area of a triangle is $5\ sq. unit$ . If two vertices of the triangle are $(2,1),(3,-2)$ and the third vertex is $(x,y)$ where $y=x+3$ , then find the coordinates of the third vertex.

Given that,

Two vertices of two triangle are

$Q(2, 1)$ and

$R(3, -2)$ .

Let the third vertex be

$P(x, y)$

Area of

$\triangle PQR = 5\ sq. units$

Area of

$\triangle PQR = \dfrac {1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$5 = \dfrac {1}{2} [x(1 - (-2)) + 2(-2 - y) + 3(y -1)]$

$10 = 3x - 4 - 2y + 3y - 3$

$3x + y - 7 = 10$

$3x + y - 17 = 0 ..... (1)$

Given that the equation

$y = x + 3 ..... (2)$

Put the value of

$y$ in equation (1) and (2) we get,

$3x + y - 17 = 0$

$3x + x + 3 - 17 = 0$

$4x - 14 = 0$

$4x = 14$

$x = \dfrac {14}{4}$

$x = \dfrac {7}{2}$

Put the value of

$x$ in equation (2)

$y = x + 3$

$y = \dfrac {7}{2} + 3$

$y = \dfrac {13}{2}$

Hence, point

$P(x, y) = \left (\dfrac {7}{2}, \dfrac {13}{2}\right )$ the third vertex.

Find the nature of the triangle formed by the points $A\left(4,\,0\right),\,B\left(-1,\,-1\right),C\left(3,\,5\right)$

1338022_1274998_ans_13dcbfd4f5fd49c1a58267df411bdf6d.PNG

The point $A\left( {0,3} \right),B\left( { - 2,a} \right)$ and $C\left( { - 1,4} \right)$ are the vertices of a right angled triangle at $A$ , find the value of $a$ .

1352008_1270120_ans_205d1024077941758e8c0d1299b62386.PNG

Find the point on the X-axis which is equidistant from A(-3,4) and B(1,-4)

1302402_1277576_ans_d71f21852dd64d0dadff63cab90e1ce3.jpg

Let $P\left(-5,1\right),\,Q\left(3,\,5\right)$ and $A$ divide $PQ$ in the ratio $k:1$ .Find the value of $k$ for which the area of the triangle $ABC$ , where $B=\left(1,\,5\right)$ and $C=\left(7,-2\right)$ is $2$

1338060_1276780_ans_251a882208d443f3a7e6977441db8d88.PNG

Find the distance of point $(2,\ 3, -5)$ from the plane $x+2y-2z-9=0$

We have,

$x+2y-2z-9=0$

Distance from the given point

$(2, 3, -5)$

$=\dfrac{2\times 1+3\times 2-5\times -2-9}{\sqrt{1+4+4}}$

$=\dfrac{2+6+10-9}{\sqrt{9}}$

$=\dfrac{9}{3}$

$=3\ units$

Find a relation between x and y such that the point (x,y) is equidistant from the points (7,1) and (3,5).

$\textbf{Step 1: Find the relation between x and y}.$

$\text{Let P(x, y) be equidistant from the points A(7,1) and B(3,5).}$

$\text{So, AP = BP}$

$\text{Squaring on both sides, we get}$

$\Rightarrow (AP)^2=(BP)^2$

$\text{Using, distance formula, }$

$\text{Distance between}\ (x_1,y_1)\ \text{and} \ (x_2,y_2)= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.....(1)$

$\text{Now,}$

$\Rightarrow (AP)^2=(BP)^2$

$\Rightarrow (x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2$

$\textbf{[Using eq(1)]}$

$\Rightarrow x^2+49-14x+y^2+1-2y=x^2+9-6x+y^2+25-10y$

$\Rightarrow -14x+50-2y=-6x+34-10y$

$\Rightarrow -7x+25-y=-3x+17-5y$

$\Rightarrow -4x+8+4y=0$

$\Rightarrow 4x-4y=8$

$\Rightarrow 4(x-y)=8$

$\Rightarrow x-y=2$

$\textbf{Hence, this is the required relation between x and y . }$

If equation of line is $\left( {y - 2\sqrt 3 } \right) = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}\left( {x - 2} \right)$ , then find the slope.

We have,

$\left( {y - 2\sqrt 3 } \right) = \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}\left( {x - 2} \right)$

We know that

$\left( {y - y_1 } \right) = m\left( {x - x_1} \right)$

Where

$m$ is a slope.

So,

$m=\dfrac{\sqrt 3+1}{\sqrt 3-1}$

Hence, this is the answer.

The point $A(2,3), B(4,-1), C(-1,2)$ are the vertices of $\triangle ABC$ . Find the length of perpendicular from $C$ on $AB$ and hence find area of $\triangle ABC$ .

Equation of

$AB$ is

$\dfrac {y-3}{x-2}=\dfrac {-4}{2}\Rightarrow 2x+y-7=0$

Formula for perpendicular distance from point to line

$D=\cfrac { ax_{ 1 }+by_{ 1 }+c }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } }$

Length of perpendicular from

$C(-1,2)$ to

$AB$ with equation

$2x+y-7=0$ is equal to

$\dfrac {|2\times (-1)+2-7|}{\sqrt {2^2 +1^2}}=\dfrac {7}{\sqrt 5}$ units.

$ar (\triangle ABC)=\dfrac{2} \times \text{base}\times \text{corresponding height} \\=\dfrac {1}{2}\times AB \times \dfrac {7}{\sqrt 5}\\=\left(\dfrac {1}{2}\times \sqrt {20}\times \dfrac {7}{\sqrt 5}\right)=7$

$\therefore$ Area of

$\triangle ABC = 7$ sq. units.

If the lines through the point $(4,1,2)$ ans $(5,k,0)$ is parallel to the line through $(2,1,1)$ and $(3,3,-1)$ , then value of k is?

1212257_1293965_ans_96c559c6b31c4dbd942682dd202d3f19.JPG

The area of a triangle whose vertices are (2, 3), (3, 5) and (7, 5) is :

Given: Vertices of triangle:

$(2,3),\ (3,5)$ and

$(7,5)$

Area of triangle

$=\dfrac{1}{2} \bigg| x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \bigg|$

$=\dfrac{1}{2}\bigg | 2(5-5)+3(5-3)+7(3-5) \bigg|$

$=\dfrac12 \bigg| 0+6+14 \bigg|$

$=\dfrac{1}{2}\times 20$

$=10$ sq. units

What are the points on the $y-$ axis whose distance from the line $\dfrac{x}{3}+\dfrac{y}{4}=1$ is $4$ units.

The given line is

$\dfrac{x}{3}+\dfrac{y}{4}=1$

i.e.,

$4x+3y-12=0$ .......

$(1)$

Let

$\left(0,k\right)$ be a point on the

$y-$ axis whose perpendicular distance from the line

$(1)$ is

$4$ units, then

$\dfrac{\left|4\times 0+3\times k-12\right|}{\sqrt{{4}^{2}+{3}^{2}}}=4$

$\Rightarrow\,\left|3k-12\right|=20$

$\Rightarrow \,3k-12=\pm 20$

$\Rightarrow\,3k=12\pm 20$

$\Rightarrow\,3k=32,\,-8$

$\therefore\,k=\dfrac{32}{3},\,-4$

Find the joint equation of lines passing through the origin, each of which making angle of measure $150^{o}$ with the line $x-y=0$ .

Slope of the first line

$={m}_{1}=\tan{{150}^{\circ}}=\tan{\left({180}^{\circ}-{30}^{\circ}\right)}=-\tan{{30}^{\circ}}=\dfrac{-1}{\sqrt{3}}$

The equation of the first line is

$y=\dfrac{-1}{\sqrt{3}}x$

$\sqrt{3}y+x=0$

$\therefore$ equations of the lines are

$\sqrt{3}y+x=0$ and

$x-y=0$

The joint equation is

$\left(x+\sqrt{3}y\right)\left(x-y\right)=0$

$={x}^{2}-xy+\sqrt{3}xy-\sqrt{3}{y}^{2}=0$

${x}^{2}+\left(\sqrt{3}-1\right)xy-\sqrt{3}{y}^{2}=0$ is the required joint equation.

Show that the tangent of an angle between the lines $\frac{x}{a}+\frac{y}{b}=1 and \frac{x}{a}-\frac{y}{b}= 1\ is \frac{2ab}{a^{2}-b^{2}}$

Given

$\dfrac{x}{a}+\dfrac{y}{b}=1$ …………

$(1)$

$\dfrac{x}{a}-\dfrac{y}{b}=1$ ………….

$(2)$

Equation

$(1)$ can be written as

$y=-\dfrac{b}{a}x+1$

where slope

$\left[m_1=\dfrac{-b}{a}\right]$

Equation

$(2)$ can be written as

$y=\dfrac{b}{a}x-1$

where slope

$[m_2=\dfrac{b}{a}]$

now angle between line

$(1)$ &

$(2)$ is

$\tan\theta =\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|=\left|\dfrac{\dfrac{-b}{a}-\dfrac{b}{a}}{1-\dfrac{b^2}{a^2}}\right|$

$\tan\theta =\left|\dfrac{-\dfrac{2b}{a}}{\dfrac{a^2-b^2}{a^2}}\right|=\left|\dfrac{-2ba}{a^2-b^2}\right|=\left|\dfrac{-2ab}{-b^2+a^2}\right|$

Hence tangent of an angle between line

$(1)$ &

$(2)$

$\dfrac{2ab}{a^2-b^2}$ .

1182996_1357234_ans_fc9eab4860444145a4393e0ce0e3ad40.JPG

Prove that the points $\left(0,5\right), \left(0,-9\right)$ and $\left(3,6\right)$ are non-collinear.

Given are coordinates of three points.

To prove: The three points are non-collinear.

Let the given points be

$A(0,5),\ B(0,-9)\ and\ C(3,6)$ .

We know that, three points are collinear if the area of the triangle they form is

$0$ . Conversely, if the given points make a triangle having a non-zero area, then they are non-collinear.

We also know that, the area of the triangle formed by points

$A(x_1,y_1),\ B(x_2,y_2)\ and\ C(x_3,y_3)$ is:

$\dfrac12\left| x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right|$

$\therefore \$ Area of

$\triangle ABC=\dfrac12\left| 0(-9-6)+0(6-5)+3(5+9) \right|$

$\Rightarrow \dfrac12\left| 0+0+42 \right|$

$\Rightarrow \dfrac{42}{2}=21$

Hence, Area of

$\triangle ABC=21$

Since, area of

$\triangle ABC\ne0$ , points

$A,\ B\ and\ C$ are non-collinear.

$\quad [Hence\ proved]$

Find the foot point of $1$ from the point $(-2,3)$ to the line $2x-y-3=0$

We have to find the foot of perpendicular drawn from

$(-2,3)$ to the line

$2x-y-3=0$

So, the equation of a line perpendicular to

$2x-y-3=0$ is given by

$x+2y+k=0$

This line should pass through

$(-2,3)$

so,

$-2+2(3)+k=0\\\implies k=-4$

So, the equation of the perpendicular is

$x+2y-4=0$

Now the foot of perpendicular is the intersection point of two lines

$2x-3y=0$ and

$x+2y-4=0$

Solving these two equation we get

$x=2$ and

$y=-1$

Hence the coordinate of foot of perpendicular is

$(2,-1)$

Find a point on the Y - axis which is equidistant from the points A(-4,3) and B(6,5).

Let the point on y-axis be (0,y).

By, Distance formula we have,

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Therefore,

$\sqrt{(4)^{2}+(y-3)^{2}} = \sqrt{(-6)^{2}+(y-5)^{2}}$

$16+y^{2}+9-6y=36+y^{2}+25-10y$

$25-6y=61-10y$

$4y=36$

$y=9$

Hence, the required point is

$(0,9)$ .

Find the equations of lines with slope $-2$ and length of perpendicular distance from origin is equal to $\sqrt{3}$ units.

The equation of the family of lines with slope

$-2$ is

$y=-2x+c$

$2x-y-c=0$ .......

$(1)$

For the required members of the family, the perpendicular distance from the origin

$\left(0,0\right)$ is

$\sqrt{3}$ units.

$\Rightarrow\,\dfrac{\left|2\times 0-0-c\right|}{\sqrt{{\left(2\right)}^{2}+{\left(-1\right)}^{2}}}=\sqrt{3}$

$\Rightarrow \dfrac{\left|c\right|}{\sqrt{4+1}}=\sqrt{6}$

$\Rightarrow \,c=\pm\,\sqrt{30}=\pm\sqrt{30}$

$\therefore\,$ the equations of the lines are

$2x-y+\sqrt{30}=0$ or

$2x-y-\sqrt{30}=0$

Find a point on x-axis which is equidistant from (2, -5) and (-2, 9)

Let the required point on the x-axis be

$(x,0).$

Therefore,

$\sqrt{(x-2)^{2}+(5)^{2}} = \sqrt{(x+2)^{2}+(-9)^{2}}$

$(x-2)^{2}+25=(x+2)^{2}+81$

$x^{2}+4-4x+25=x^{2}+4+4x+81$

$8x=25-81=-56$

$x=-7$

hence, the required point is

$(-7,0)$ .

Find the points on $x-$ axis that is equidistant from $(2,7)$ & $(5,-2)$ .

Let

$\left( h, 0 \right)$ be the point on

$x$ -axis whis is equidistant from

$\left( 2, 7 \right)$ and

$\left( 5, -2 \right)$ .

Therefore,

$\sqrt{{\left( 2 - h \right)}^{2} + {\left( 7 - 0 \right)}^{2}} = \sqrt{{\left( 5 - h \right)}^{2} + {\left( -2 - 0 \right)}^{2}}$

$\sqrt{\left( 4 + {h}^{2} - 4h \right) + 49} = \sqrt{25 + {h}^{2} - 10h + 4}$

Squaring both sides, we get

${h}^{2} - 4h + 53 = {h}^{2} - 10h + 29$

$\Rightarrow 10h - 4h = 29 - 53$

$\Rightarrow h = \cfrac{-24}{6} = -4$

Hence the point is

$\left( -4, 0 \right)$ .

If the points $A(4,3)$ and $B(x,5)$ are on the circle with centre $O(2,3)$ , find the value of $x$ .

$\textbf{Step 1: Apply distance formula between two points.}$

$\text{The coordinates of the centre }$

$O=(2,3)$

$\text{And the coordinates of two points A and B on the circle are }$

$(4,3)$

$\text{and }$

$(x,5)$

$\text{respectively. }$

$\text{Distance between point O and point A =}$

$\sqrt{(4-2)^2+(3-3)^2}\ units$

$=\sqrt{2^2+0^2}\ units$

$=\sqrt{4+0}\ units$

$=2\ units$

$\text{And. distance between point O and point B =}$

$\sqrt{(x-2)^2+(5-3)^2}\ units$

$=\sqrt{x^2-4x+4+4}\ units$

$=\sqrt{x^2-4x+8}\ units$

$\textbf{Step 2: Apply radii of a circle condition.}$

$\text{We know, the lengths of all the radii of a circle are equal.}$

$\therefore OB=OA$

$\Rightarrow OB^2=OA^2$

$\Rightarrow x^2-4x+8=4$

$\Rightarrow x^2-4x+4=0$

$\Rightarrow x^2-2\times x\times 2+2^2=0$

$\Rightarrow (x-2)^2=0$

$\Rightarrow x-2=0$

$\Rightarrow x=2$

$\textbf{Hence, the required value of x is 2.}$

Find the coordinates of points on the x-axis which are at a distance of 17 units from the point (11,-8)

Let the required point be (x,0).

$\sqrt{(x-11)^{2}+(8)^{2}} = 17$

$(x-11)^{2}+64=289$

$x^{2}+121-22x+64=289$

$x^{2}-22x-104=0$

$x^{2}-26x+4x-104=0$

$x(x-26)+4(x-26)=0$

$(x-26)(x+4)=0$

$x=26,-4$

Hence, the required points are

$(26,0)$ and

$(-4,0)$ .

The area of a trapezium is $90$ sq.cm and its height is $6$ cm.If one of the parallel sides is twice that of the others.Find the two parallel sides.

Let the length of the smaller parallel side be

$\,x$ cm

Then length of the other parallel sides

$=\,2x$ cm

Area of the trapezium

$=\dfrac{1}{2}\times\,sum\,of\,parallel\,sides\times\,height$

$\Rightarrow\,90=\dfrac{1}{2}\times\left(x+2x\right)\times 6$

$\Rightarrow\,90=9x$

$\Rightarrow \,x=\dfrac{90}{9}=10$ cm

What are the points on the $y-$ axis whose distance from the line $\dfrac{x}{6}-\dfrac{y}{8}=1$ is $-3$ units.

The given line is

$\dfrac{x}{6}-\dfrac{y}{8}=1$

$8x-6y-48=0$ ........

$(1)$

Let

$\left(0,k\right)$ be a point on the

$y-$ axis whose perpendicular distance from the line

$(1)$ is

$-3$ units, then

$\dfrac{\left|8\times 0-6\times k-48\right|}{\sqrt{{8}^{2}+{6}^{2}}}=-3$

$\Rightarrow\,\dfrac{\left|-6k-48\right|}{\sqrt{64+36}}=-3$

$\Rightarrow\,6k+48=\pm\left(-30\right)$

$\Rightarrow\,6k+48=\mp\,30$

$\Rightarrow \,2k+16=\mp\,10$

$\Rightarrow\,2k=-16\mp\,10$

$\Rightarrow\,2k=-26,\,-6$

$\therefore\,k=\dfrac{-26}{2},\,\dfrac{-6}{2}$

$\therefore\,k=\dfrac{-26}{2},\,-3$

Hence the required points are

$\left(0, -13\right)$ and

$\left(0,-3\right)$

The distance between the points $\left(8,9\right)$ and $\left(-7,4\right)$ is

Let

$A\left({x}_{1},{y}_{1}\right)=\left(8,9\right)$ and

$B\left({x}_{2},{y}_{2}\right)=\left(-7,4\right)$

$\therefore$ Distance

$AB=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$

$\therefore \,AB=\sqrt{{\left(-7-8\right)}^{2}+{\left(4-9\right)}^{2}}$

$=\sqrt{225+25}=\sqrt{250}=\sqrt{25\times 10}=5\sqrt{10}$ units

The points A(4,7) B(p,3) and C(7,3) are the vertices of a right triangle, right -angled at B, Find of P.

REF.Image

A(4,7), B(p,3) & c (7,3) are vertices of right angle triangle then u

using Pythagoras theorem

$(AC)^{2}=(AB)^{2}+(BC)^{2}$

$(7-4)^{2}+(3-7)^{2}=(p-4)^{2}+(3-7)^{2}+(p-7)^{2}+(3-3)^{2}$

$9+16=(p-4)^{2}+16+(p-7)^{2}$

$9=p^{2}+16-8p+p^{2}+49-14p$

$2p^{2}-22p+56=0$

$p^{2}-11p+28=0$

$p^{2}-7p-4p+28=0$

(p-4)(p-7)=0

Hence p-4=0 & p-7=0

[p=4,7]

1194420_1373238_ans_75392471f56243cd9496b7575bdd5f4f.JPG

Find the distance between the points $\left(2,1\right)$ and $\left(3,2\right)$

Let

$A\left({x}_{1},{y}_{1}\right)=\left(2,1\right)$ and

$B\left({x}_{2},{y}_{2}\right)=\left(3,2\right)$

$\therefore$ Distance

$AB=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$

$\therefore \,AB=\sqrt{{\left(3-2\right)}^{2}+{\left(2-1\right)}^{2}}$

$=\sqrt{1+1}=\sqrt{2}$ units

If the three vertices of a parallelogram are $\left(-1,3\right),\left(2,4\right)$ and $\left(3,5\right)$ ,find the fourth vertex.

Let

$A\left(-1,3\right),B\left(2,4\right), C\left(3,5\right)$ and

$D\left(x,y\right)$ be the vertices of the parallelogram.

Midpoint of

$AC=\left(\dfrac{-1+3}{2},\,\dfrac{3+5}{2}\right)=\left(\dfrac{2}{2},\,\dfrac{8}{2}\right)=\left(1,4\right)$

Also midpoint of

$BD=\left(\dfrac{2+x}{2},\,\dfrac{4+y}{2}\right)=\left(\dfrac{x+2}{2},\,\dfrac{y+4}{2}\right)$

Since the diagonals of a parallelogram bisect each other, the mid-points of

$AC$ and

$BD$ are same.

$\Rightarrow \dfrac{x+2}{2}=1$ and

$\dfrac{y+4}{2}=4$

$\Rightarrow x+2=1$ and

$y+4=8$

$\therefore\,x=2-2=0,\,\,y=8-4=4$

Hence the fourth vertex is

$\left(0,4\right)$

Find the area of the shaded triangle in the given figure.

This figure consist of

$1$ cm rectangle,

$3$ unshaded and

$1$ shaded triangle.

Area of rectangle

$=8\times\,6=48$ sq.cm

Area of triangle

$A=\dfrac{1}{2}\times 4\times 5=10$ sq.cm

Area of triangle

$B=\dfrac{1}{2}\times 6\times 3=9$ sq.cm

Area of triangle

$C=\dfrac{1}{2}\times 8\times 2=8$ sq.cm

Area of shaded triangle

$=$ Area of rectangle

$-$ Area of unshaded triangle

$=48-\left(10+9+8\right)=48-27=21$ sq.cm

Find the co-ordinates of the incentre of the triangle whose vertices are $\left(-2,4\right),\left(5,5\right)$ and $\left(4,-2\right)$

Let

$A\left(-2,4\right),B\left(5,5\right)$ and

$C\left(4,-2\right)$ be the vertices of the given

$\triangle{ABC}$ then

$a=\left|BC\right|=\sqrt{{\left(4-5\right)}^{2}+{\left(-2-5\right)}^{2}}=\sqrt{1+49}=\sqrt{50}=5\sqrt{2}$

$b=\left|CA\right|=\sqrt{{\left(4+2\right)}^{2}+{\left(-2-4\right)}^{2}}=\sqrt{36+36}=\sqrt{36\times 2}=6\sqrt{2}$

and

$c=\left|AB\right|=\sqrt{{\left(5+2\right)}^{2}+{\left(5-4\right)}^{2}}=\sqrt{49+1}=\sqrt{50}=5\sqrt{2}$

$\therefore\,$ The coordinates of the incentre of

$\triangle{ABC}$ is

$\left(\dfrac{a{x}_{1}+b{x}_{2}+c{x}_{3}}{a+b+c},\dfrac{a{y}_{1}+b{y}_{2}+c{y}_{3}}{a+b+c}\right)$

$=\left(\dfrac{5\sqrt{2}\times -2+6\sqrt{2}\times 5+5\sqrt{2}\times 4}{5\sqrt{2}+6\sqrt{2}+5\sqrt{2}},\dfrac{5\sqrt{2}\times 4+6\sqrt{2}\times 5+5\sqrt{2}\times -2}{5\sqrt{2}+6\sqrt{2}+5\sqrt{2}}\right)$

$=\left(\dfrac{-10\sqrt{2}+30\sqrt{2}+20\sqrt{2}}{16\sqrt{2}},\dfrac{20\sqrt{2}+30\sqrt{2}-10\sqrt{2}}{16\sqrt{2}}\right)$

$=\left(\dfrac{40\sqrt{2}}{16\sqrt{2}},\dfrac{40\sqrt{2}}{16\sqrt{2}}\right)$

$=\left(\dfrac{5}{2},\dfrac{5}{2}\right)$

Find the area of the triangle whose vertices are $(2,5),(-1,0),(2,-4)$ .

Given vertices are

$A(2,5)$

$B(-1,0)$

$C(2,-4)$

$(x_{1},y_{1})$

$(x_{2},y_{2})$

$(x_{3},y_{3})$

Area of

$\triangle ABC =\dfrac{1}{2}\left [ x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right ]$

$=\dfrac{1}{2}\left [ 2(0+4)+(-1)(-4-5)+2(5-0) \right ]$

$=\dfrac{1}{2}\left [ 2(4)+(-1)(-9)+2(5) \right ]$

$=\dfrac{1}{2}\left [ 8+9+10 \right ]$

$=\dfrac{1}{2}\left [ 27 \right ]$

$\therefore$ Area of triangle

$ABC=13.5 \ cm^2$

Find the shortest distance between the $x^2+y^2=9$ and $(6,8)$ is

The center of circle is

$(0,0)$

The distance from

$(6,8)$ is

$\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt {100}=10$

The radius of circle is

$3$

So the shortest distance is given as

$10-3=7$ units

The distance between $(4,0)$ and $(7,4)$

Points $(4,0),(7,4)$

Distance between 2 points is given as $\sqrt{(x_2-x_1)^2+(y_1-y_2)^2}\\\sqrt{(7-4)^2+(4-0)^2}\\\sqrt{9+16}\\\sqrt {25}=5$

Find the area of a triangle whose vertices are
$(3, 6), (-1, 3), (2, -1)$

The vertices of the triangle are

$(3,6), (-1,3), (2,-1)$ .

We know,

Area of triangle =

$|\dfrac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]|$

$=|\dfrac{1}{2}[3(4)-1(-7)+2(3)]|$ =

$|\dfrac{1}{2}[12+7+6]|$ =

$\dfrac{25}{2}$ sq. units

Find the slope of the line, which makes an angle of $30^{o}$ with the positive direction of $y-axis$ measured anticlockwise.

REF.Image

The line make angle of

$30^{\circ}$ from positive y axis

Hence it make angle of

$(90^{\circ}+30^{\circ})$ from positive x axis, i.e,

$120^{\circ}$ from positive x axis

Hence

$\theta =120^{\circ}$

Slope =

$m= tan \theta = atn 120^{\circ}= tan (180-60^{\circ})=tan 60^{\circ}=-\sqrt{3}$

Slope of line =

$-\sqrt{3}$

1210850_1379777_ans_6eef5e2f9f964975941b6282fe015b9c.jpg

The distance between the points $\left(\dfrac{1}{2},\dfrac{3}{2}\right)$ and $\left(\dfrac{3}{2},\dfrac{-1}{2}\right)$ is

Let

$A\left({x}_{1},{y}_{1}\right)=\left(\dfrac{1}{2},\dfrac{3}{2}\right)$ and

$B\left({x}_{2},{y}_{2}\right)=\left(\dfrac{3}{2},\dfrac{-1}{2}\right)$

$\therefore$ Distance

$AB=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$

$\therefore \,AB=\sqrt{{\left(\dfrac{3}{2}-\dfrac{1}{2}\right)}^{2}+{\left(\dfrac{-1}{2}-\dfrac{3}{2}\right)}^{2}}$

$=\sqrt{{\left(\dfrac{3-1}{2}\right)}^{2}+{\left(\dfrac{-1-3}{2}\right)}^{2}}$

$=\sqrt{{\left(\dfrac{2}{2}\right)}^{2}+{\left(\dfrac{-4}{2}\right)}^{2}}$

$=\sqrt{{\left(1\right)}^{2}+{\left(-2\right)}^{2}}$

$=\sqrt{1+4}=\sqrt{5}$ units.

Find the area of the triangle formed by the points $\left(5,2\right),\,\left(-9,-3\right)$ and $\left(-3,-5\right)$

Area of

$\triangle{ABC}=\dfrac{1}{2}\begin{vmatrix} {x}_{1} & {x}_{2} & {x}_{3} & {x}_{1} \\ {y}_{1} & {y}_{2} & {y}_{3} & {y}_{1}\end{vmatrix}$

Let

$A\left({x}_{1},{y}_{1}\right)=\left(5,2\right),\,\,\, B\left({x}_{2},{y}_{2}\right)=\left(-9,-3\right)$ and

$C\left({x}_{3},{y}_{3}\right)=\left(-3,-5\right)$ be the points of a triangle.

Area of

$\triangle{ABC}=\dfrac{1}{2}\begin{vmatrix} 5 & -9 & -3 & 5 \\ 2 & -3 & -5 & 2\end{vmatrix}$

$=\dfrac{1}{2}\left[\left(5\times -3-2\times -9\right)+\left(-9\times -5-\left(-3\times -3\right)\right)+\left(-3\times 2-\left(-5\times 5\right)\right)\right]$

$=\dfrac{1}{2}\left[\left(-15+18\right)+\left(45-9\right)+\left(-6+25\right)\right]$

$=\dfrac{1}{2}\left[3+36+19\right]$

$=\dfrac{1}{2}\times 58=29$ sq.units.

If $A(a^{2},2a)B(\frac{1}{a^{2}},\frac{-2}{a})$ and S(1,0) then prove that $\frac{1}{SA}+\frac{1}{SB}=1$

$SA=\sqrt{(1-a^2)^2+(-2a)^2}$

$=\sqrt{1+a^4-2a^2+4a^2}$

$=\sqrt{1+a^4+2a^2}$

$=\sqrt{(1+a^2)^2}$

$=(1+a^2)$ ..........1

$SB=\sqrt{(1-\dfrac{1}{a^2})^2+(\dfrac{2}{a})^2}$

$=\sqrt{1+\dfrac{1}{a^4}-\dfrac{2}{a^2}+\dfrac{4}{a^2}}$

$=\sqrt{1+\dfrac{1}{a^4}+\dfrac{2}{a^2}}$

$=\sqrt{(1+\dfrac{1}{a^2})^2}$

$=1+\dfrac{1}{a^2}$

$=\dfrac{a^2+1}{a^2}$ ......2

Then,

$\dfrac{1}{SA}+\dfrac{1}{SB}=\dfrac{1}{1+a^2}+\dfrac{a^2}{1+a^2}$

$=\dfrac{1+a^2}{1+a^2}=1$

A line is of length 10 units and one of its ends is (-2,3). If the ordinate of the order end is 9, prove that the abscissa of the other end is 6 or -10.

The length of line is

$10cm$

let the abscissa be

$x$

So the points are

$(-2,3);(x,9)$

Distance formula is

$\sqrt{(x+2)^2+(9-3)^2}=10\\x^2+4x+4+36=100\\x^2+4x-60=0\\x^2+10x-6x-60=0\\x(x+10)-6(x+10)\\(x+10)(x-6)\\x=6,-10$

Find the distance between $\left(3,6\right)$ from the origin.

Distance between

$\left(3,6\right)$ from the origin

$=$ Distance between

$\left(3,6\right)$ from

$\left(0,0\right)$

$=\sqrt{ { \left( {x_2}-{ { x_1 } } \right) }^{ 2}+{ \left( {y_2 }-{ y_1 } \right) }^{ 2 } }$

$=\sqrt{{\left(3-0\right)}^{2}+{\left(6-0\right)}^{2}}$

$=\sqrt{9+36}$

$=\sqrt{45}=3\sqrt{5}$ units.

Find the distance between $\left(x+3,x-3\right)$ from the origin.

Distance between

$\left(x+3,x-3\right)$ from the origin

$=$ Distance between

$\left(x+3,x-3\right)$ from

$\left(0,0\right)$

$=\sqrt{ { \left( {x_2}-{ { x_1 } } \right) }^{ 2}+{ \left( {y_2 }-{ y_1 } \right) }^{ 2 } }$

$=\sqrt{{\left(x+3-0\right)}^{2}+{\left(x-3-0\right)}^{2}}$

$=\sqrt{{x}^{2}+9+6x+{x}^{2}+9-6x}$

$=\sqrt{2{x}^{2}+18}$ units

If the point $A(2,-4)$ is equidistance from the points $P(3,8)$ and $Q(-10,y)$ find the value of $y$ . Also find the value of $PQ$ .

As we know that distance between the two points

$\left( {x}_{1}, {y}_{1} \right)$ and

$\left( {x}_{2}, {y}_{2} \right)$ is given as-

$d = \sqrt{{\left( {x}_{2} - {}_{1} \right)}^{[2} + {\left( {y}_{2} - {y}_{1} \right)}^{2}}$

Therefore,

$AP = \sqrt{{\left( 3 - 2 \right)}^{2} + {\left( 8 - \left( -4 \right) \right)}^{2}} = \sqrt{145} \text{ units}$

$AQ = \sqrt{{\left( -10 - 2 \right)}^{2} + {\left( y - \left( - 4 \right) \right)}^{2}} = \sqrt{144 + {\left( y + 4 \right)}^{2}} \text{ units}$

$PQ = \sqrt{{\left( -10 - 3 \right)}^{2} + {\left( y - 8 \right)}^{2}} = \sqrt{169 + {\left( y - 8 \right)}^{2}} \text{ units}$

Now,

$AP = AQ \quad \left( \text{Given} \right)$

$\therefore \sqrt{145} = \sqrt{144 + {\left( y + 4 \right)}^{2}}$

Squaring both sides, we have

$145 = 144 + {\left( y + 4 \right)}^{2}$

${y}^{2} + 16 + 8y = 1$

$\Rightarrow {y}^{2} + 8y + 15 = 0$

$\Rightarrow \left( y + 5 \right) \left( y + 3 \right) = 0$

$\Rightarrow y = -5$ OR

$y = -3$

Case I:-

$y = -3$

$PQ = \sqrt{169 + {\left( -3 - 8 \right)}^{2}} = \sqrt{169 + 121} = \sqrt{290} \text{ units}$

Case II:-

$y = -5$

$PQ = \sqrt{169 + {\left( -5 - 8 \right)}^{2}} = \sqrt{169 + 169} = 13 \sqrt{2} \text{ units}$

How are the points $(1, -1), (2, 2)$ and $(-1, 2)$ situated with respect to the circle ${x}^{2}+{y}^{2}+2x–4y-5=0$

Given equation is

${x}^{2}+{y}^{2}+2x–4y-5=0$

Let

$A=(1, -1), B=(3, 1)$ and

$C=(1, 3)$

For point

$A(0, 1),\, {x}^{2}+{y}^{2}+2x–4y-5=0+1+0–4-5=-8<0$

Hence point

$A$ lies inside the circle

For point

$B=(2, 2),\, {x}^{2}+{y}^{2}+2x–4y-5=4+4+4–8-5=-1<0$

Hence point

$B$ lies inside the circle

For point

$C=(-1, 2),\, {x}^{2}+{y}^{2}+2x–4y-5=1+4-2–8-5=-10<0$

Hence point

$C$ lies inside the circle

Prove that the points $A(0,0), B(0,2)$ and $C(2,0)$ are the vertices of an equilateral triangle.

Using distance formula,

$AB=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$

$AB=\sqrt{{\left(0-0\right)}^{2}+{\left(2-0\right)}^{2}}=\sqrt{4}=2$ units

$BC=\sqrt{{\left(0-2\right)}^{2}+{\left(0-0\right)}^{2}}=\sqrt{4}=2$ units

$CA=\sqrt{{\left(0-0\right)}^{2}+{\left(2-0\right)}^{2}}=\sqrt{4}=2$ units

$\therefore AB=BC=CA$

Hence the given triangle form an equilateral triangle.

Write the perpendicular distance of point $(-3, 2)$ from x-axis.

1309482_1450562_ans_877b55de10ee4c0296d5c10d2b65189f.jpeg

Find the point on the curve ${ y=x }^{ 3 }-{ 2x }^{ 2 }-x,$ where the tangents are parallel to $3x-y+1=0$ .

Let

$P(x_{1},y_{1})$ be required point

Given curve is,

$y=x^{3}-2x^{2}-x$

$\frac{dy}{dx}=3x^{2}-4x-1$

$(\frac{dy}{dx})_{x_{1},y_{1}}=3x^{2}_{1}-4x_{1}-1$

Since tangent at

$(x_{1},y_{1})$ is parallel to line

$y=3x-1$

Slope of tangent at

$(x_{1},y_{1})$ = slope of linr,

$y=3x-1$

$(\frac{dy}{dx})_{(x_{1},y_{1})}=3$

$3x_{1}^{2}-4x_{1}-1=3$

$3x_{1}^{2}-4x_{1}-4=0$

$(3x_{1}+2)(x_{1}-2)=0$

$x_{1}=2,\frac{-2}{3}$

Since

$(x_{1},y_{1})$ lies on curve,

$y_{1}=x_{1}^{3}-2x_{1}^{2}-x_{1}$

$x_{1}=2, y_{1}=2^{3}-2\times 4-2=-2$

$x_{1}=\frac{-2}{3}, y_{1}=(\frac{-2}{3})^{2}-2\times (\frac{-2}{3})^{2}+\frac{2}{3}=\frac{-14}{27}$

Required points are (2,-2) and

$(\frac{-2}{3}$ and

$\frac{-14}{27})$

1230705_1502322_ans_5887f69c98ef4023a344b54545f7454a.jpg

Write the formula for area of a triangle where A $\left( { x }_{ 1 },{ y }_{ 1 } \right) ,B\left( { x }_{ x },{ y }_{ 2 } \right)$ and $C\left( { x }_{ 3 },{ y }_{ 3 } \right)$ are the vertices of a triangle $ABC$ .

$A(x_{1},y_{1}) ,B(x_{2},y_{2}), C(x_{3},y_{3})$

Area of

$\Delta \Rightarrow$

$\dfrac{1}{2}|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|$

Find the area of the triangle with vertices at the points :
$(-1,-8),(-2,-3) \ and \ (3,2)$

Given Points are

$(-1,-8),(-2,-3),(3,2)$

comparing with

$(x_1,y_1),(x_2,y_2),(x_3,y_3)$

$\therefore$ Area of triangle

$:-$

$=\dfrac 12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) |$

$=\dfrac 12|-1(-3-2)-2(2+8)+3(-8+3) |$

$=\dfrac 12|5-20-15|$

$=15$

Find the area of the triangle with vertices at the points :
$(0,0),(6,0) \ and \ (4,3)$

Given Points are

$(0,0),(6,0),(4,3)$

comparing with

$(x_1,y_1),(x_2,y_2),(x_3,y_3)$

$\therefore$ Area of triangle

$:-$

$=\dfrac 12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) |$

$=\dfrac 12|0(0-3)+6(3-0)+4(0-0) |$

$=\dfrac 12|0+18+0|=9$

If the coordinates of the points $A, B, C, D$ be $(1, 2, 3), (4, 5, 7), (-4,3, -6)$ and $(2, 9, 2)$ respectively, then find the angle between the lines $AB$ and $CD$ .

The direction ratios of

$AB$ and

$CD$ are proportional to

$3, 3, 4$ and

$6, 6, 8$ respectively.
Let

$\theta$ be the angle between

$AB$ and

$CD$ . Then,

$\cos \theta = \dfrac {6\times 3 + 6\times 3 + 8\times 4}{\sqrt {9 + 9 + 16} \sqrt {36 + 36 + 64}} = \dfrac {68}{\sqrt {34}\sqrt {136}} = 1\Rightarrow \theta = 0^{\circ}$ .

If the area of the triangle with vertices at the points:
$(2,7),(1,1) \ and \ (10,8)$

is $\left (\dfrac {b}{2}\right )$ sq. uts.

Then what is the value of b?

Given Points are

$(2,7),(1,1),(10,8)$

comparing with

$(x_1,y_1),(x_2,y_2),(x_3,y_3)$

$\therefore$ Area of triangle

$:-$

$=\dfrac 12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) |$

$=\dfrac 12|2(1-8)+1(8-7)+10(7-1) |$

$=\dfrac 12|-14+1+60|$

$=\dfrac {47}2$

$\therefore A=\dfrac{47}{2}$

but,

$Area=\dfrac{b}{2}$ [given]

$\therefore b=47$

Find the area of the triangle with vertices at the points:
(3,8),(-4,2) and (5,-1). If the area is $\left ( \dfrac a2 \right )$ sq. units, then what will be the value of $a$ ?

Given Points are

$(3,8),(-4,2),(5,-1)$

comparing with

$(x_1,y_1),(x_2,y_2),(x_3,y_3)$

$\therefore$ Area of triangle

$:-$

$\dfrac 12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) |$

$=\dfrac 12|3(2+1)-4(-1-8)+5(8-2) |$

$=\dfrac 12|9+36+30|$

$\therefore Area=\dfrac {75}2$

but

$Area=\dfrac{a}{2}$ given,

$\therefore a=75$

Find the area of the triangle with vertices as given below:
$(0,0),(6,0)$ and $(4,3)$

Given Points are

$(0,0),(6,0),(4,3)$

comparing with

$(x_1,y_1),(x_2,y_2),(x_3,y_3)$

$\therefore$ Area of triangle

$=\dfrac 12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) |$

$=\dfrac 12|0(0-3)+6(3-0)+4(0-0) |$

$=\dfrac 12|0+18+0|=9$

Show that the points $(0,7,10),(-1,6,6)$ and $(-4,9,6)$ are the vertices of an isosceles right-angled triangle.

1412717_1667716_ans_60bf9ce65b7b4724bd8125264c5905ce.jpg

Verify the following:
$(0,7,10),(-1,6,6)$ and $(-4,9,6)$ are vertices of a right angled traingle.

1413434_1667802_ans_b18abe643709495c9b46a13b3f7f72b9.jpg

Verify the following:
$(0,7,-10),(1,6,-6)$ and $(4,9,-6)$ are vertices of an isosceles triangle.

1413433_1667798_ans_611d3d6bb3664830ac00732455f87d5e.jpg

Prove that the triangle formed by joining the three points whose coordinates are $(1,2,3), (2,3,1)$ and $(3,1,2)$ is an equilateral triangle.

1412715_1667715_ans_eea9f162d71b4e998e1cfca25e6f6c53.jpg

Find the slope of the line which make the following angle with the positive direction of $x-$ axis :
$\dfrac{2\pi}{3}$

Slope

$=\tan \theta$

$=\tan (\dfrac {2 \pi} {3})=\tan (\pi- \dfrac {\pi} {3})=- \tan (\dfrac \pi 3)$

$=-\sqrt 3$

Ans

1433748_1667202_ans_34bd4c824254412bb0f1a8630536cb89.png

Find the slope of the lines which make the following angles with the positive direction of $x-$ axis :
$\dfrac{\pi}{3}$

Slope,

$m=\tan \theta$

$=\tan \dfrac {\pi} 3 =\sqrt 3$ Ans

1433746_1667204_ans_478ed88fc78d48508f9927c5c81530af.png

Show that the product of perpendiculars on the line $\dfrac {x}{a}\cos \theta +\dfrac {y}{b}\sin \theta =1$ from the points $(\pm \sqrt {a^2 -b^2}, 0)$ is $b^2$ .

Let

${p}_{1}$ and

${p}_{1}$ be the lengths of perpendicular draw from point

$P\left(\sqrt{{a}^{2}-{2}^{2}},0\right)$

$Q\left(-\sqrt{{a}^{2}-{2}^{2}},0\right)$ on the line

$\dfrac{x}{a}\cos \theta +\dfrac{y}{b} \sin \theta =1$ .

Then,

${ P }_{ 2 }=$

$\left| \dfrac { \dfrac { \sqrt { { a }^{ 2 }-{ b }^{ 2 } } }{ a } \cos { \theta } +\dfrac { 0 }{ b } \sin { \theta } -1 }{ \sqrt { \dfrac { \cos ^{ 2 }{ \theta } }{ { a }^{ 2 } } +\dfrac { \sin ^{ 2 }{ \theta } }{ { b }^{ 2 } } } } \right|$

Show that the points $(a,b,c),(b,c,a)$ and $(c,a,b)$ are the vertices of a equilateral triangle.

1413221_1667789_ans_be47dbaba5974acc8ef754d5e589cf8f.jpg

Find the inclination of a line whose slope is
$\sqrt{3}$

Given: Slope of line

$\sqrt{3}$

slope of line

$= tan \theta$

$tan \theta = \sqrt{3}$

$\theta = 60^\circ$

Find the slope of a line which passes through the points
$(0, -3)$ and $(2, 1)$

slope of line

$= \dfrac{y_2-y_1}{x_2-x_1}$

Put values in formula

Slope of line

$=\dfrac{2-0}{1-(-3)}$

$=\dfrac{2}{4}=\dfrac{1}{2}$

Find the inclination of a line whose slope is $-1$

$\textbf{Step 1: Find inclination of line whose slope is -1}$

$\text{Given,}$

$\text{Slope}=-1$

$\tan\theta=-1$

$[\because\textbf{Slope}\boldsymbol{=\tan\theta}]$

$\tan\theta=\tan 135^\circ$

$\theta=135^\circ$

$\textbf{Hence, the inclination is 135}^\circ$

Find the area of triangle $PQR$ formed by the points $P(-5, 7), Q(-4,-5)$ and $R(4,5)$

Given data :-

$P(-5, 7) \equiv (x_1 , y_1)$

$Q (-4, -5) \equiv (x_2, y_2)$

$R (4, 5) \equiv (x_3 , y_3)$

To find :- Area of triangle

$PQR$

Solution :-

Ref. image

$P(x_1,y_1),\,Q(x_2,y_2),\,R(x_3,y_3)$ are vertices of a triangle

$ar(\triangle PQR)=\dfrac12[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$= \dfrac{1}{2} [-5 (-5-5) -4 (5-7) + 4 (7+5)]$

$= \dfrac{1}{2} [50 +8 + 48]$

$= \dfrac{1}{2} [50 + 56]$

$= \dfrac{1}{2} \times 106$

$= 53 \, sq. unit$

1501296_1700971_ans_801cb7f7d9cd490cb89add807868734d.png

Find the inclination of a line whose slope is: $\sqrt {3}$ .

We know that, the slope of a line is given by the tan of its inclination.

Slope

$=\tan \theta=\sqrt3$

$\theta=60^o$

Find the slope of a line which passes through the points
$(0, 0)$ and $(4, -2)$

1981553_1710123_ans_970135307b964f6f902aff790e1fbff7.jpg

Find the slope of a line which passes through the points
$(-2, 3)$ and $(4, -6)$

1973232_1710128_ans_cf4f0b352d86481585144146ba3b3a98.jpg

Find the slope of a line which passes through the points
$(2, 5)$ and $(-4, -4)$

1981555_1710125_ans_9492921156a74c67967d9c8dcc203b0f.jpg

Find the equation of a line which is equidistant from the lines $x=-2$ and $x=6$

Midpoint of

$A(-2,0)$ and

$B(6,0)$ is

$M\left(\dfrac{-2+6}{2},0\right)$ , i.e.

$M(2,0)$

So, the required line is

$x=2$

Find the equation of a line which is equidistant from the lines $y=8$ and $y=-2$ .

Midpoint of

$A(0,8)$ and

$B(0,-2)$ is

$M\left(0,\dfrac{8-2}{2}\right)$ , i.e.

$M(0,3)$

So , the required line is

$y=3$

Using slopes, show that the points $A(6, -1), B(5, 0)$ and $C(2, 3)$ are collinear.

1849048_1710156_ans_659c55b7bf264ff0bdff33cd2ee0e61e.jpeg

Show that the line through the points $(5, 6)$ and $(2, 3)$ is parallel to the line through the points $(9,-2)$ and $(6, -5)$ .

1849018_1710136_ans_d78c6234ff034df3ad1a6a10160a89f0.jpeg

Show that the line through the points $(-2, 6)$ and $(4, 8)$ is perpendicular to the line through the points $(3, -3)$ and $(5, -9)$ .

1849021_1710144_ans_266468321a4a48bf98ec3c820242b9e8.jpeg

Using slopes, find the value of $x$ for which the points $A(5, 1), B(1, -1)$ and $C(x, 4)$ are collinear.

1981562_1710165_ans_6c7499100618457f86d8905e2e81b4bf.jpg

Show that the points $A(2,-2), B(8,4), C(5,7)$ and $D(-1, 1)$ are the angular points of a rectangle.

1973350_1710160_ans_0bf03dfb6982418abf3626158b1d8b61.jpg

Show that $A(3,2), B(0,5), C(-3,2)$ and $D(0,-1)$ are the vertices of a square.

1973356_1710164_ans_e933f268a9854f6f8dee10626b12d2c7.jpg

If $A(2, -5), B(-2, 5), C(x, 3)$ and $D(1, 1)$ be four points such that $AB$ and $CD$ are perpendicular to each other, find the value of $x$ .

1981559_1710149_ans_299739472eae443f9e260b7a66f71375.jpg

If the slope of the line joining the points $A(x, 2)$ and $B(6, -8)$ is $\dfrac{-5}{4}$ , find the value of $x$ .

1973233_1710132_ans_ffbd002626c94ab5b62fd3644dfc2fea.jpg

Using slopes, prove that the points $A(-2, -1), B(1, 0), C(4, 3)$ and $D(1, 2)$ are the vertices of a parallelogram.

1849060_1710174_ans_739bbd0487e64c319529f27f22b08c5f.jpeg

Find the slope and the equation of the line passing through the points:
$(3,-2)$ and $(-5,-7)$

1973371_1710170_ans_52737cd23f0746bf914eff6571588e3b.jpg

Show that the points $A(2,-1), B(3,4), C(-2,3)$ and $D(-3,-2)$ are the vertices of rhombus.

1849059_1710173_ans_fd295b2ec92a4954bf9af4c72db7c968.jpeg

If the points $A(h, k), B(x_1, y_1)$ and $C(x_2, y_2)$ lies on a line then show that $(h-x_1)(y_2-y_1)=(k-y_1)(x_2-x)$ .

Points A, B and C lie on the same line.

So, Slope of

$AB=$ slope of

$BC$

$\Rightarrow \dfrac{(y_1-k)}{(x_1-h)}=\dfrac{y_2-y_1)}{(x_2-x_1)}\Rightarrow \dfrac{(k-y_1)}{(h-x_1)}=\dfrac{(y_2-y_1)}{(x_2-x_1)}$

$\Rightarrow (h-x_1)(y_2-y_1)=(k-y_1)(x_2-x_1)$ .

If the points $A(-2,-1), B(1,0), C(x,3)$ and $D(1,y)$ are the vertices of a parallelogram , find the value of $x$ and $y$ .

1973228_1710180_ans_621c214c3cfc4a7fa84454b746e2be09.jpg

Find the slope and the equation of the line passing through the points:
$(a,b)$ and $(-a,b)$

1973227_1710177_ans_571eb58bff864fb1a257c39eec1a61da.jpg

Show that the points $A(-5,1), B(5,5)$ and $C(10,7)$ are collinear.

1972459_1710189_ans_ccefba8dbe024293b1cd7002b6742979.jpg

Find the slope and the equation of the line passing through the points:
$(-1,1)$ and $(2,-4)$

1973375_1710172_ans_03db3d33b35e48209dc152de7e80daee.jpg

Show that $A(1,-2), B(3,6), C(5,10)$ and $D(3,2)$ are the vertices of a parallelogram.

1973359_1710167_ans_44ae498d9d754a8fbf66ec598c430bd4.jpg

Find the area of $\triangle ABC$ whose vertices are $A(-3,-5), B(5,2)$ and $C(-9,-3)$ .

1973229_1710184_ans_832ec0a83e8a4eae91c2e096625a1791.jpg

Find the value of $k$ for which the points $A(-2,3), B(1,2)$ and $C(k, 0)$ are collinear.

If three points

$(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are collinear then

$\Rightarrow$

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{y_3-y_2}{x_3-x_2}$

The given points are

$A(-2,3), B(1,2)$ and

$C(k,0)$

If these points are collinear then

$\Rightarrow$

$\dfrac{2-3}{1-(-2)}=\dfrac{0-2}{k-1}$

$\Rightarrow$

$\dfrac{-1}{3}=\dfrac{-2}{k-1}$

$\Rightarrow$

$\dfrac{1}{3}=\dfrac{2}{k-1}$

$\Rightarrow$

$k-1=6$

$\therefore$

$k=7$

Find the area of the triangle, the equation of whose sides are $y=x, y=2x$ and $y-3x=4$

1981642_1710324_ans_41fd127ade994867bfbc60e859c5df7d.jpg

Find the area of quadrilateral whose vertices are $A(-4,5), B(0,7), C(5,-5)$ and $D(-4,-2)$ .

1972477_1710195_ans_a5657175615a4f82aaa93ad27c5ea2b4.jpg

Find the area of the triangle formed by the lines $x=0, y=1$ and $2x+y=2$ .

1981640_1710320_ans_1e3e608d5cc6457db05c76f068512287.jpg

Find the area of $\Delta ABC$ whose vertices are $A(1, 2), B(-2, 3)$ and $C(-3, -4)$ .

Area of

$\Delta ABC$ whose vertices are is

$(x_1, y_1), (x_2, y_2)$ and

$(x_3, y_3)$ are

Area of

$\Delta ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$\Delta ABC$ , vertices are

$A(1, 2), B(-2, 3)$ and

$C(-3, -4)$

Area of triangle

$=1/2(1(-2+3)-2(-4-2)-3(2-3))$

$=1/2(1+12+3)$

$=8$ sq units.

If the points $A(a, 0), B(0, b)$ and $P(x, y)$ are collinear, using slopes, prove that $\dfrac{x}{a}+\dfrac{y}{b}=1$ .

Since the points

$A, B, P$ are collinear, we have

(slope of

$AB=$ slope of

$BP$ )

$\Rightarrow \dfrac{b-0}{0-a}=\dfrac{y-b}{x-0}$

$\Rightarrow bx+ay=ab\Rightarrow \dfrac{x}{a}+\dfrac{y}{b}=1$ .

A line passes through the points $A(4, -6)$ and $B(-2, -5)$ . Show that the line $AB$ makes and obtuse angle with $x-$ axis.

1972462_1710193_ans_716c47add0b54e7390180353aefc7d15.jpg

Find the area of $\triangle ABC$ , the midpoints of whose sides $AB, BC$ and $CA$ are $D(3,-1), E(5,3)$ and $F(1,-3)$ respectively.

1981564_1710200_ans_84417ebf98fa47bb90869d003058068a.jpg

Find the area of the triangle formed by the lines $x+y=6, x-3y=2$ and $5x-3y+2=0$ .

1981636_1710316_ans_f7473d743ade44558e16b71193133543.jpg

Find the area of $\Delta ABC$ whose vertices are $A(-5, 7), B(-4, -5)$ and $C(4, 5)$ .

Area of

$\Delta ABC$ whose vertices are is

$(x_1, y_1), (x_2, y_2)$ and

$(x_3, y_3)$

$=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$\Delta ABC$ , vertices are

$A(-5, 7), B(-4, -5)$ and

$C(4, 5)$

Area of triangle

$=\dfrac{1}{2}[-5(-5-5)-4(5-7)+4(7+5)]$

Area of triangle

$=$

$\dfrac{10}{2}$

$=5$ sq units.

$A(6, 1), B(8, 2)$ and $C(9, 4)$ are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of $\Delta ADE$ .

$A(6, 1), B(8, 2)$ and

$C(9, 4)$ are the three vertices of a parallelogram ABCD.

E is the midpoint of DC.

Join AE, AC and BD which intersects at O, where Q is midpoint of AC.

Midpoint formula:

$(x, y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

Find coordinates of midpoints D and E:

$\dfrac{x+8}{2}=\dfrac{15}{2}\Rightarrow x+8=15$

$x=15-8=7$

and

$\dfrac{y+2}{2}=\dfrac{5}{2}\Rightarrow y+2=5$

$y=5-2=3$

Coordinate of D are

$(7, 3)$

And

$=\left(\dfrac{7+9}{2}, \dfrac{3+4}{2}\right)$

$=\left(8, \dfrac{7}{2}\right)$

Coordinate of E are

$(7, 3)$

Now, We know that:

Area of

$\Delta ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Area of

$\Delta ADC$

$=\dfrac{1}{2}[6\left(3-\dfrac{7}{2}\right)+7\left(\dfrac{7}{2}-1\right)+8(1-3)]$

$=\dfrac{1}{2}[6\times \left(\dfrac{-1}{2}\right)+7\times \dfrac{5}{2}+8(-2)]$

$=\dfrac{5}{2}[-3+\dfrac{35}{2}-16]$

$=\dfrac{1}{2}\times \dfrac{3}{2}=\dfrac{3}{4}$ sq. units.

1602243_1714464_ans_bb47054c4af14c28b9bcb0b6b0d413d7.png

If the vertices of $\Delta$ ABC be $A(1, -3), B(4, p)$ and $C(-9, 7)$ and its area is $15$ square units, find the values of p.

We know that:

Area of

$\Delta ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Vertices of a

$\Delta ABC$ are

$A(1, -3), B(4, p)$ and

$C(-9, 7)$ and

Area

$=15$ sq. units

$15=\dfrac{1}{2}[1(p-7)+4(7+3)+(-9)(-3-p)]$

$30=(p-7+40+27+9p)$

$30=10p+60$

$10p=30-60=-30\Rightarrow p=\dfrac{-30}{10}=-3$ .

Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are $A(2, 1), B(4, 3)$ and $C(2, 5)$ .

Given: A triangle whose vertices are

$A(2, 1), B(4, 3)$ and

$C(2, 5)$

Let D, E and F are the midpoints of the sides CB, CA and AB respectively of

$\Delta ABC$ , as shown in the figure.

Find vertices of D, E and F:

Midpoint formula:

$(x, y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

Vertices of D:

$=\left(\dfrac{4+2}{2}, \dfrac{3+5}{2}\right)$

$=\left(\dfrac{6}{2}, \dfrac{8}{2}\right)=(3, 4)$

Vertices of E:

$=\left(\dfrac{2+2}{2}, \dfrac{5+1}{2}\right)$

$=\left(\dfrac{4}{2}, \dfrac{6}{2}\right)=(2, 3)$

Vertices of F:

$=\left(\dfrac{2+4}{2}, \dfrac{1+3}{2}\right)$

$=\left(\dfrac{6}{2}, \dfrac{4}{2}\right)=(3, 2)$

Area of triangle DEF:

We know that:

Area of

$\Delta ABC=\dfrac{1}{2}[x_1(y_2-y_2)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Area of triangle DEF

$=$

$=\dfrac{1}{2}[3(3-2)+2(2-4)+3(4-3)]$

$=\dfrac{1}{2}[3\times 1+2\times (-2)+3\times 0]$

$=\dfrac{1}{2}\times 2=1$ sq. units.

1602207_1714457_ans_03eb821f4f6c4ce2abcaad511bebe715.png

Find the area of $\Delta ABC$ whose vertices are $A(3, 8), B(-4, 2)$ and $C(5, -1)$ .

Area of

$\Delta ABC$ whose vertices are is

$(x_1, y_1), (x_2, y_2)$ and

$(x_3, y_3)$ are

Area of

$\Delta ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$\Delta ABC$ , vertices are

$A(3, 8), B(-4, 2)$ and

$C(5, -1)$ .

Area of triangle

$=\dfrac12(3(2+1)-4(-1-8)+5(8-2))$

$=\dfrac12(9+36+30)$

$=\dfrac12(75)$

$=37.5$ sq units.

The area of a triangle is $5$ sq units. Two of its vertices are $(2, 1)$ and $(3, -2)$ . If the third vertex is $(7/2, y)$ , find the value of y.

We know that:

Area of

$\Delta ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$5=\dfrac{1}{2}[2(-2-y)+3(y-1)+\dfrac{7}{2}(1+2)]$

$10=\left[-4-2y+3y-3y+\dfrac{7}{2}+7\right]$

$10=\left[y+\dfrac{7}{2}\right]\Rightarrow 10-\dfrac{7}{2}=y\Rightarrow y=\dfrac{13}{2}$ .

Find the area of $\Delta ABC$ with $A(1, -4)$ and midpoints of sides through A being $(2, -1)$ and $(0, -1)$ .

Given: A

$\Delta ABC$ with

$A(1, -4)$

Let F and E are the midpoints of AB and AC respectively

Let Coordinates of F are

$(2, -1)$ and Coordinates of E are

$(0, -1)$

Let Coordinates of B be

$(x_1, y_1)$ and Coordinates of C be

$(x_2, y_2)$

Find coordinate of B:

using section formula:

$2=\dfrac{1+x_1}{2}\Rightarrow x_1=4-1=3$

$-1=\dfrac{4+y_1}{2}\Rightarrow y_1=-2+4=2$

Coordinate of B are

$(3, 2)$

Find coordinate of C:

using section formula:

$0=\dfrac{1+x_2}{2}\Rightarrow 1+x_2=0\Rightarrow x_2=-1$

$-1=\dfrac{-4+y_2}{2}\Rightarrow -4+y_2=-2$

$\Rightarrow y_2=-2+4=2$

Coordinate of C are

$(-1, 2)$

Now,

Area of

$\Delta ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Area of triangle ABC:

$=\dfrac{1}{2}[1(2-2)+3(2+4)+(-1)(-4-2)]$

$=\dfrac{1}{2}[1\times 0+3\times 6+(-1)\times (-6)]$

$=\dfrac{1}{2}[0+18+6]=\dfrac{24}{2}=12$ sq. units.

1602229_1714462_ans_cb8e6dffe7c74608911db6c2de3ec0b3.png

Find the area of $\Delta ABC$ whose vertices are $A(10, -6), B(2, 5)$ and $C(-1, 3)$ .

Area of

$\Delta ABC$ whose vertices are is

$(x_1, y_1), (x_2, y_2)$ and

$(x_3, y_3)$ are

Area of

$\Delta ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$\Delta ABC$ , vertices are

$A(10, -6), B(2, 5)$ and

$C(-1, 3)$

Area of triangle

$=1/2(10(5-3)+2(3+6)-1(-6-5))$

$=1/2(20+18+11)$

$=1/2(49)$

$=24.5$ sq units.

Find the value of k so that the area of teh triange with vertices $A(k+1, 1), B(4, -3)$ and $C(7, -k)$ is $6$ square units.

Area of

$\Delta ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$\Rightarrow 6=\dfrac{1}{2}[(k+1)(-3+k)+4(-k-1)+7(1+3)]$

$\Rightarrow 6\times 2=[-3k+k^2-3+k-4k-4+28]$

$\Rightarrow 12=[k^2-6k+21]$

$\Rightarrow k^2-6k+21-12=0$

$\Rightarrow k^2-6k+9=0$

$\Rightarrow (k-3)^2=0\Rightarrow k-3=0$

$k=3$

The value of k is

$3$ .

Find the area of quadrilateral ABCD whose vertices are $A(3, -1), B(9, -5), C(14, 0)$ and $D(9, 19)$ .

Vertices of quadrilateral

$ABCD$ are

$A(3, -1), B(9, -5), C(14, 0)$ and

$D(9, 19)$

Construction: Join diagonal

$AC$ .

We know that:

Area of

$\Delta ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Area of triangle

$ABC$ :

$=\dfrac{1}{2}[3(-5-0)+9(0+1)+14(-1+5)]$

$=\dfrac{1}{2}[-15+9+14(4)]$

$=\dfrac{1}{2}[-15+9+56]$

$=\dfrac12\times 50$

$=25$ sq. units.

Area of triangle

$ADC$ :

$=\dfrac{1}{2}[3(0-19)+14(19+1)+9(-1+0)]$

$=\dfrac{1}{2}[3(-19)+14\times 20+9\times (-1)]$

$=\dfrac{1}{2}[-57+280-9]$

$=\dfrac12\times 214$

$=107$ sq. units.

Now, Area of quadrilateral

$ABCD=$ Area of triangle

$ABC+$ Area of triangle

$ADC$

$=(25+107)$ sq. units.

$=132$ sq.units.

1602177_1714445_ans_e35911f17f514f55957c5cc2031e93bd.png

Find the area of the triangle the coordinates of whose vertices are $(5 , 2) , (-9 , -3)$ and $(-3 , -5)$ .

1563867_1740534_ans_8de5f4f43bb948ff9993406729e05305.png

For what value of $k(k > 0)$ is the area of the triangle with vertices $(-2, 5), (k, -4)$ and $(2k+1, 10)$ equal to $53$ square units?

Area of triangle

$=53$ square units

Area of

$\Delta ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$53=\dfrac{1}{2}[-2(-4-10)+k(10-5)+(2k+1)(5+4)]$

$53=\dfrac{1}{2}[-2\times (-14)+k\times 5+(2k+1)\times 9]$

$106=23k+37$

$k=\dfrac{69}{23}=3$

The value of k is

$3$ .

Find the area of the triangle the coordinates of whose angular points are respectively
$(0 , 4) , (3 , 6)$ and $(-8 , -2)$

1563853_1740533_ans_3520bbba1bbc4cd1a620c51a9191c074.png

Plot the points $\displaystyle A (2, 5), B (-2, 2)$ and $\displaystyle C (4, 2)$ on a graph paper. Join $\displaystyle AB, BC$ and $\displaystyle AC$ .
Calculate the area of $\displaystyle \Delta ABC$ .

Construct a line

$\displaystyle AM$ which is perpendicular to

$\displaystyle BC$ .

Area of

$\displaystyle \Delta ABC = \frac{1}{2} \times b \times h = \frac{1}{2} \times BC \times AM$ .

So on further calculation we get

Area of

$\displaystyle \Delta ABC = \frac{1}{2} \times 6 \times 3$

Area of

$\displaystyle \Delta ABC = 9$ sq. units

1565080_1715359_ans_09ed38ea847648ebb908336db9fc6f49.png

Find the area of the triangle the coordinates of whose angular points are respectively
$(a , b + c) , (a , b - c)$ and $(-a , c)$

1563871_1740535_ans_cda00da6876a48d29e67277c551988e9.png

Find the area of the triangle the coordinates of whose angular points are respectively
$(1 , 3) , (-7 , 6)$ and $(5 , -1)$

1563862_1740532_ans_0119e5c685ee4d9495e2aa7346327f5d.png

The axes being inclined at angle of $60^{o}$ , find the inclination to the axis of $x$ the straight lines whose equations are
(1) $y=2x+5$ ,
And
(2) $2y=\left(\surd3-1\right)x+7$

1564852_1740511_ans_b0715b2ebfa64cfa87c86bb2907b7638.jpg

Find the area of $\Delta ABC$ with vertices $A(0, -1), B(2, 1)$ and $C(0, 3)$ . Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of these two triangles is $4:1$ .

Vertices of

$\Delta ABC$ are

$A(0, -1), B(2, 1)$ and

$C(0, 3)$

Area of

$\Delta ABC=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Area of triangle ABC:

$=\dfrac{1}{2}[0(1-3)+2(3+1)+0(-1-1)]$

$=\dfrac{1}{2}[0+2\times 4+0]=\dfrac{1}{2}\times 8=4$ sq. units.

From figure: Points D, E and F are midpoints of sides BC, CA and AB respectively.

Now the coordinates of D, E and F :

Coordinates of D

$=\left(\dfrac{2+0}{2}, \dfrac{1+3}{2}\right)$

$=(1, 2)$

Coordinates of E

$=\left(\dfrac{0+0}{2}, \dfrac{3-1}{2}\right)$

$=(0, 1)$

Coordinates of F

$=\left(\dfrac{0+2}{2}, \dfrac{-1+1}{2}\right)$

$=(1, 0)$

Area of triangle DEF using the same formula:

$=\dfrac12[1(1-0)+0(0-2)+1(2-1)]$

$=1$ sq. units

Therefore,

The ratio of the area of triangles ABC and DEF

$=\dfrac41=4:1$ .

1602549_1714489_ans_cee7c1aa82a742788d6636d45eab73af.png

If the point $P(4,-2)$ is the one end of the focal chord $P Q$ Of the $y^{2}=x,$ then find the slope of the tangent at $Q$ .

$y^{2}=x \quad \therefore 4 a=1$

$P\left(a t_{1}^{2}, 2 a t_{1}\right)=(4,-2)$

$\therefore t_{1}=-4$

Also

$t_{1} t_{2}=-1$ as

$P Q$ is a focal chord.

Slope of tangent at

$t_{2}$ is

$\dfrac{1}{t_{2}}=-t_{1}=4$

Prove (by showing that the area of the triangle formed by them is zero) that the following sets of three points are in a straight line:
$(1, 4), (3, -2),$ and $(-3, 16).$

1563870_1740537_ans_5258ee8baadb404d9bb24a1feaaf083f.png

Find the areas of the triangles the coordinates of whose angular points are
$(-3 , -30^\circ) , (5 , 150^\circ) ,$ and $(7 , 210^\circ).$

1563918_1740581_ans_f21ecc43979c4d729fb790be3db8201e.png

Find the areas of the triangles the coordinates of whose angular points are
$\left(-a , \dfrac {\pi}{6}\right) , \left(a , \dfrac {\pi}{2}\right) ,$ and $\left(-2a , -\dfrac {2\pi}{3}\right).$

1563892_1740583_ans_6afc12142d1d4e4f96ed5bcbbf94615d.png

A median of a triangle divides it into two triangles of equal areas. Verify this result for $\Delta \mathrm{ABC}$ whose vertices are $A(4,-6), B(3,-2),$ and $C(5,2)$

Since

$A D$ is the median of

$\Delta A B C,$ therefore,

$D$ is the mid-point of

$B C$ .

Therefore, by section formula,

Coordinates of

$D$ are

$\left(\dfrac{3+5}{2}, \dfrac{-2+2}{2}\right)=(4,0)$

Now, area of

$\Delta {ABD}=\dfrac{1}{2}[4(-2-0)+3(0+6)+4(-6-2)]$

$=\dfrac{1}{2}(-8+18-16)=\dfrac{1}{2}\times(-6)=-3$

Since the area is a measure, it cannot be negative.

$\therefore ar (\Delta {ABD})=3$ sq. units

And area of

$\Delta{ADC}=\dfrac{1}{2}[4(0-2)+4(2+6)+5(-6-0)]$

$=\dfrac{1}{2}(-8+32-30)$

$=\dfrac{1}{2}(-6)=-3, \text { which cannot be negative. }$

$\therefore ar(\Delta ADC)=3 \text{sq units }$

Here,

$ar(\Delta ABD)=ar(\Delta {ADC})$

Hence, the median divides it into two triangles of areas.

$B(5,0) ?$

We know that,

Area of a triangle formed by

$(x_1,y_1) , \ (x_2,y_2)$ and

$(x_3,y_3)$ is given by,

Area

$=\dfrac12[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$\therefore$ Area of

$\triangle OAB=\dfrac{1}{2}[0(0-0)-3(0-0)+5(0-0)]=0$

$\Rightarrow$ Given points are collinear.

A line through the origin intersects the parabola $5 y=2 x^{2} -9 x+10$ at two points whose $x$ -coordinates add up to 17 then find the slope of the line.

Let the line be

$y=m x \qquad\qquad\qquad(1)$

Solving it with

$5 y=2 x^{2}-9 x+10, \text { we get }$

$5 m x=2 x^{2}-9 x+10$

$2 x^{2}-(9+5 m) x+10=0$

sum of the roots

$=\dfrac{9+5 m}{2}=17$

$\Rightarrow 9+5 m=34$

$\Rightarrow \quad 5 m=25$

$\Rightarrow \quad m=5$

The area of the $\Delta ABC$ , coordinates of whose vertices are $A(2,0), \ B(4,5)$ and $C(6,3)$ is..........

1730813_1778704_ans_f134516f4a144bd8b274ec2b5ce19895.jpg

Find the inclination of a line whose gradient is
(i) $\, 1$
(ii) $\sqrt{3}$
(iii) $1/\sqrt{3}$

Given,

(i)

$tan \theta = 1$

$\Rightarrow$

$\theta = 45^o$

(ii)

$tan\theta = \sqrt{3}$

$\Rightarrow$

$\theta = 60^o$

(i)

$tan\theta = 1\sqrt{3}$

$\Rightarrow$

$\theta = 30^o$

Prove that the area of the triangle formed by the normal's to the parabola at the points $\left(at_{1}^{2}, 2at_{1}\right), \left(at_{2}^{2}, 2at_{2}\right)$ and $\left(at_{3}^{2}, 2at_{3}\right)$ is $\dfrac{a^{2}}{2}\left(t_{2}-t_{3}\right)\left(t_{3}-t_{1}\right)\left(t_{1}-t_{2}\right)\left(t_{1}+t_{2}+t_{3}\right)^{2}$ .

1759924_1741181_ans_c209501a854d4c9c892c42397134b9f8.jpg

Find the area of the triangle whose vertices are $(-8 , 4) (-6 , 6)$ and $(-3 , 9)$ .

Vertices of

$\delta ABC$ are

$A(-8 , 4) , B(-6 , 6)$ and

$C(-3 , 9)$ .

$\therefore \,$ Area of

$\Delta ABC = \dfrac{1}{2} [x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3(y_1 - y_2)]$

$\Rightarrow \,$ Area of

$\Delta = \dfrac{1}{2} [-8 (6 - 9) - 6 (9 - 4) - 3(4 - 6)]$

$= \dfrac{1}{2} [-8(-3) - 6(5) - 3(-2)]$

$= \dfrac{1}{2} [24 - 30 + 6] = 0$
Hence , the area of given triangle is zero.

If $A(4,2), B(7,6)$ and $C(1,4)$ are the vertices of a $\Delta A B C$ and $A D$ is its median, prove that the median AD divides $\Delta A B C$ into two triangles of equal areas.

Given:

$AD$ is the median on

$BC$ .

$\Rightarrow B D=D C$

To prove:

$AD$ divides

$\triangle{ABC}$ into two equal areas.

Proof:

By mid-point formula,

The coordinates of mid-point D are given by,

$\left(\dfrac{x_{2}+x_{1}}{2}, \dfrac{y_{2}+y_{1}}{2}\right) \\ =\left(\dfrac{1+7}{2}, \dfrac{4+6}{2}\right)\\ =(4,5)$

Now, Area of triangle

$A B D=\dfrac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

$=\dfrac{1}{2}[4(6-5)+7(5-2)+4(2-6)]$

$=\dfrac{1}{2}[4+21-16]$

$=\dfrac{9}{2} \text { sq. units }$

Area of triangle

${ACD}=\dfrac{1}{2}[4(4-5)+1(5-2)+4(2-4)]$

$= \dfrac{1}{2}[-4+3-8]$

$=-\dfrac{9}{2}$

$=\dfrac{9}{2} \text { sq. units }$

$\therefore$ Area of

$\triangle A B D=$ Area of

$\triangle ACD$

Hence,

$AD$ divides

$\triangle{ABC}$ into two equal areas.

Find the area of the triangle whose vertices are $(3, -4), (7, 5), (-1, 10)$

Given:

$(3, -4), (7, 5), (-1, 10)$

Let us assume

$A(x_1, y_1) = (3, -4)$

Let us assume

$B(x_2, y_2) = (7, 5)$

Let us assume

$C(x_3, y_3) = (-1, 10)$

Area of triangle

$= \dfrac{1}{2} \left [ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right ]$

Now, Area of triangle

$= \dfrac{1}{2} \left [ \left \{ 3(5 - (10)) \right \}+ (7)\left \{ (10 + 4) \right \}- 1(-4 - 5) \right ]$

$= \dfrac{1}{2} \left \{ -15 + 98 + 9 \right \}$

$= \dfrac{92}{2}$

$= 46$ square units

Draw the graph of $4x-3y+12 = 0$ and use it to find the area of the triangle formed by the line and the co-ordinate axes. Take $2 cm = 1$ unit on both axes.

1111

Find the area of the triangle whose vertices are
$(-1.5, 3), (6, -2), (-3, 4)$

Given

$(-1. 5, 3), (6, -2), (-3, 4)$
Let us assume

$A(x_1, y_1) = (-1.5, 3)$
Let us assume

$B(x_2, y_2) = (6, -2)$
Let us assume

$C(x_3, y_3) = (-3, 4)$
Now,
Area of given triangle

$= \dfrac{1}{2} \left [ \left \{ x_1 (y_2 - y_3) \right \} + x_2 \left \{ (y_3 - y_1) \right \}+x_3 (y_1 - y_2) \right ]$

$= \dfrac{1}{2} \left [ \left \{ -1.5 (-2 - (4)) \right \} + (6) \left \{ (4 - 3) \right \}- 3 (3 + 2) \right ]$

$= \dfrac{1}{2} \left \{ +9 + 6 - 15 \right \}$

$= \dfrac{0}{2}$

$= 0$ square units

In the given figure, $PQR$ is equilateral. If the coordinates of the points $Q$ and $R$ are $(0, 2)$ and $(0, -2)$ respectively, find the coordinates of the point $P.$

Given

$PQR$ is an equilateral triangle in which

$Q(0,2)$ and

$R(0,-2).$
Let

$(x,0)$ be the coordinates of

$P$ . [

$∵P$ lies on

$x$ axis. So y-coordinate is zero.]

$PQ = PR = QR = 2+2 = 4$

$OQ = 2$ [from figure]
In

$POQ,$

$PQ^2 = OP^2+OQ^2$ [Pythagoras theorem]

$4^2 = OP^2+2^2$

$OP^2 = 4^2-2^2$

$OP^2 = 16-4$

$OP^2 = 12$

$OP = √(4×3) = 2√3$
Hence the coordinates of

$P$ are

$(2√3,0).$

In Figure, the vertices of $\Delta$ ABC are $A(4,6), B(1,5)$ and $C(7,2) .$ A line segment $DE$ is drawn to intersect the sides $A B$ and $A C$ at $D$ and $E$ respectively such that $\dfrac{A D}{A B}=\dfrac{A E}{A C}=\dfrac{1}{3}$ . Calculate the area of $\Delta{ADE}$ and compare it with area of $\Delta{ABC}$ .

Since

$\dfrac{A D}{A B}=\dfrac{A E}{A C}=\dfrac{1}{3}$

$\Rightarrow D$ and

$E$ divide

$AB$ and

$AC$ respectively in the ratio

$1: 2$

Therefore,

Coordinates of

$D$ are

$\left(\dfrac{1(1)+2(4)}{3}\right),\left(\dfrac{1(5)+2(6)}{3}\right)$ i.

$e_{\cdot,}\left(3, \dfrac{17}{3}\right)$

Coordinates of

$E$ are

$\left(\dfrac{1(7)+2(4)}{3}, \dfrac{1(2+2(6))}{3}\right)$ i.e.,

$\left(5,\dfrac{14}{3}\right)$

Area of

$\Delta A D E =\dfrac{1}{2}\left[4\left(\dfrac{17}{3}-\dfrac{14}{3}\right)+3\left(\dfrac{14}{3}-6\right)+5\left(6-\dfrac{17}{3}\right)\right]$

$=\dfrac{1}{2}\left(4+(-4)+\dfrac{5}{3}\right)=\dfrac{5}{6}\text { sq unit }$

Area of

$\Delta {ABC} =\dfrac{2}{2}[4(3)+1(-4)+7(1)]=\dfrac{15}{2} \text { sq unit }$

Hence,

$ar(\Delta ADE) : ar(\Delta ABC)=\dfrac{5}{6}: \dfrac{15}{2}= 1: 9$

The adjoining figure shows an equilateral triangle $OAB$ with each side $= 2a$ units. Find the coordinates of the vertices.

Given equilateral triangle

$OAB.$

$OA = OB = AB = 2a$ units.
Draw

$AD$

$OB.$

$AD = √(AO^2-DO^2)$

$= √((2a)^2-a^2)$

$= √(4a^2-a^2)$

$= √(3a^2)$

$= √3 a$
Co-ordinates of

$O$ are

$(0,0).$
Co-ordinates of

$A$ are

$(a, √3 a)$
Co-ordinates of

$B$ are

$(2a,0).$

1721760_1811921_ans_84ecbcdab34847dcb38c21303833cbcb.PNG

Find the equation of the line which passes through $\left( 2, 2\sqrt{3} \right)$ and is inclined with the x-axis at an angle of $75^{0}$ .

The slope of the line
that inclines with the x-axis at an angle of $75^{0}$ is $m = \tan 75^{0}$

$m= \tan \left ( 45^{0} + 30^{0} \right ) = \frac{ \tan 45^{0} + \tan 30^{0}} {1 -\tan 45^{0} \cdot \tan 30^{0} } = \frac{1 + \frac{1} {\sqrt{3}}} {1 +1 \cdot \frac{1} {\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1} {\sqrt{3}}} {\frac{\sqrt{3} - 1} {\sqrt{3}}} = \frac{\sqrt{3} + 1} {\sqrt{3} - 1}$

We know that the
equation of the line passing through point $\left (x_0, y_0\right )$ , whose
slope is m, is $\left( y – y_0\right ) = m\left( x – x_0\right )$ .

$\left (y - 2\sqrt{3} \right ) = \frac{\sqrt{3} + 1} {\sqrt{3} - 1} \left (x - 2 \right )$

$\left (y - 2\sqrt{3} \right ) \left (\sqrt{3} - 1 \right ) = \left (\sqrt{3} + 1\right ) \left (x - 2 \right )$

$y\left (\sqrt{3} - 1 \right ) - 2\sqrt{3} \left ( \sqrt{3} - 1 \right ) = x \left ( \sqrt{3} + 1 \right ) - 2\left (\sqrt{3} + 1 \right )$

$\left ( \sqrt{3} + 1 \right ) x - \left ( \sqrt{3} - 1 \right ) y = 2\sqrt{3} + 2 - 6 + 2\sqrt{3}$

$\left ( \sqrt{3} + 1 \right ) x - \left ( \sqrt{3} - 1 \right ) y = 4\sqrt{3} - 4$

i.e. $\left ( \sqrt{3} + 1 \right ) x - \left ( \sqrt{3} - 1 \right ) y = 4\left (\sqrt{3} - 1 \right)$

Three vertices of a triangle are $A(1, 2), B(-3, 6)$ and $C(5, 4).$ If D, E, and C, respectively, show that the area of triangle ABC is four times the area of triangle DEF.

Given: ABC is a triangle with points

$(1, 2), (-3, 6), (5, 4)$
To prove: the area of triangle ABC is four times the area of triangle DEF
We know that
Area of triangle

$= \dfrac{1}{2} \left [ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right ]$
Then, Area of triangle ABC

$= \dfrac{1}{2} |(1) \left \{ (6 - 4) \right \} - 3(4 - 2) + 5(2 - 6) |$

$= \dfrac{1}{2} |2 - 6 - 20|$

$= \dfrac{-24}{2} = 12$
Now we have to find point D, E, F
Hence D is the midpoint of side BC then,
Coordinates of

$D = \left [ \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right ]$

$= \left [ \dfrac{-3 + 5}{2}, \dfrac{6 + 4}{2} \right ]$

$= (1, 5)$
Hence E is the midpoint of side AC then,
Coordinates of

$E = \left [ \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right ]$

$= \left [ \dfrac{1 + 5}{2}, \dfrac{2 + 4}{2} \right ]$

$= (3, 3)$
Hence F is the midpoint of side AB then,
Coordinates of

$F = \left [ \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right ]$

$= \left [ \dfrac{1 - 3}{2}, \dfrac{2 + 6}{2} \right ]$

$= (-1, 4)$
Area of triangle

$= \dfrac{1}{2} \left [ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right ]$
Now Area of triangle

$DEF = \dfrac{1}{2} |1 (3 - 4) + 3(4 - 5) - 1(5 - 3)|$

$= \dfrac{1}{2} [-1 - 3 - 2]$

$= \dfrac{1}{2} |-6|$

$= 3$
Therefore Area of

$\Delta ABC = 4$ Area of

$\Delta DEF.$
Hence, Proved.
1795260_1816253_ans_c958b537057343809387e9352ce2ebe6.png

Find the area of the quadrilateral whose vertices taken in order are $(-4, -2), (-3, -5), (3, -2)$ and $(2, 3).$

Given: The vertices of the quadrilateral be

$A(-4, -2), B(-3, -5), C(3, -2)$ and

$D(2, 3).$

Let join AC to form two triangles,

Now, We know that

Area of triangle

$= \dfrac{1}{2} \left [ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right ]$

Then, Area of triangle ABC

$= \dfrac{1}{2} |(-4) \left \{ (-5 + 2) \right \} + (-3) \left \{ ((-2) + 2) \right \} + 3 \left \{ ((-2) + 2) \right \} |$

$= \dfrac{1}{2} [12 + 0 + 9]$

$= \dfrac{21}{2}$ Square Units

Now, Area of triangle

$ACD = \dfrac{1}{2} | [ (-4) \left \{ (-2 + 3) \right \} + (3) \left \{ ((3) + 2) \right \} + 2 \left \{ ((-2) + 2) \right \} ]|$

$= \dfrac{1}{2} [20 + 15 + 0]$

$= \dfrac{35}{2}$ Square Units

Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

$= \dfrac{21}{2} + \dfrac{35}{2}$

$= \dfrac{56}{2}$

Hence, Area of quadrilateral

$ABCD = 28$ square units

Find the area of a triangle ABC if the coordinates of the middle points of the sides of the triangle are $(-1, -2), (6, 1)$ and $(3, 5).$

Given: Coordinates of middle points are

$D(-1, -2), E(6, 1)$ and

$F(3, 5).$

We know, area of triangle

$= \dfrac{1}{2} \left [ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right ]$

Now Area of triangle DEF

$= \dfrac{1}{2} | -1 (1 - 5) + 6(5 + 2) + 3(-2 - 1)|$

$= \dfrac{1}{2} [4 + 42 - 9]$

$= \dfrac{1}{2} [37] = \dfrac{37}{2}$ square units

Hence the area of

$\Delta ABC$ is four times the area of

$\triangle DEF$

$= 4\times \dfrac{37}{2}$

$= 74$ square units

Find the area of the triangle whose vertices are
$(-5, -1), (3, -5), (5, 2)$

Given

$(-1. 5, 3), (6, -2), (-3, 4)$
Let us assume

$A(x_1, y_1) = (-1.5, 3)$
Let us assume

$B(x_2, y_2) = (6, -2)$
Let us assume

$C(x_3, y_3) = (-3, 4)$
Now,
Area of given triangle

$= \dfrac{1}{2} \left [ \left \{ -1.5 (-2 - (-4)) \right \} + (6) \left \{ (4 - 3) \right \}- 3 (3 + 2) \right ]$

$= \dfrac{1}{2} \left \{ +9 + 6 - 15 \right \}$

$= \dfrac{0}{2}$

$= 0$ square units

Find the area of the triangle whose vertices are
$((a + 1)(a + 2), (a + 2)), ((a + 2)(a + 3), (a + 3))$ and $((a, 3)(a + 4), (a + 4))$

Given, A triangle whose vertices are

$A((a + 1)(a + 2), (a + 2)), B((a + 2)(a + 3), (a + 3))$ and

$C= ((a + b)(a + 4), (a + 4)).$
To find: Find the area of a triangle.
Since, Area of triangle

$= \dfrac{1}{2} \left [ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right ]$
Then,

$\dfrac{1}{2} [(a + 1)(a + 2) \left \{ (a + 3) - (a + 4) \right \} + (a + 2)(a + 3) \left \{ (a + 4) - (a + 2) \right \} + (a + 3)(a + 4) \left \{ (a + 2) - (a + 3) \right \} ]$

$= \dfrac{1}{2} [(a^2 + 3a + 2)(-1) + (a^2 + 5a + 6)(2) + (a^2 + 7a + 12)(-1) ]$

$= \dfrac{1}{2} [-a^2 - 3a - 2 + 2a^2 + 10a + 12 - a^2 + 7a + 12]$
Common terms will be canceled out

$= \dfrac{1}{2}[2]$
Hence,

$= 1$ sq unit

Find the area of the triangle whose vertices are
$(-5, 7), (-4, -5), (4, 5)$

Given

$(-5, 7), (-4, -5), (4, 5)$
Let us Assume

$A(x_1, y_1) = (-5, 7)$
Let us Assume

$B(x_2, y_2) = (-4, -5)$
Let us Assume

$C(x_3, y_3) = (4, 5)$
Area of triangle

$= \dfrac{1}{2} \left [ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right ]$
Now,
Area of given triangle

$= \dfrac{1}{2} \left [ \left \{ -5 (-5 - (5) ) \right \} + (-4) \left \{ (5 - 7) + 4 (7 + 5) \right \} \right ]$

$= \dfrac{1}{2} \left \{ 50 + 8 + 48 \right \}$

$= \dfrac{106}{2}$

$= 53$ Square units

Find the area of the triangle whose vertices are
$(2, 3), (-1, 0), (2, -4)$

Given

$(2, 3), (-1, 0), (2, -4)$
Let us Assume

$A(x_1, y_1) = (2, 3)$
Let us Assume

$B(x_2, y_2) = (-1, 0)$
Let us Assume

$C(x_3, y_3) = (2, -4)$
Area of triangle

$= \dfrac{1}{2} \left [ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right ]$
Now,
Area of given triangle

$= \dfrac{1}{2} [ \left \{ 2(0 - (-4)) \right \} + (-1) \left \{ (-4 - 3) + 2(3 - 0) \right \} ]$

$= \dfrac{1}{2} \left \{ 8 + 7 + 6 \right \}$

$= \dfrac{21}{2}$ square units

Find the area of the triangle whose vertices are
$(1, -1), (-4, 6), (-3, -5)$

Given that, the coordinates of the vertices of a triangle are

$(1, -1), (-4, 6), (-3, -5)$

Let us Assume

$A(x_1, y_1) = (1, -1)$ ,

$B(x_2, y_2) = (-4, 6)$ and

$C(x_3, y_3) = (-3, -5)$

We know that,

Area of triangle

$= \dfrac{1}{2} \left [ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right ]$
Therefore,

Area of the given triangle

$= \dfrac{1}{2} [ \left \{ 1(6 - (-5)) \right \} + \left \{(-4) (-5 - (-1))\right \} +\left \{(- 3)((-1)- 6) \right \}]$

$\Rightarrow \dfrac{1}{2} \left \{ 11 + 16 + 21 \right \}$

$\Rightarrow \dfrac{48}{2}=24$ square units

Hence, the area of the triangle with given vertices is

$24$ square units

Find the area of the triangle formed by joining the mid-points of the sides of the triangles whose vertices are $(0, -1), (2, 1)$ and $(0, 3).$ Find the ratio of this area to the area of the given triangle.

Let ABC is a triangle with points

$(0, -1), (2, 1), (0, 3)$
To Find: Ratio of area of triangle ABC to triangle DEF
We know that
Area of triangle

$= \dfrac{1}{2} \left [ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right ]$
Then, Area of triangle ABC

$= \dfrac{1}{2} |0 (1 - 3) + 2(3 + 1) + 0(-1 - 1)|$

$= \dfrac{1}{2} [8] = 4$
Now we have to find point D, E and F.
Hence D is the midpoint of side BC then,
Coordinates of

$D = \left [ \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right ]$

$= \left [ \dfrac{2 + 0}{2}, \dfrac{1 + 3}{2} \right ]$

$= (1, 2)$
Hence E is the midpoint of side AC then,
Coordinates of

$E = \left [ \dfrac{x_1 + x_2}{2}, \dfrac{y_1}{y_2}{2} \right ]$

$= \left [ \dfrac{0 + 0}{2}, \dfrac{1 + 3}{2} \right ]$

$= (0, 1)$
Hence F is the midpoint of side AB then,
Coordinates of

$F = \left [ \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right ]$

$= \left [ \dfrac{0 + 2}{2}, \dfrac{-1 + 1}{2} \right ]$

$= (1, 0)$
Area of triangle

$= \dfrac{1}{2} \left [ x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right ]$
Now Area of triangle

$DEF = \dfrac{1}{2} |x_1 (y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

$= \dfrac{1}{2} [1 + 1]$

$= \dfrac{1}{2} [2] = 1$
Therefore Area of

$\Delta ABC = 4$ Area of

$\Delta DEF.$
Then, The ratio of

$\Delta DEF$ and

$\Delta ABC = 1 : 4$

The vertices of $\Delta ABC$ are $A(3, 0), B(0, 6)$ and $(6, 9).$ A straight line DE divides AB and AC in the ratio $1 : 2$ at D and E respectively, prove that $\dfrac{\Delta ABC}{\Delta ADE} = 9$

Find the inclination of the line whose slope is $\dfrac {1}{\sqrt {3}}$ .

We know that, if

$\tan \theta$ is the slope of a line; then inclination of the line is

$\theta$

When slope of line is

$\dfrac {1}{\sqrt {3}}$ ; then

$\tan \theta = \dfrac {1}{\sqrt {3}}$

$\tan \theta = \tan 30^{\circ}$

$\theta = 30^{\circ}$

So, the inclination of the given line is

$\theta = 30^{\circ}$ .

The sides of a triangle are given by the equations $y - 2 = 0; y + 1 = 3(x - 2)$ and $x + 2y = 0$ . Find, graphically:
(i) the area of triangle;
(ii) the co-ordinates of the vertices of the triangle.

$y - 2 = 0$

$\Rightarrow y = 2$

$y + 1 = 3(x - 2)$

$\Rightarrow y + 1 = 3x - 6$

$\Rightarrow y = 3x - 6 - 1$

$\Rightarrow y = 3x - 7$
The table for

$y + 1 = 3(x - 2)$ is

$x$	$1$	$2$	$3$
$y$	$-4$	$-1$	$2$

Also we have

$x + 2y = 0$

$\Rightarrow x = -2y$
The table for

$x + 2y = 0$ is

$x$	$-4$	$4$	$-6$
$y$	$2$	$-2$	$3$

The area of the triangle

$ABC = \dfrac {1}{2}\times AB \times CD$

$= \dfrac {1}{2} \times 7\times 3$

$= \dfrac {21}{2}$

$= 10.5\ sq. units$ .
(ii) The coordinates of the vertices of the triangle

$= (-4, 2), (3, 2)$ and

$(2, -1)$ .
1800778_1843804_ans_a8884cea24c5410e87286ab8054413af.png

The points $A(3,0),B(a,-2)$ and $C(4,-1)$ are the vertices of triangle $ABC$ right angled at vertex $A$ . Find the value $a$

$AB=\sqrt { { \left( a-3 \right) }^{ 2 }+{ \left( -2-0 \right) }^{ 2 } } =\sqrt { { a }^{ 2 }+9-6a+4 } =\sqrt { { a }^{ 2 }-6a+13 } \quad$

$BC=\sqrt { { \left( 4-a \right) }^{ 2 }+{ \left( -1=2 \right) }^{ 2 } } =\sqrt { { a }^{ 2 }+16-8a+1 } =\sqrt { { a }^{ 2 }-8a+17 }$

$CA=\sqrt { { \left( 3-4 \right) }^{ 2 }+{ \left( 0+1 \right) }^{ 2 } } =\sqrt { 1+1 } =\sqrt { 2 }$
Since triangle

$ABC$ is a right angled at

$A$ we have

${ AB }^{ 2 }+{ AC }^{ 2 }={ BC }^{ 2 }\Rightarrow { a }^{ 2 }-6a+13+2={ a }^{ 2 }-8a+17\Rightarrow 2a=2\Rightarrow a=1$

Write the slope of the line whose inclination is $30^{\circ}$ .

We know that, the slope of the line is

$\tan \theta$ if

$\theta$ is the inclination of a line.

Slope is usually denoted by letter

$m$ .

Here the inclination of a line is

$30^{\circ}$ , then

$\theta = 30^{\circ}$

Hence, the slope of the line is

$m = \tan \theta = 30^{\circ} = \dfrac {1}{\sqrt {3}}$ .

Show that points P(2, -2), Q(7, 3), R(11, -1) and S (6, -6) are vertices of a parallelogram.

The given points are P(2, -2), Q(7, 3), R(11, -1) and S (6, -6).

$PQ=\sqrt{(3-(-2))^2+(7-2)^2}=\sqrt{25+25}=5\sqrt{2}$

$QR=\sqrt{(7-11)^2+(3-(-1))^2}=\sqrt{16+16}=4\sqrt{2}$

$RS=\sqrt{(11-6)^2+(-1-(-6))^2}=\sqrt{25+25}=5\sqrt{2}$

$PS=\sqrt{(2-6)^2+(-2-(-6))^2}=\sqrt{16+16}=4\sqrt{2}$
So, PQ = RS and QR = PS
Thus, opposite sides are equal.
Hence, the given points form a parallelogram.

Find the angle $P$ of the triangle whose vertices are $P(0,-1,-2), Q(3,1,4)$ and $R(5,7,1)$ .

The position vectors

$\bar{p}, \bar{q}$ , and

$\bar{r}$ of the points

$P(0,-1,-2), Q(3,1,4)$ and

$R(5,7,1)$ are

$\bar{p}=-\hat{j}-2\hat{k}$

$\bar{q}=3\hat{i}+\hat{j}+4\hat{k}$ ,

$\bar{r}=5\hat{i}+7\hat{j}+\hat{k}$

$\therefore \bar{PQ}=\bar{q}-\bar{p}$

$=(3\hat{i}+\hat{j}+4\hat{k})-(-\hat{j}-2\hat{k})$

$=3\hat{i}+2\hat{j}+6\hat{k}$

and

$\bar{PR}=\bar{r}-\bar{p}$

$=(5\hat{i}+7\hat{j}+\hat{k})-(-\hat{j}-2\hat{k})$

$=5\hat{i}+8\hat{j}+3\hat{k}$

$=\bar{PQ}.\bar{PR}=(3\hat{i}+2\hat{j}+6\hat{k}).(5\hat{i}+8\hat{j}+3\hat{k})$

$=(3)(5)+(2)(8)+(6)(3)$

$=15+16+18=49$

$|\bar{PQ}|=\sqrt{3^{2}+2^{2}+6^{2}}=\sqrt{9+4+36}=\sqrt{49}=7$

$|\bar{PR}|=\sqrt{5^{2}+8^{2}+3^{2}}=\sqrt{25+64+9}=\sqrt{98}=7\sqrt{2}$

Using the formula for angle between two vectors,

$\cos P=\dfrac{\bar{PQ}.\bar{PR}}{|\bar{PQ}||\bar{PR}|}$

$=\dfrac{49}{7\times 7\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\cos 45^{o}$

$\therefore P=45^{o}$ .

Write the slope of the line whose inclination is $60^{\circ}$ .

We know that, the slope of the line is

$\tan \theta$ if

$\theta$ is the inclination of a line.

Slope is usually denoted by letter

$m$ .

Here the inclination of a line is

$60^{\circ}$ , then

$\theta = 60^{\circ}$

Hence, the slope of the line is

$m = \tan 60^{\circ} = \sqrt {3}$ .

Prove that the points $A(1,-3),B(-3,0)$ and $C(4,1)$ are the vertices of an isosceles right-angled triangle. Find the area of the triangle.

$AB=\sqrt { { \left( -3-1 \right) }^{ 2 }+{ \left( 0+3 \right) }^{ 2 } } =\sqrt { 16+9 } =\sqrt { 25 } =5$

$BC=\sqrt { { \left( 4+3 \right) }^{ 2 }+{ \left( 1-0 \right) }^{ 2 } } =\sqrt { 49-1 } =\sqrt { 50 } =5\sqrt { 2 }$

$CA=\sqrt { { \left( 1-4 \right) }^{ 2 }+{ \left( -3-1 \right) }^{ 2 } } =\sqrt { 9+16 } =\sqrt { 25 } =5$

$\because$

$AB=CA$

$A,B,C$ are the vertices of an isosceles triangle.

${ AB }^{ 2 }+{ CA }^{ 2 }=25+25=50\quad$

${ BC }^{ 2 }={ (5\sqrt { 2 } ) }^{ 2 }=50$

${ AB }^{ 2 }+{ CA }^{ 2 }={ BC }^{ 2 }$
Hence

$A,B,C$ are the vertices of a right-angled triangle
Hence,

$\triangle ABC$ is an isosceles right-angled triangle
Area of

$\triangle ABC=\cfrac { 1 }{ 2 } \times AB\times CA=\cfrac { 1 }{ 2 } \times \times 5=12.5sq.units$

Verify that points P(-2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled triangle.

$PQ=\sqrt{(-2-2)^2+(2-2)^2}$

$=\sqrt{(-4)^2+0}$

$\sqrt{16}=4$

$OR=\sqrt{(2-2)^2+(2-7)^2}$

$=\sqrt{0+(-5)^2}$

$=\sqrt{25}$

$=5$

$PR=\sqrt{(-2-2)^2+(2-7)^2}$

$\sqrt{(-4)^2+(-5)^2}$

$\sqrt{16+25}$

$=\sqrt{41}$

$PQ^2+QR^2=PR^2$

$\Rightarrow 4^2+5^2=16+25=41=PR$
Thus, the square of the third is equal to the sum of the squares of the other two sides.
Thus, they are the vertices of the right angled triangle.

For the linear equation, given above, draw the graph and then use the graph drawn to find the area of a triangle enclosed by the graph and the co-ordinates axes:
$3x - (5 - y) = 7$ .

Given equation is

$3x - (5 - y) = 7$ .

$3x+y=12$

$\dfrac x 4+\dfrac y{12}=1$

Here

$x$ intercept is

$4$ and

$y$ intercept is

$12$

The required triangle formed will be right triangle having base and height equal to

$4$ and

$12$ respectively.

Area of the right triangle obtained will be:

$= \dfrac {1}{2}\times base\times altitude$

$= \dfrac {1}{2}\times 4\times 12$

$= 24\ sq. units$ .

1805031_1843629_ans_ae3e58a489cd456c9536e7eb097873ec.png

Find the slope of the lines passing through the given point. L (-2, -3) , M (-6, -8)

L (-2, -3) , M (-6, -8)
Slope=

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-8-(-3)}{-6-(-2)}=\dfrac{-5}{-4}=\dfrac{5}{4}$

If A (1, -1), B (0, 4), C (-5, 3) are vertices of a triangle then find the slope of each side.

A (1, -1), B (0, 4), C (-5, 3) from a triangle.
Slope of AB=

$\dfrac{4-(-1)}{0-1}=\dfrac{5}{-1}=-5$
Slope of BC=

$\dfrac{3-4}{-5-0}=\dfrac{-1}{-5}=\dfrac{1}{4}$
Slope of AC=

$\dfrac{3-(-1)}{-5-1}=\dfrac{4}{-6}=\dfrac{-2}{3}$

Find the slope of the lines passing through the given point. C (5, -2) , $\triangle$ (7, 3)

C (5, —2) ,

$\triangle$ (7, 3)
Slope =

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{3-(-2)}{7-5}=\dfrac{5}{2}$

Angles made by the line with the positive direction of X-axis is given. Find the slope of these line. $90^0$

$90^0$

$m = tan\, 90^0$ = undefined
Thus, the slope cannot be defined.

Show that points A(-4, -7), I3(-1, 2), C(8, 5) and D(5, -4) are vertices of a rhombus ABCD.

The given points are A(-4, -7), I3(-1, 2), C(8, 5) and D(5, -4).

$AB=\sqrt{(-4-(-1))^2+(-7-2)^2}=\sqrt{9+81}=\sqrt{90}=3\sqrt{10}$

$BC=\sqrt{(-1-8)^2+(2-5)^2}=\sqrt{81+9}=\sqrt{90}=3\sqrt{10}$

$CD=\sqrt{(8-5)^2+(5-(-4))^2}=\sqrt{9-81}=\sqrt{90}=3\sqrt{10}$

$DA=\sqrt{(5-(-4))^2+(-4-(-7))^2}=\sqrt{81+9}=\sqrt{90}=3\sqrt{10}$
Thus, all the sides are equal.

$AC=\sqrt{(-4-8)^2+(-7-5)^2}=\sqrt{144+144}=12\sqrt{2}$

$DB=\sqrt{(8-5)^2+(2-(-4))^2}=\sqrt{9+36}=3\sqrt{5}$
Thus, the diagonals are not equal.
So, the given vertices form a rhombus ABCD.

Find the slope of the lines passing through the given point. R(0,-3),S(0,4)

R(0,-3),S(0,4)
Slope=

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{4-(-3)}{0-0}=\dfrac{7}{0}$ =slope not defined.

Find the slope of the lines passing through the given point. P (-3, 1) , Q (5, -2)

P (-3, 1) , Q (5, -2)
Slope=

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-2-1}{5-(-3)}=\dfrac{-3}{8}$

Angles made by the line with the positive direction of X-axis is given. Find the slope of these line. $60^0$

$60^0$

$m = tan\, 60^0 =\sqrt{3}$
Thus, slope =

$\sqrt{3}$

Angles made by the line with the positive direction of X-axis is given. Find the slope of these line. $$45^0$$

$45^0$

$m = tan 45^0 = 1$
Thus, slope = 1

Find the slope of the lines passing through the given point.E(-4,-2),F(6,3)

E(-4,-2),F(6,3)
Slope=

$\dfrac{y_2-y-1}{x_2-x_1}=\dfrac{3-(-2)}{6-(-4)}=\dfrac{5}{10}=\dfrac{1}{2}$

Derive the formula for finding the angle between the lines in terms of their slopes

1976786_1865030_ans_4e5ec0718a594a439f06c4ad4d3bae98.jpg

Show that A(-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are the vertices of a parallelogram.

The given points are A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4).
Slope of AB =

$\dfrac{2-(-7)}{-1-(-4)}=\dfrac{9}{3}=3$
Slope of BC =

$\dfrac{5-2}{8-(-1)}=\dfrac{3}{9}=\dfrac{1}{3}$
Slope of CD =

$\dfrac{-4-5}{5-8}=\dfrac{-9}{-3}=3$
Slope of AD =

$\dfrac{-4-(-7)}{5-(-4)}=\dfrac{3}{9}=\dfrac{1}{3}$
Slope of AB = Slope of CD
Slope of BC = Slope of AD
So, AB II CD and BC II AD
Hence, ABCD is a parallelogram.

Show that points P(1, -2), Q(5, 2), R(3, -1), S(-1, -5) are the vertices of a parallelogram.

The given points are P(1, -2), Q(5, 2), R(3, -1), S(-1, -5).

Slope of PQ =

$\dfrac{2+2}{5-1}=\dfrac{4}{4}=1$

Slope of QR=

$\dfrac{-1-2}{3-5}=\dfrac{-3}{-2}=\dfrac{3}{2}$

Slope of RS =

$\dfrac{-5+1}{-1-3}=\dfrac{-4}{-4}=1$

Slope of PS =

$\dfrac{-5+2}{-1-1}=\dfrac{-3}{-2}=\dfrac{3}{2}$

Slope of PQ = Slope of RS and Slope of QR = Slope of PS

So, PQ II RS and QR ||PS

Thus, PQRS is a parallelogram.

Find the area of $ABC$ , where vertices are $A(4,1)B(6,6),C(8,4)(CBSE\ 2010)$

$Equation\space of$

$AB$

$\dfrac{y-6}{6-1}=\dfrac{x-6}{6-4}\Rightarrow \dfrac{y-6}{5}=\dfrac{x-6}{2}$

$2y-12=5x-30\Rightarrow y=\dfrac{5}{2}x-9$

$Similarly\space equation\space of$

$BC\Rightarrow y=12-x$

$Equation\space of$

$AC\Rightarrow y=\dfrac{3}{4}x-2$

$\therefore$

$Required\space Area =$

$\displaystyle\int_{4}^{6}\left(\dfrac{5}{2}x-9\right)dx+\int_{6}^{8}(12-x)dx-\int_{4}^{8}\left(\dfrac{3}{4}x-2\right)dx$

$=\left[\dfrac{5}{2}\dfrac{x^{2}}{2}-9x\right]_{4}^{6}+[12x-x^{2}]_{6}^{8}-\left[\dfrac{3x^{2}}{8}-2x\right]_{4}^{8}$

$=7+10-10=7\ sq.units$ .
1878458_1858345_ans_d594cac9b9cf4671963631946be094d8.png

Find k, if PQ II RS and P(2, 4), Q (3, 6), R(3, 1), S(5, k).

Given P(2, 4), Q (3, 6), R(3, 1), S(5, k)
PQ II RS so, slope of PQ = slope of RS

$\dfrac{6-4}{3-2}=\dfrac{k-1}{5-3}$

$\Rightarrow \dfrac{2}{1}=\dfrac{k-1}{2}$

$\Rightarrow k-1=4$

$\Rightarrow k=5$

The base of an equilateral triangle with side 2a lies along y -axis such that the mid- point of the base it at the origin . Find the vertices of the triangle

As 'O" is the midpoint of the side BC of an equilateral triangle

$ABC, ' A'$ should lie on the x -axis
Given that

$BV = 2a,$ then

$B = ( 0 , - a )$ and

$C = ( 0 , a )$
To find A :
Consider

$\bigtriangleup OAC$ in which

$OC = a , AC = 2a$

$\therefore OA^{2} = AC^{2} - OC^{2} = (2a)^{2} - (a)^{2}$

$\therefore oA^{2} = 4a^{2} - a^{2}= 3a^{2} \therefore OA = \pm \sqrt{3}a$

$\therefore$ vertices of the triangle are

$(0,a) , (0 , -a)$ and

$(-\sqrt{3}a, 0)$

Find k if the line passing through points P(-12,-3) and Q(4, k) has slope $\dfrac{1}{2}$ .

Slope

$\dfrac{1}{2}$ .
Given points are P(-12,-3) and Q(4, k)
Slope of PQ =

$\dfrac{k+3}{4+12}=\dfrac{1}{2}$

$\Rightarrow 2k+6=16$

$\Rightarrow 2k=10$

$\Rightarrow k=5$

Show that the line joining the points A(4, 8) and B(5, 5) is parallel to the line joining the points C(2, 4) and DO, 7).

Slope of the line joining the points A(4, 8) and B(5, 5) will be -

$=\dfrac{5-8}{5-4}=\dfrac{-3}{1}=-3$
Slope of the line joining the points C(2, 4) and D(1, 7) will be

$\dfrac{7-4}{1-2}=\dfrac{3}{-1}=-3$
Since the slope of AB = slope of CD
so, the given lines are parallel.

A line passes through $(x_{1} , y_{1} )$ and $( h,k)$ . If slope of the line is m , show that $k - y_{1} = m (h -x_{1} )$

1975252_1865522_ans_62fe5e57f82f421c8da8db7d535ad592.jpg

Find the area of the triangle whose vertices are:
$(-5 , -1) , (3 , -5) , (5 , 2)$

Let

$A (-5 , -1) = (x_1 , y_1)$

$B (3 , -5) = (x_{2} , y_{2})$

$C(5 , 2 ) = (x_{3} , y_{3})$

$Area\space of\space the\space triangle\space from\space the\space given\space data:$

$= \dfrac{1}{2} [x_1 (y_2 - y_3) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$= \dfrac{1}{2} [-5 (-5 - 2) + 3 (2 + 1) + 5 (-1 + 5)]$

$= \dfrac{1}{2} [-5 (-7) + 3 (3) + 5 (4)]$

$= \dfrac{1}{2} [+ 35 + 9 + 20]$

$= \dfrac{1}{2} \times 64$

$= 32$

$sq. units$

$\therefore$

$Area \space of$

$\triangle ABC = 32$

$sq. units$

If $(1 , 2) , (4 , y) , (x , 6)$ and $(3 , 5)$ are the vertices of a parallelogram taken in order , find $x$ and $y$ .

$Let\space A(1,2)\space B(4,y)\space C(x,6) \space D(3,5)$

$AC$

$and$

$BD$

$are\space the\space diagonals\space of\space a\space parallelogram\space$

$ABCD$

$and\space diagonals\space bisect\space at\space 'O'$ .

$\therefore$

$Coordinate\space of\space mid\space point\space of\space$

$AC$ =

$Coordinate\space of\space mid\space point\space$

$BD$ .

$As\space per\space Mid-Point\space formula\space,$

$= \left ( \dfrac{x_{1} + x_{2}}{2} , \dfrac{y_{1} + y_{2}}{2} \right )$

$= \left ( \dfrac{1 + x}{2} , \dfrac{2 + 6}{2} \right ) = \left (\dfrac{4 + 3}{2} , \dfrac{y + 5}{2} \right )$

$= \left ( \dfrac{x + 1}{2} , \dfrac{8}{2} \right ) = \left ( \dfrac{7}{2} , \dfrac{y + 5}{2} \right )$

$\therefore \, \dfrac{x + 1}{2} = \dfrac{7}{2}$

$2 (x + 1) = 7 \times 2$

$2x + 2 = 14$

$2x = 14 - 2$

$2x = 12$

$x = \dfrac{12}{2}$

$\therefore \, x = 6$

$\therefore \dfrac{8}{2} = \dfrac{y + 5}{2}$

$2 \times 8 = 2 (y + 5)$

$16 = 2y + 10$

$2y + 10 = 16$

$2y = 16 - 10$

$2y = 6$

$\therefore \, y = \dfrac{6}{2}$

$\therefore \, y = 3$

$\therefore x = 6 , y = 3$

Line through the points $(-2, 6)$ and $(4,8)$ is perpendicular to the line through the points $(8,12)$ and $( x , 24 )$ . Find the value of x

Slope of the line passing through the points

$(-2,6)$ and

$(4,8)$ is

$m_{1} = \dfrac{8-6}{4-(-2)}= \dfrac{2}{6} = \dfrac{1}{3}$ Slope of the line passing through the point

$(8,12)$ and

$(x,25)$ is

$m_{2} = \dfrac{24-21}{x-8} = \dfrac{12}{x-8}$
Since the lines are perpendicular then

$m_{1} \times m_{2} = -1$

$i.e, \dfrac{1}{3} \times \dfrac{12}{x-8} = -1$

$i.e, 4 = =x + 8 \Rightarrow x = 4$

Without using pythagoras theorem . Show that the points $( 4 , 4 ) , (3,5)$ and $( -1 , 1)$ are the vertices of a right angled triangle

1976792_1865451_ans_1d1abcb8cfde4ec8b9441e618df1304c.jpg

Find the angle between the x-axis and the line joining the point $(3,-1)$ and $( 4 , -2)$

1975272_1865495_ans_0f5fead22f924e89b7f99ec860107f5c.jpg

Find the area of the triangle whose vertices are:
$(2, 3) , (-1 , 0) , (2 , -4)$

$Let$

$A (2 , 3) = (x_1 , y_1)$

$B (-1 , 0) = (x_{2} , y_{2})$

$C(2 , -4) = (x_{3} , y_{3})$

$Area\space of\space the\space triangle\space from\space the\space given\space data:$

$= \dfrac{1}{2} [x_1 (y_2 - y_3) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$= \dfrac{1}{2} [ 2 (0 + 4) + 1 (-4 - 3) + 2 (3 - 0)]$

$= \dfrac{1}{2} [2(4) - 1 (-7) + 2 (3)]$

$= \dfrac{1}{2} [8 + 7 + 6]$

$= \dfrac{1}{2} \times 21$

$= \dfrac{21}{2}$

$sq. units$

$= 10.5$

$sq. units$

$\therefore$

$Area\space of\space$

$\triangle ABC = 10.5$

$sq. units.$

Prove that each of the set of co-ordinates are the vertices of parallelograms.
$(-5 , -3) , (1 , -11) , (7 , -6) , (1 , 2 )$

$d = \sqrt{(x_{2} - x_{1})^{2} - (y_{2} - y_{1})^{2}}$

$AB = \sqrt{(-5 - 1)^2 + (-3 + 11)^{2}}$

$= \sqrt{(-6)^{2} + (8)^{2}} = \sqrt{36 + 64} = \sqrt{100}$

$\boxed{AB = 10}$

$BC = \sqrt{(1 - 7)^{2} + (-11 - (-6))^{2}}$

$= \sqrt{(-6)^{2} + (-5)^{2}} = \sqrt{36 + 25}$

$\boxed{BC = \sqrt{61}}$

$CD = \sqrt{(7 - 1)^{2} + (-6 - 2)^{2}}$

$= \sqrt{(6)^{2} + (-8)^{2}} = \sqrt{36 + 64}$

$CD = \sqrt{100}$

$\boxed{CD = 10}$

$DA = \sqrt{(1 + 5)^{2} + (2 + 3)^{2}}$

$= \sqrt{(6)^{2} + (5)^{2}}$

$= \sqrt{36 + 25} = \sqrt{61}$

$\therefore \, AD = BC\ \&\ AB = CD$

$\therefore$ Given vertices from a parallelogram.
1844359_1868657_ans_2abbffd31e4f4f58ba8d0075c30af457.png

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are $(0 , -1) , (2 , 1)$ and $(0 , 3)$ . Find the ratio of this area to the area of the given triangle.

$Let\space the\space mid-points\space of\space$

$\triangle ABC$

$are\space P ,\space Q\space , R\space and\space also\space the\space midpoints\space of \space AB\space , BC \space and \space AC.$

$As\space per\space Mid-point \space formula,$

$Coordinates \space of$

$P = \left ( \dfrac{0 + 2}{2} , \dfrac{1 + 1}{2} \right )$

$= (1 , 1)$

$Coordinates \space of$

$Q = \left ( \dfrac{0 + 2}{2} , \dfrac{1 + 3}{2} \right )$

$= \left ( \dfrac{2}{2} , \dfrac{4}{2} \right )$

$= (1 , 2)$

$Coordinates \space of\space$

$R = \left ( \dfrac{0 + 0}{2} , \dfrac{1 + 3}{2} \right )$

$= \left ( \dfrac{0}{2} , \dfrac{4}{2} \right )$

$= (0 , 2)$

$\therefore$

$P(1 , 1) , Q( 1 , 2)$ and

$R(0 , 2)$
Area of

$\triangle PQR$

$= \dfrac{1}{2} [ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$= \dfrac{1}{2}[1 (2 - 2) + 1 (2 - 1) + 0 (1 - 2)]$

$= \dfrac{1}{2}[ 1 (0) + 1 (+1) + 0 (-1)]$

$= \dfrac{1}{2}[0 + 1 + 0]$

$= \dfrac{1}{2} \times 1$

$\therefore$

$Area$

$\triangle PQR = \dfrac{1}{2}$

$sq. units$ .

$Area \space of$

$\triangle ABC$

$= \dfrac{1}{2} [ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$= \dfrac{1}{2} [0 (1 - 3) + 2(3 - 1) + 0(1 - 1)]$

$= \dfrac{1}{2} [ 0 (-2) + 2 (2) + 0(0)]$

$= \dfrac{1}{2} [0 + 4 + 0]$

$= \dfrac{1}{2} \times 4$

$= 2$

$sq. units.$

$\therefore$

$\dfrac{{\text{$$Area of$$}} \triangle PQR}{{\text{Area of}} \triangle ABC} = \dfrac{1/2}{2}$

$= \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$

$Area\space of$

$\triangle PQR$ :

$Area\space of$

$\triangle ABC$

$\therefore \, 1 : 4$
1836183_1867961_ans_495119f79e064936918d567525c3f01b.png

Check whether $4^n+15n-1$ is divisible by $9$ or not.

Proving by induction

for

$n=1, 4^n+5n-1=18$

let for

$n=R, 4^R+15R-1=9 \lambda_1$

for

$n=RH=4^{RH}+15 (TH)-1$

$=4.4^R+15R+14$

$=4(9 \lambda_1+1-15R)+15R+14$

$=36 \lambda_1 +4-60R +15R+14=36\lambda_1 -45R+18$

$=9(4 \lambda_1-5R+2)=9 \lambda_2$

Find area of triangle, whose vertices are:
$(2,5),(-2,-3)$ and $(6,0)$

$\triangle =\begin{vmatrix} 2 & 5 & 1 \\ -2 & -3 & 1 \\ 6 & 0 & 1 \end{vmatrix}$

$=\dfrac{1}{2} \left\{ 2 (-3-0 -5 (-2-6)+1 (0+18)\right\}$

$=\dfrac{1}{2} (-6+40+18)$

$=\dfrac{1}{2} (52)=25\ sq. unit$

1914252_1871919_ans_98c3030b28f3424c8a75039821f5d52a.png

Find area of triangle, whose vertices are:
$(0,0),(5,0)$ and $(3,4)$

Area of triangle
area

$(\triangle) = \dfrac{1}{2}\begin{vmatrix} 0 & 0 & 1 \\ 5 & 0 & 1 \\ 3 & 4 & 1 \end{vmatrix}$

$=\dfrac{1}{2} \left\{ 0 (0-4) -0 (5-3)+1 (20-0) \right\}$

$=\dfrac{1}{2} (0-0+20)$

$=\dfrac{20}{2} (10)$

$=\dfrac{11}{2} \ sq. unit$

1914257_1871922_ans_2920e349ef134ec391569c4bd7845d62.png

Find area of $\triangle ABC$ , if vertices are:
$A(2,7), B(2,2),C(10,8)$

Vertices of

$\triangle ABC$

$A=(2,7), B=(2,1),C=(10,8)$

Area of

$\triangle ABC$

$=\dfrac{1}{2} \begin{vmatrix} 2 & 7 & 1 \\ 2 & 2 & 1 \\ 10 & 8 & 1 \end{vmatrix}$

$=\dfrac{1}{2} \left\{2 (2-8)-7(2-10)+1(16-20)\right\}$

$=\dfrac{1}{2} (-12+56-4)$

$=\dfrac{1}{2} (40)=20\ sq. unit$

Show that $\triangle ABC$ with vertices $A(-2 , 0) , B(2 , 0)$ and $C(0 , 2)$ is similar to $\triangle DEF$ with vertices $D(-4 , 0) , E(4 , 0)$ and $F(0 , 4)$

Given vertices of

$\triangle ABC$ and

$\triangle DEF$ are

$A(-2 , 0) , B(2 , 0) , C (0 , 2) , D (-4 , 0) , E(4 , 0)$ and

$F(0 , 4)$

$AB = \sqrt{(0 + 2)^{2} + (2 - 0)^{2}}$

$= \sqrt{4 + 4} = 2 \sqrt{2}$ units.

$BC = \sqrt{(2 - 0)^{2} + (0 - 2)^{2}}$

$= \sqrt{4 + 4} = 2 \sqrt{2}$ units

$CA = \sqrt{(-2 - 2)^{2} + (0 - 0)^{2}}$

$= \sqrt{(-4)^{2} + 0} = 4$ units

$FD = \sqrt{(0 + 4)^{2} + (0 - 4)^{2}}$

$= \sqrt{4^{2} + (-4)^{2}} = 4 \sqrt{2}$ units

$FE = \sqrt{(4 - 0)^{2} + (0 - 4)^{2}}$

$= \sqrt{4^{2} + (-4)^{2}} = 4 \sqrt{2}$ units

$DE = \sqrt{(-4 - 4)^{2} + (0 - 0)^{2}}$

$= \sqrt{(-8)^2} = 8$ units.

Here, we see that sides of

$\triangle DEF$ are twice sides of a

$\triangle ABC$ .
Hence, both triangle are similar.
1844404_1868677_ans_5c14f51710444500b8e60fa47178a0e6.png

Find area of $\triangle ABC$ , if vertices are:
$A(-3,5), B(3,-6),C(7,2)$

Vertices of

$\triangle ABC$

$A=(-3,5), B=(3,-6),C=(7,2)$

Area of

$\triangle ABC$

$=\dfrac{1}{2} \begin{vmatrix} -3 & 5 & 1 \\ 3 & -6 & 1 \\ 7 & 2 & 1 \end{vmatrix}$

$=\dfrac{1}{2} \left\{-3 (-6-2)-5(3-7)+1(6+42)\right\}$

$=\dfrac{1}{2} (24+20+48)$

$=\dfrac{1}{2} (92)=46\ sq. unit$

Find area of triangle, whose vertices are:
$(3,8),(2,7)$ and $(5,-1)$

$\triangle \dfrac{1}{2}=\begin{vmatrix} 3 & 8 & 1 \\ 2 & 7 & 1 \\ 5 & -1 & 1 \end{vmatrix}$

$=\dfrac{1}{2} \left\{ 3 (7+1) -8 (2-5)+1 (-2-35) \right\}$

$=\dfrac{1}{2} (25+25-37)$

$=\dfrac{1}{2} (11)=\dfrac{11}{2} \ sq. unit$

1914254_1871921_ans_c9f7db3618164872aa08f5ec9bbef4f2.png

If $A(4 , 2) , B = (7 , 6)$ and $C(1 , 4)$ are the vertices of $\triangle ABC$ and $AD$ is its median. P.T that median $AD$ divides $\triangle ABC$ into two triangle of equal areas.

D is the median on BC.
The Co-ordinates of midpoint D are given by

$\left [\dfrac{x_{2} + x_{1}}{2} , \dfrac{y_{2} + y_{1}}{2} \right ]$ i.e.,

$\left [\dfrac{1 + 7}{2} , \dfrac{4 + 6}{2} \right ]$

Co-ordinates of D are

$(4 , 5)$
Now, Area of triangle

$ABD = \dfrac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})$

$= \dfrac{1}{2} [4(6 - 5) + 7(52) + 4(2 - 6)]$

$= \dfrac{1}{2} [ 4 + 21 - 16] = \dfrac{9}{2}$ sq. units

Area of

$\triangle ACD = \dfrac{1}{2}[4 - 5] + 1[5 - 2] + 4[2 - 4]]$

$= \dfrac{1}{2} [-4 + 3 - 8] = \dfrac{-9}{2} = \dfrac{9}{2}$ sq. units

∴ Area of

$△ABD=$ Area of

$△ACD$

Hence, AD divides △ABC into two equal areas.

If coordinates of $P$ and $Q$ are $(3,4)$ and $(12,4)$ respectively, then find $\angle POQ$ where $O$ is origin.

Coordinates of point

$P=(3.4)$
Then vector

$\vec{OP}=3\hat{i}+4\hat{j}$
Again coordinates of

$Q=(12,9)$
Then vector

$OQ=12\hat{i}+9\hat{j}$
Let angle between

$\vec{OP}$ and

$\vec{OQ}$ is

$\angle POQ=\theta$

$\cos \theta=\dfrac{\vec{OP}.\vec{OQ}}{|\vec{OP}|.|\vec{OQ}|}$

$\Rightarrow \cos \theta=\dfrac{3\hat{i}+4\hat{j}).(12\hat{i}+9\hat{j})}{|3\hat{i}+4\hat{j}||12\hat{i}+9\hat{j}|}$

$\Rightarrow \cos \theta=\dfrac{(3\times 12)+(4\times 9)}{\sqrt{(3)^{2}+(4)^{2}}\sqrt{(12)^{2}+(9)^{2}}}$

$\Rightarrow \cos \theta=\dfrac{36+36}{\sqrt{25}\sqrt{225}}$

$\Rightarrow \cos \theta=\dfrac{72}{5\times 15}$

$\Rightarrow \theta=\cos^{-1}\left(\dfrac{72}{75}\right)$

$\Rightarrow \angle POQ=\cos^{-1}\left(\dfrac{24}{25}\right)$
1860268_1872859_ans_505b33c8b17643d1824f35a1ea77f1e3.png

The vertices of a $\Delta ABC$ are $A(3, 8), B(-1, 2)$ and $C(6, -6)$ . Find Slope of BC.

Vertices of a

$\Delta ABC$ are

$A(3,8),B(-1,2)$ and

$C(6,-6)$

Slope of the line

$=\dfrac{y_2-y_1}{x_2-x_1}$

Slope of

$BC=\dfrac{-6-2}{6+1}=\dfrac{-8}{7}$

Let the length of the sides of a triangle $\triangle ABC$ be integers with $A$ as the origin. $(2, -1)$ and $(3, 6)$ are points on the line $AB$ and $AC$ respectively (line $AB$ and $AC$ may be extended to contain these points), and lengths of exactly two sides are primes that differ by $50$ . If $a$ is least possible length of the third side and $S$ is the least possible perimeter of the triangle, then $aS$ is equal to

Since

$(AB)^{2}+(AC)^{2}=50=(BC)^{2}$ the lines AB and AC are at right angles.
It is given that all sides are integers and since the triangle is right angled, they form a pythagorean triplet,
so one leg must be even. As two sides are primes, both cannot be the legs of the right angled triangle.
So, let

$AC=p$ , a prime,

$BC=p+50$ and

$AB=a$ (even)
then

$a^{2}=(p+50)^{2}-p^{2}=100(p+25)$

$\Rightarrow a=10\sqrt{p+25}$ .
Since a is least possible integer and p is a prime
we get

$a=60$ when

$p=11$ .
So smallest value of the third side is 60.

$S=60+11+61=132$

$aS=60\times 132=7920$

Find the area of the figure bounded by the following curves
y = $x^3, \, y^2$ = x.

$y={ x }^{ 3 }\quad \quad { y }^{ 2 }=x$

${ x }^{ 6 }=x$

${ x }^{ 6 }-x=0$

$x\left( { x }^{ 5 }-1 \right) =0$

$x=0$ or

$x=1$

So,

$y=0$ at

$x=0$ and

$y=1$ at

$x=1$

$=\int _{ 0 }^{ 1 }{ \left( { x }^{ 3 }-\sqrt { x } \right) } dx$

$=\dfrac { { x }^{ 4 } }{ 4 } -\dfrac { { \left( x \right) }^{ 3/2 } }{ 3/2 } { ] }_{ 0 }^{ 1 }$

$=\dfrac { 1 }{ 4 } -\dfrac { 2 }{ 3 } -0$

$=\dfrac { 3-8 }{ 12 } =\left| -\dfrac { 5 }{ 12 } \right| =\dfrac { 5 }{ 12 }$ sq units.

Find if possible, the slope of the line through the points $(1/2,3/4)$ and $(-1/3,5/4)$

Since slope of line passing through two points

$\left( { x }_{ 1 },{ y }_{ 1 } \right)$ and

$\left( { x }_{ 2 },{ y }_{ 2 } \right)$ is

${ m }=\dfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$

We now find the slope of the line passing through the points

$\left( \dfrac { 1 }{ 2 } ,\dfrac { 3 }{ 4 } \right)$ and

$\left( -\dfrac { 1 }{ 3 } ,\dfrac { 5 }{ 4 } \right)$ as shown below:

${ m }=\dfrac { \frac { 5 }{ 4 } -\frac { 3 }{ 4 } }{ -\frac { 1 }{ 3 } -\frac { 1 }{ 2 } } =\dfrac { \frac { 2 }{ 4 } }{ -\frac { 5 }{ 6 } } =\dfrac { \frac { 1 }{ 2 } }{ -\frac { 5 }{ 6 } } =\dfrac { 1 }{ 2 } \times -\dfrac { 6 }{ 5 } =-\dfrac { 6 }{ 10 } =-\dfrac { 3 }{ 5 }$

Therefore, the slope of the line is

$-\dfrac { 3 }{ 5 }$ .

Write an equation of the vertical line through the point $(3,0)$

Here the point lie on

$x-axis$ and equation of

$x-axis$ is

$y=0$

where SLope is Zero

$m=0$

The slope of vertical line means perpendicular line to

$x-axis$ is

$m_{1}\times 0=-1$

$m_{1}=\infty$

The equation of line from

$(3,0)$ and slope

$m_{1}=\infty$

$y=\dfrac{1}{0}(x-3)$

$0=x-3$

$x=3$

A balloon moving in a straight line passes vertically above two points $A$ and $B$ on a horizontal plane $1000\ ft.$ When above $A$ it has an altitude of ${60}^{o}$ as seen from $B$ and when above $B$ , ${30}^{o}$ as seen from $A$ . Find the distance from $A$ at which it will strike the plane.

A balloon passes vertically above A nd B on a horizontal plane of

$1000$ ft, When above A its angle is

$60^o$ from B. When above B its angle is

$30^o$ from A.

To find: distance form A at which it will strike the plane.

solution : Let the balloon is at P and Q while, passing above A and B respectively.

Let R be the point at which it will strike the plane and let

$BR=xft.$

In triangle APB,

$\dfrac{AP}{AB}=\tan 60^o$

$\therefore AP=1000\sqrt 3$

$\triangle APB \sim \triangle AQB$

$$\therefore \dfrac{AB}{BQ}=\dfrac{AP}{AB}$$

Or,

$BQ=\dfrac{1000\times 1000}{1000\sqrt 3}[\because AB=1000 and \, AP =1000\sqrt 3]$

$\therefore BQ=\dfrac{1000}{\sqrt 3}$

Again

$\triangle ABP \sim \triangle RBQ$

$\therefore \dfrac{AP}{BQ}=\dfrac{AR}{BR}$

Or ,

$\dfrac{1000\sqrt 3}{\dfrac{1000}{\sqrt 3}}=\dfrac{1000+x}{x}$

$3x=1000+x$

$\therefore x=500$

$\therefore AR=AB+BR=1000+500$

$1500ft$

Hence it will strike at

$1500ft$ from A.

2114641_1006460_ans_25885476e4844531b9114855f782bdde.png

If two vertices of an equilateral triangle are $(0,0),(3,\sqrt {3})$ , find the third vertices of the triangle.

Two vertices of an equilateral triangle are

$(0, 0) and (3, √3).$

Let the third vertex of the equilateral triangle be

$(x, y)$

Distance between

$(0, 0) and (x, y)$ = Distance between

$(0, 0) and (3, √3)$ = Distance between

$(x, y) and (3, √3)$

$\sqrt { (x^2+y^2) } =\sqrt { (3^2+3) } =\sqrt{(x-3)^2+(y-\sqrt{3})^2}$

${ x }^{ 2 }+{ y }^{ 2 }=12$

${ x }^{ 2 }+9-6x+{ y }^{ 2 }+3-2\sqrt { 3y } =12$

$24-6x-2\sqrt { 3y } =12$

$-6x-2\sqrt { 3y } =-12$

$3x+\sqrt { 3y } =6$

$x=\frac { 6-\sqrt { 3y } }{ 3 }$

$\frac { (6-\sqrt { 3y } ) }{ 3 } 2+{ y }^{ 2 }=12$

$\frac { (36+3y2-12\sqrt { 3y } ) }{ 9 } +{ y }^{ 2 }=12$

$36+3{ y }^{ 2 }-12\sqrt { 3y } +9y2=108$

$-12\sqrt { 3y } +12{ y }^{ 2 }-72=0$

$\sqrt { -3y } +y2-6=0$

$(y-2\sqrt { 3 } )(y+\sqrt { 3 } )=0$

$y=2\sqrt { 3 } or\sqrt { -3 }$

$Ify=2\sqrt { 3 } ,x=(6-6)/3=0$

$Ify=-\sqrt { 3 } ,x=\frac { 9 }{ 3 } =3$

So,the third vertex of the equilateral triangle

$=(0,2\sqrt { 3 } )or(3,-\sqrt { 3 } ).$

The base AB of the two equilateral triangles ABC and ABC' with side $2a$ lies along the X-axis such that the mid-point of AB is at the origin. Find the coordinates of the vertices C and C' of the triangles.

Since the mid-point of AB is at the origin O and AB=2a

$\therefore$ OA=OB=a.
Thus, the coordinates of A and B are

$(a,0)$ and

$(-a,0)$ respectively.
Since triangles ABC. and ABC' are equilateral. Therefore, their third vertices C and C'
lie on the perpendicular bisector of base AB. Clearly, YOY is the perpendicular bisector
of AB. Thus, C and C' lie on Y-axis. Consequently, their x-coordinates are equal to Zero.
In

$\triangle AOC,\quad we\quad have$

$O{ A }^{ 2 }+{ OC }^{ 2 }={ AC }^{ 2 }$

$\Rightarrow \quad \quad { a }^{ 2 }+{ OC }^{ 2 }={ \left( 2a \right) }^{ 2 }$

$\Rightarrow \quad \quad { OC }^{ 2 }={ 4a }^{ 2 }-{ a }^{ 2 }$

$\Rightarrow \quad \quad { OC }^{ 2 }={ 3a }^{ 2 }$

$OC=\sqrt { 3a}$
Similarly, by applying Pythagoras theorem in

$\triangle$ AOC; we have

$OC'=\sqrt { 3a}$
Thus, the coordinates of C and C' are

$\left( 0,\sqrt { 3 } a \right) and\quad \left( 0,-\sqrt { 3 } a \right) respectively.$

Find the area of a triangle : $y=x,y=2x$ and $y=3x+4$ ?

First, we will find out the co-ordinates of the triangle’s three vertices. These occur at the points that a pair of lines intersect.

Vertex1:Thisiswherethelines

$y=x$ and

$y=2x$ intersect.Thishappenswhen

$x=2x⇒x=0⇒y=0.$ Thusthepointis

$(0,0).$

Vertex2:Thisiswherethelines

$y=x$ andy=3x+4intersect.Thishappenswhenx=3x+4⇒x=−2⇒y=−2.Thusthepointis(-2,-2).

Vertex3:Thisiswherethelinesy=2xand

$y=3x+4$ intersect.Thishappenswhen

$2x=3x+4⇒x=−4⇒y=−8.$ Thusthepointis

$(-4,-8)$ .

The circumcentre of a triangle is the point

$(a,b)$ such that a circle of radius r centre on

$(a,b)$ passes through the three vertices.

Every point on this circle satisfies the generic equation:

${ (x−a) }^{ 2 }+{ (y−b) }^{ 2 }={ r }^{ 2 }$ .Expanding this equation,we have

${ x }^{ 2 }−2ax+{ a }^{ 2 }+{ y }^{ 2 }−2by+{ b }^{ 2 }={ r }^{ 2 }.$

At vertex1,

$x=0$ and

$y=0$ ,thus we have Equation

$A:{ a }^{ 2 }+{ b }^{ 2 }={ r }^{ 2 }$

At vertex2,

$x=−2$ and

$y=−2$ ,thus we have Equation

$B:4+4a+{ a }^{ 2 }+4+4b+{ b }^{ 2 }={ r }^{ 2 }$

Atvertex3,

$x=−4$ and

$y=−8$ ,thuswehaveEquation

$C:16+8a+{ a }^{ 2 }+64+16b+{ b }^{ 2 }={ r }^{ 2 }$

Subtracting Equation A from Equation B,we have Equation

$D:8+4a+4b=0$

Subtracting Equation B from Equation C,we have Equation

$E:72+4a+12b=0$

Subtracting Equation D from Equation E,we have Equation

$F:64+8b=0⇒b=−8$

Substituting this into Equation

$D:8+4a−32=0⇒a=6$

So,thecircumcentreis

$(6,−8).$

As a bonus,using Equation A,the radius of the circle is 10.

If the vertices of a triangle be $(am_{1}^{2},2am_{1}),(am_{2}^{2},2am_{2})$ and $(am_{3}^{2},2am_{3})$ , then the area of the triangle is

Distance,

$d=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } }$

Here\quad

${ x }_{ 1 }=a,{ x }_{ 2 }=-a,{ y }_{ 1 }=b,{ y }_{ 2 }=-b$

$d=\sqrt { { \left( -a-a \right) }^{ 2 }+{ \left( -b-b \right) }^{ 2 } }$

$d=\sqrt { { 4a }^{ 2 }+4{ b }^{ 2 } }$

$d=2\sqrt { { a }^{ 2 }+{ b }^{ 2 } }$

If $A({x}_{1},{y}_{1}),\ B({x}_{2},{y}_{2})$ and $C({x}_{3},{y}_{3})$ are vertices of an equilateral triangle whose each side is equal to $a$ , the prove that ${ \begin{bmatrix} { x }_{ 1 } & { y }_{ 1 } & 2 \\ { x }_{ 2 } & { y }_{ 2 } & 2 \\ { x }_{ 3 } & { y }_{ 3 } & 2 \end{bmatrix} }^{ 2 }={ 3a }^{ 4 }$ .

The area of triangle with vertices

$(x_{1}y_{1})(x_{2},y_{2})$ and

$(x_{3}, y_{3})$ is

$\triangle = \dfrac{1}{2}$

${ \begin{bmatrix} { x }_{ 1 } & { y }_{ 1 } & 1 \\ { x }_{ 2 } & { y }_{ 2 } & 1 \\ { x }_{ 3 } & { y }_{ 3 } & 1\end{bmatrix} }...(1)$

The area of equilateral triangle with side 'a' is

$\dfrac{\sqrt{3}}{4}a^{2}...(2)$

We have to prove

${ \begin{bmatrix} { x }_{ 1 } & { y }_{ 1 } & 2 \\ { x }_{ 2 } & { y }_{ 2 } & 2 \\ { x }_{ 3 } & { y }_{ 3 } & 2 \end{bmatrix} }^{ 2 }={ 3a }^{ 4 }$

From (1) & (2)

$\dfrac{1}{2}$

${ \begin{bmatrix} { x }_{ 1 } & { y }_{ 1 } & 1 \\ { x }_{ 2 } & { y }_{ 2 } & 1 \\ { x }_{ 3 } & { y }_{ 3 } & 1 \end{bmatrix} }$

$= \dfrac{\sqrt{3}}{4} a^{2}$

$\triangle^{2} = \dfrac{1}{4}$

${ \begin{bmatrix} { x }_{ 1 } & { y }_{ 1 } & 1 \\ { x }_{ 2 } & { y }_{ 2 } & 1 \\ { x }_{ 3 } & { y }_{ 3 } & 1 \end{bmatrix} }^{2}$

$= \dfrac{3}{16} a^{4}$

$\dfrac{1}{16}$

${ \begin{bmatrix} { x }_{ 1 } & { y }_{ 1 } & 2 \\ { x }_{ 2 } & { y }_{ 2 } & 2 \\ { x }_{ 3 } & { y }_{ 3 } & 2 \end{bmatrix} }^{2} = \dfrac{3}{16} a^{4}$

${ \begin{bmatrix} { x }_{ 1 } & { y }_{ 1 } & 2 \\ { x }_{ 2 } & { y }_{ 2 } & 2 \\ { x }_{ 3 } & { y }_{ 3 } & 2 \end{bmatrix} }^{ 2 }={ 3a }^{ 4 }$ .

1118024_1041305_ans_8f52cef12a60403d8e527d0a94282f36.jpg

If $P,Q,R$ are collinear points such that $P(3,4),Q(7,7)$ and $PR=10$ , find $R$

Given ,

$p\left(3,4\right), Q\left(7,7\right)$ and

$R\left(x,y\right) \left(say\right)$ are

Colliuear

$\therefore \dfrac{1}{2}\left| {x}_{1}\left({y}_{2}-{y}_{3}\right)+ {x}_{2}\left({y}_{3}-{y}_{1}\right)+ {x}_{3}\left({y}_{1}-{y}_{2}\right)\right|=0$

Or,

$\therefore \dfrac{1}{2}\left|3\left(7-y\right)+ 7\left( y-4\right)+ x \left(4-7\right)\right|=0$

Or,

$21-3y+7y-28+4x-7x=0$

Or,

$4y-3x-7=0 \Rightarrow 3x=4y-7 \Rightarrow \boxed{x=\dfrac{4y-7}{3}}$

Again ,

$PR=10$

$\sqrt{{\left({x}_{1}-{x}_{3}\right)}^{2}}+\sqrt{{\left({y}_{1}-{y}_{3}\right)}^{2}}=10$

Squaring both the side ,

${\left(3-x\right)}^{2}+{\left(4-y\right)}^{2}=100$

Or,

$9-6x+{x}^{2}+16+{y}^{2}-8y=100$

Or,

${x}^{2}+{y}^{2}-6x-8y=75$

Or,

$\left({\dfrac{4y-7}{3}}\right)^{2}+{y}^{2}-6\left(\dfrac{4y-7}{3}\right)^{2}-8y=75$

Or,

$\dfrac{16{y}^{2}-56y+49}{9}+{y}^{2}-\left(8y-14\right)-8y=75$

Or,

$16{y}^{2}-56y+49+9{y}^{2}-144y+126=675$

Or,

$25{y}^{2}-200y-500=0$

Or,

${y}^{2}-8y-20=0$

Or,

${y}^{2}-10y+2y-20=0$

Or,

$y\left(y-10\right)+2\left(y-10\right)=0$

Or,

$\left(y+2\right) \left(y-10\right)=0 \quad \Rightarrow \boxed{y=-2,10}$

When

$y=-2 \quad \quad \quad \quad \quad$ when

$y=10$

$x=\dfrac{4\left(-2\right)-7}{3}=-5. \quad \ \ \ x=\dfrac{4\left(10\right)-7}{3}=-11.$

$\therefore$ coordinates of

$R$ are

$\left(-5,-2\right)$ and

$\left(11,10\right)$

The coordinate of points $A,\ B,\ C$ and $D$ are $(-3,\ 5),\ (4,\ -2), (x,\ 3x)$ and $(6,\ 3)$ respectively and $\dfrac {\triangle ABC}{\triangle BCD}=\dfrac {2}{3}$ , find $x$

Given

$A(-3,5)\: B(4,-2)\: C(x,3x), D(6,3)$

Area of

$\triangle$

$ABC = \dfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right |$

$= \dfrac{1}{2}[-3(-2-3x)+4(3x-5)+x(5+2)]$

$= \dfrac{1}{2}[ 6+9x+12x-20+5x+2x]= \dfrac{28x-14}{2}= 14x-7$

Area of

$\triangle$

$BCD=\dfrac{1}{2}[4(3x-3)+x(3+2)+6(-2-3x)]$

$=\dfrac{1}{2}[ 12x-12+3x+2x-12-18x]$

$= \dfrac{-x-24}{2}$

$\dfrac{\triangle ABC}{\triangle BCD}= \dfrac{2}{3}\Rightarrow \dfrac{14x-7}{\dfrac{-x-24}{2}}=\dfrac{2}{3}\Rightarrow 3(14x-7)=2\times \dfrac{-x-24}{2}$

$42x-21= -x-24\Rightarrow 42x-21+x+24=0$

$43x+3=0$

$x=-\dfrac{3}{43}$

Find the point on the y-axis which is equidistant from the points $(5,-2)$ and $(-3,2)$

Let the given point be P(5,-2) and Q(-3,2)

And the point required be R(0,a)

As per question,

Point R is equidistant from P & Q

Hence,

$PR=QR$

Finding PR

$x_1=5,\ y_1=-2$

$x_2=0,\ y_2=a$

$PR=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$=\sqrt{(0-5)^2+(a+2)^2}$

$=\sqrt{25+a^2+4a+4}$

$=\sqrt{a^2+4a+29}$

Finding QR

$x_1=-3,\ y_1=2$

$x_2=0,\ y_2=a$

$PR=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$=\sqrt{(0+3)^2+(a-2)^2}$

$=\sqrt{9+a^2+4-4a}$

$=\sqrt{a^2-4a+13}$

As per question

$PR=QR$

$\sqrt{a^2+4a+29}=\sqrt{a^2-4a+13}$

Squaring both sides

$a^2+4a+29=a^2-4a+13$

$4a+4a=13-29$

$8a=-16$

$a=\dfrac{-16}{8}$

Hence the required point is

$R(0,-2)$

Find the distance of the line $4x + 7y + 5 = 0$ from the point (1, 2) along the line $2x - y = 0$ .

2040857_1410081_ans_75c137fb9a6841eebb500ec420403e6e.jpg

Using ruler and compasses only, construct an angle of measure $135^{\circ}$

2039688_1413159_ans_437fb697caa54f019b08272ee823d851.jpg

If $P,Q,R$ are $\left(6,-1\right),\,\left(1,3\right),\,\left(x,8\right)$ respectively, then find the value of $x$ so that $PQ=QR$

Distance between two points

$=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$

Given:

$PQ=QR$

$\Rightarrow \sqrt{{\left(1-6\right)}^{2}+{\left(3+1\right)}^{2}}=\sqrt{{\left(x-1\right)}^{2}+{\left(8-3\right)}^{2}}$

$\Rightarrow \sqrt{{\left(-5\right)}^{2}+{\left(4\right)}^{2}}=\sqrt{{\left(x-1\right)}^{2}+{\left(5\right)}^{2}}$

$\Rightarrow \sqrt{25+16}=\sqrt{{x}^{2}-2x+1+25}$

$\Rightarrow \sqrt{41}=\sqrt{{x}^{2}-2x+26}$

squaring both sides, we get

$\Rightarrow {x}^{2}-2x+26-41=0$

$\Rightarrow {x}^{2}-2x-15=0$

$\Rightarrow {x}^{2}+3x-5x-15=0$

$\Rightarrow x\left(x+3\right)-5\left(x+3\right)=0$

$\Rightarrow \left(x+3\right)\left(x-5\right)=0$

$\therefore x=-3,\,5$

If the equation of the locus of a point equidistant from the points $\left({a}_{1},{b}_{1}\right)$ and $\left({a}_{2},{b}_{2}\right)$ is $\left({a}_{1}-{a}_{2}\right)x+\left({b}_{1}+{b}_{2}\right)y+c=0$ , then the value of $c$ is?

Let

$P\left(h,k\right)$ be the required point and let

$A\left({a}_{1},{b}_{1}\right)$ and

$B\left({a}_{2},{b}_{2}\right)$ be the given points.

Now

$PA=PB$

$\sqrt{{\left(h-{a}_{1}\right)}^{2}+{\left(k-{b}_{1}\right)}^{2}}=\sqrt{{\left(h-{a}_{2}\right)}^{2}+{\left(k-{b}_{2}\right)}^{2}}$

${\left(h-{a}_{1}\right)}^{2}+{\left(k-{b}_{1}\right)}^{2}={\left(h-{a}_{2}\right)}^{2}+{\left(k-{b}_{2}\right)}^{2}$

by squaring on both sides

$\Rightarrow {h}^{2}+{a}_{1}^{2}-2h{a}_{1}+{k}^{2}+{b}_{1}^{2}-2k{b}_{1}={h}^{2}+{a}_{2}^{2}-2h{a}_{2}+{k}^{2}+{b}_{2}^{2}-2k{b}_{2}$

on expanding

$\Rightarrow {a}_{1}^{2}-{a}_{2}^{2}+{b}_{1}^{2}-{b}_{2}^{2}-2h{a}_{1}+2h{a}_{2}-2k{b}_{1}+2k{b}_{2}=0$

$\Rightarrow \left({a}_{1}-{a}_{2}\right)h+\left({b}_{1}-{b}_{2}\right)k+\dfrac{1}{2}\left({a}_{1}^{2}-{a}_{2}^{2}+{b}_{1}^{2}-{b}_{2}^{2}\right)=0$

Replace

$h\rightarrow x$ and

$k\rightarrow y$

$\Rightarrow \left({a}_{1}-{a}_{2}\right)x+\left({b}_{1}-{b}_{2}\right)y+\dfrac{1}{2}\left({a}_{1}^{2}-{a}_{2}^{2}+{b}_{1}^{2}-{b}_{2}^{2}\right)=0$

Comparing the above equation with

$\left({a}_{1}-{a}_{2}\right)x+\left({b}_{1}-{b}_{2}\right)y+c=0$

then

$c=\dfrac{1}{2}\left({a}_{1}^{2}-{a}_{2}^{2}+{b}_{1}^{2}-{b}_{2}^{2}\right)$

If (a,b) (x $_1,y_1),(x_2, y_2)$ are vertices of triangle . Where a, x $_1, x_2$ , are in G . P . with common ratio 'r' and b, y $_1, y_2$ are in G.P with common ratio 's'. Then find area of triangle?

2045130_1469360_ans_b8a1caadfe3545dd90fa246df0efde4e.jpeg

If the points $A(0,0), B(2,2\sqrt 3)$ and $C(x,y)$ are the vertices of an equilateral triangle, find the coordinates of point C.

2044371_1563297_ans_ae19b5777114444db5668248de24a877.jpg

If (-3,2),(1,-2) and (5,6) are the mid-point of the sides of a traingle, find the coordinates of the vertices of the traingle.

2044022_1558414_ans_823673b2ad614e5f974c538d93b7de0e.jpg

The vertices of a quadrilateral are $A(-4, 2), B(2, 6), C(8, 5)$ and $D(9, -7)$ . Using slopes, show that the midpoints of the sides of the quad. $ABCD$ form a parallelogram.

1972492_1710198_ans_8350d2cf8bf241ab982cbb32de5d2209.jpg

Using slopes, show that the points $A(-4, -1), B(-2, -4), C(4, 0)$ and $D(2, 3)$ takes in order, are the vertices of a rectangle.

1981563_1710169_ans_5878239fc7d742358278850d7fc5d8f8.jpg

Find the inclination of a line whose slope is
$\dfrac{1}{\sqrt{3}}$

1981715_1710112_ans_2dfca17b2fbb4759a528296387a606bb.jpg

Find the slope and the equation of the line passing through the points:
$(5,3)$ and $(-5,-3)$

1973380_1710175_ans_886a6fe8e1cd42a1901608d92daf2303.jpg

Find the value of $x$ so that the line through $(3, x)$ and $(2, 7)$ is parallel to the line through $(-1, 4)$ and $(0, 6)$ .

1981557_1710139_ans_578a62d81007405ca96f39f2383220b8.jpg

Find the equation of a horizontal line passing through the point $(4,-2)$ .

1981550_1710119_ans_ec526c322ce248de919abc9e8a388e97.jpg

Straight Lines - Class 11 Engineering Maths - Extra Questions

Find inclination (in degrees) of a line perpendicular to x-axis.

If P is the mid point of seg AB and AB =7=7 cm, then AP is 3.53.5 cmIf true then enter 11 and if false then enter 00.

Write the slope of the line whose inclination is 0∘0^{\circ}.

Find inclination (in degrees) of a line parallel to x-axis.

Find inclination (in degrees) of a line parallel to yy-axis.

Find inclination (in degrees) of a line perpendicular to y-axis.

Find the inclination of the line whose slope is 1

Find the inclination of the line whose slope is 00.

Consider the following population and year graph, find the slope of the line ABAB and using it, find what will be the population in the year 20102010?

Find the coordinates of the foot of perpendicular from the point (-1, 3)(-1, 3) to the line 3x - 4y - 16\displaystyle = 03x - 4y - 16\displaystyle = 0

Find the point on the X-axis which is equidistant from (2, -5)(2, -5) and (-2, 9)(-2, 9).

Determine the equation of the line in the given graph.

Draw the graph of y=3x+5y=3x+5.

Determine the equation of the line in the given graph.

Determine the equation of the line in graph.

Find the area of a triangle whose vertices are A(3, 4), B(-1, 2),A(3, 4), B(-1, 2), and C(2, 3)C(2, 3).

For the angle in standard position if the Initial arm rotates 130^o130^o in anticlockwise direction, then state the quadrant in which terminal arm lies. (Draw the figure and write the answer).

Determine the equation of the line in graph.

Find the slope of the line passing through the points G(-4, 5)G(-4, 5) and H(-2, 1)H(-2, 1)

Find the slope of the line having its inclination {60}^{o}{60}^{o}.

Find the slope of line passing through the point P\left(1, -1\right)P\left(1, -1\right) and Q\left(-2, 5\right)Q\left(-2, 5\right).

Find the slope of the line passing through the points M(4, 0)M(4, 0) and N(-2, -3)N(-2, -3).

Find the area of the triangle, whose vertices are (-5, -1), (3, -5), (5, 2).(-5, -1), (3, -5), (5, 2).

Find the area of the triangle, whose vertices are (2, 0), (1, 2), (-1, 6)(2, 0), (1, 2), (-1, 6). What do you observe?

Find the area of the triangles (in sq. units) formed by the points (-2, 3), (7, 5)(-2, 3), (7, 5) and (3, -5)(3, -5).

Show that S(4, 3)S(4, 3) is the circum-centre of the triangle joining the points A(9, 3), B(7, -1)A(9, 3), B(7, -1) and C(1, -1) C(1, -1).

Find the area of the triangle formed by joining the midpoints of the sides of a triangle whose vertices are (0,-1),(2,1)(0,-1),(2,1) and (0,3)(0,3).

Find the area of triangle \triangle ABC\triangle ABC having vertices A(4, 2),B(3, 9)A(4, 2),B(3, 9) and C(10, 10)C(10, 10).

Find the slope of the straight line passing through the points (3,-2)(3,-2) and (-1,4)(-1,4).

Show that the following point taken in order form the vertices of a rhombus.(0, 0), (3, 4), (0, 8) and (-3, 4)

Show that the three points (4, 2), (7, 5)\ and\ (9, 7)(4, 2), (7, 5)\ and\ (9, 7) lie on a straight line.

Find the distance between A(a+b,b-a)A(a+b,b-a) and B(a-b,a+b)B(a-b,a+b).

Find the angle of inclination(in degrees) of the line passing through the points (1,2)(1,2) and (2,3)(2,3)

Prove that the points (-2, -1), (1, 0), (4, 3), and (1, 2) are at the vertices of a parallelogram.

A line is of length 10 and one end is at the point (2, -3); if the abscissa of the other end be 10, prove that its ordinate must be 3 or -9.

Prove (by showing that the area of the triangle formed by them is zero) that the following sets of three points are in a straight line:(1,4), (3, -2),(1,4), (3, -2), and (-3,16)(-3,16)

Find the areas of the triangles the whose coordinates of the points are respectively.(5, 2), (-9, -3) and (-3, -5)

Prove that the point \left ( -\frac{1}{14}, \frac{39}{14} \right )\left ( -\frac{1}{14}, \frac{39}{14} \right ) is the centre of the circle circumscribing the triangle whose angular points are (1, 1), (2, 3), and ( - 2, 2).

Find the areas of the triangles the whose coordinates of the points are respectively.(0, 4), (3, 6) and ( - 8, - 2)

(a, c + a), (a, c)(a, c + a), (a, c) and (-a, c - a).(-a, c - a).

Lay down in a figure the positions of the points (1, -3) and (- 2,1), and prove that the distance between them is 5.

Find the areas of the triangles the whose coordinates of the points are respectively.(1, 3), ( - 7, 6) and (5, - 1)

Prove that the straight line x + y = 1 touches the parabola y = x - x^2 . y = x - x^2 .

Find the slope of the line joining the points (4, -8)(4, -8) and (5, -2)(5, -2).

Find the perpendicular distance of the point (-1,1)(-1,1) from the line 12(x+6)=5(y-2)12(x+6)=5(y-2).

Find the areas of the triangles the coordinates of whose angular points are (1, 30^o), (2, 60^o) (1, 30^o), (2, 60^o) and (3, 90^o)(3, 90^o)

Find the areas of the triangles the coordinates of whose angular points are\left ( -a, \dfrac{\pi}{6} \right ), \left ( a, \dfrac{\pi}{2} \right )\left ( -a, \dfrac{\pi}{6} \right ), \left ( a, \dfrac{\pi}{2} \right ), and \left ( -2a, -\dfrac{2\pi}{2} \right )\left ( -2a, -\dfrac{2\pi}{2} \right )

Find the equation of the straight line which passes through the point (2,-3)(2,-3) and the point of intersection of the lines x+y+4=0x+y+4=0 and 3x-y-8=03x-y-8=0.

Show that the line joining the points A(1, -1, 2)A(1, -1, 2) and B(3, 4, -2)B(3, 4, -2) is perpendicular to the line joining the points C(0, 3, 2)C(0, 3, 2) and D(3, 5, 6)D(3, 5, 6).

(2, 30^o)(2, 30^o) and (4, 120^o)(4, 120^o)

If be the origin, and if the coordinates of any two points P_1P_1 and P_2P_2 respectively (x_1, y_2)(x_1, y_2) and(x_2, y_2)(x_2, y_2), prove thatOP_1 \cdot OP_2\cdot cos P_1OP_2=x_1x_2+y_1y_2OP_1 \cdot OP_2\cdot cos P_1OP_2=x_1x_2+y_1y_2

Find the area of the triangle formed by the linesy^{2}-9\ xy+18x^{2}=0y^{2}-9\ xy+18x^{2}=0 and y=9y=9

If the point (x,y)(x,y) is equidistant from the points (a+b,b-a)(a+b,b-a) and (a-b,a+b),(a-b,a+b), prove that bx=aybx=ay

If the area of triangle formed by points A(x,y), B(1,2)A(x,y), B(1,2) and C(2,1)C(2,1) is 6868 units then show that x+y=15x+y=15.

Find the angle which the straight line y = \sqrt 3 x - 4y = \sqrt 3 x - 4 makes with y-axis.

Find the area of a triangle whose vertices are (3,8), (-4,2) and (5,-1).

Find the acute angle between the lines 2x-y+3=02x-y+3=0 and x-3y+2=0x-3y+2=0.

Find the perpendicular distance of (0,0)(0,0) from 3x+4y=123x+4y=12.

How many equilateral triangles of side 2a with one vertex at origin and side along the x-axis is possible.

AA is a point on x-x-axis with abscissa -8-8 and BB is point on y-y-axis with coordinate 1515. Find distance ABAB.

Find the length of sides of a triangle whose vertices are (1, -1), (0, 4)(1, -1), (0, 4) and (-5, 3)(-5, 3).

Find the distance of a point P(x,y) from the origin.

Determine the ratio in which the point P(3, 5)P(3, 5) divides the join of A(1, 3)A(1, 3) & B(7, 9)B(7, 9).

Find the area of rhombus if its vertices are (3, 0), (4, 5), (-1, 4) (3, 0), (4, 5), (-1, 4) and (-2, -1) (-2, -1) taken in order. [Hint: Area of rhombus =\dfrac{1}{2}\times=\dfrac{1}{2}\times product of its diagonals.]

If the point A(-2,1),\ B(a,b)A(-2,1),\ B(a,b) and C(4,-1)C(4,-1) are collinear of a-b=1a-b=1. Find the value of aa of bb.

The length of line PQ is 10 units and the coordinates of P are \left( {2, - 3} \right)\left( {2, - 3} \right); calculate the coordinates of point Q, if its abscissa is 10.

Find the value of a , if the distance between points P(a,-8,4)P(a,-8,4) and Q(-3,-5,4)Q(-3,-5,4) is 5.

Find the equation of the line passing through the point (2, 3)(2, 3) and making intercept of length 33 unit between the lines y + 2x = 2y + 2x = 2 and y + 2x = 5y + 2x = 5.

If the point P(x,y) is equidistant from the points A(5,1) and B(-1,5), prove that 3x=2y3x=2y.

Find the values of x for which the distance between the points P(4,-5) and Q(12,x) is 10 units.

Using slopes, find the value of xx so that the points (6,\ -1),\ (5,\ 0)(6,\ -1),\ (5,\ 0) and (x,\ 3)(x,\ 3) are collinear.

The co-ordinate of the mid-points of the sides of a triangle ABC are D(2, 1), E(5, 3)D(2, 1), E(5, 3) and F(3, 7)F(3, 7). Find the lengths and equations of its sides.

Find the angle between the lines joining the points (0,\ 0),\ (2,\ 3)(0,\ 0),\ (2,\ 3) and (2,\ -2),\ (3,\ 5)(2,\ -2),\ (3,\ 5).

Find the area of \triangle PQR\triangle PQR whose vertices are P(-1, 1) Q(0, 5) R(3, 2)P(-1, 1) Q(0, 5) R(3, 2)

Find the point on y-axis which is equidistant from the points (5,-2) and (-3,2).

Prove that area if triangle ABCABC with vertices A(3,2), B(11,8), C(8,12)A(3,2), B(11,8), C(8,12) is 2525 square units.

Prove that the distance between the origin and the point (-6,\ -8)(-6,\ -8) is twice the distance between the points (4,\ 0)(4,\ 0) and (0,\ 3)(0,\ 3).

The medians AD and BE of a triangle with vertices A(0,b),B(0,0)A(0,b),B(0,0) and C(a,0)C(a,0) are perpendicular to each other, if

Find the area of the triangle formed by the point (0, 0) (4, 0) (0, 3)(0, 0) (4, 0) (0, 3).

If P is the mid point of seg AB and AB $=7$ cm, then AP is $3.5$ cm
If true then enter $1$ and if false then enter $0$ .

Write the slope of the line whose inclination is $0^{\circ}$ .

Find inclination (in degrees) of a line parallel to $y$ -axis.

Find the inclination of the line whose slope is $0$ .

Consider the following population and year graph, find the slope of the line $AB$ and using it, find what will be the population in the year $2010$ ?

Find the coordinates of the foot of perpendicular from the point $(-1, 3)$ to the line
$3x - 4y - 16\displaystyle = 0$

Find the point on the X-axis which is equidistant from $(2, -5)$ and $(-2, 9)$ .

Draw the graph of $y=3x+5$ .

Find the area of a triangle whose vertices are $A(3, 4), B(-1, 2),$ and $C(2, 3)$ .

For the angle in standard position if the Initial arm rotates $130^o$ in anticlockwise direction, then state the quadrant in which terminal arm lies. (Draw the figure and write the answer).

Find the slope of the line passing through the points $G(-4, 5)$ and $H(-2, 1)$

Find the slope of the line having its inclination ${60}^{o}$ .

Find the slope of line passing through the point $P\left(1, -1\right)$ and $Q\left(-2, 5\right)$ .

Find the slope of the line passing through the points $M(4, 0)$ and $N(-2, -3)$ .

Find the area of the triangle, whose vertices are $(-5, -1), (3, -5), (5, 2).$

Find the area of the triangle, whose vertices are $(2, 0), (1, 2), (-1, 6)$ . What do you observe?

Find the area of the triangles (in sq. units) formed by the points $(-2, 3), (7, 5)$ and $(3, -5)$ .

Show that $S(4, 3)$ is the circum-centre of the triangle joining the points $A(9, 3), B(7, -1)$ and $C(1, -1)$ .

Find the area of the triangle formed by joining the midpoints of the sides of a triangle whose vertices are $(0,-1),(2,1)$ and $(0,3)$ .

Find the area of triangle $\triangle ABC$ having vertices $A(4, 2),B(3, 9)$ and $C(10, 10)$ .

Find the slope of the straight line passing through the points $(3,-2)$ and $(-1,4)$ .

Show that the following point taken in order form the vertices of a rhombus.
(0, 0), (3, 4), (0, 8) and (-3, 4)

Show that the three points $(4, 2), (7, 5)\ and\ (9, 7)$ lie on a straight line.

Find the distance between $A(a+b,b-a)$ and $B(a-b,a+b)$ .

Find the angle of inclination(in degrees) of the line passing through the points $(1,2)$ and $(2,3)$

Prove (by showing that the area of the triangle formed by them is zero) that the following sets of three points are in a straight line:
$(1,4), (3, -2),$ and $(-3,16)$

Find the areas of the triangles the whose coordinates of the points are respectively.
(5, 2), (-9, -3) and (-3, -5)

Prove that the point $\left ( -\frac{1}{14}, \frac{39}{14} \right )$ is the centre of the circle circumscribing the triangle whose angular points are (1, 1), (2, 3), and ( - 2, 2).

Find the areas of the triangles the whose coordinates of the points are respectively.
(0, 4), (3, 6) and ( - 8, - 2)

$(a, c + a), (a, c)$ and $(-a, c - a).$

Find the areas of the triangles the whose coordinates of the points are respectively.
(1, 3), ( - 7, 6) and (5, - 1)

Prove that the straight line x + y = 1 touches the parabola $y = x - x^2 .$

Find the slope of the line joining the points $(4, -8)$ and $(5, -2)$ .

Find the perpendicular distance of the point $(-1,1)$ from the line $12(x+6)=5(y-2)$ .

$(1, 30^o), (2, 60^o)$ and $(3, 90^o)$

$\left ( -a, \dfrac{\pi}{6} \right ), \left ( a, \dfrac{\pi}{2} \right )$ , and $\left ( -2a, -\dfrac{2\pi}{2} \right )$

Find the equation of the straight line which passes through the point $(2,-3)$ and the point of intersection of the lines $x+y+4=0$ and $3x-y-8=0$ .

Show that the line joining the points $A(1, -1, 2)$ and $B(3, 4, -2)$ is perpendicular to the line joining the points $C(0, 3, 2)$ and $D(3, 5, 6)$ .

$(2, 30^o)$ and $(4, 120^o)$

If be the origin, and if the coordinates of any two points $P_1$ and $P_2$ respectively $(x_1, y_2)$ and $(x_2, y_2)$ , prove that
$OP_1 \cdot OP_2\cdot cos P_1OP_2=x_1x_2+y_1y_2$

Find the area of the triangle formed by the lines
$y^{2}-9\ xy+18x^{2}=0$ and $y=9$

If the point $(x,y)$ is equidistant from the points $(a+b,b-a)$ and $(a-b,a+b),$ prove that $bx=ay$

If the area of triangle formed by points $A(x,y), B(1,2)$ and $C(2,1)$ is $68$ units then show that $x+y=15$ .

Find the angle which the straight line $y = \sqrt 3 x - 4$ makes with y-axis.

Find the acute angle between the lines $2x-y+3=0$ and $x-3y+2=0$ .

Find the perpendicular distance of $(0,0)$ from $3x+4y=12$ .

$A$ is a point on $x-$ axis with abscissa $-8$ and $B$ is point on $y-$ axis with coordinate $15$ . Find distance $AB$ .

Find the length of sides of a triangle whose vertices are $(1, -1), (0, 4)$ and $(-5, 3)$ .

Determine the ratio in which the point $P(3, 5)$ divides the join of $A(1, 3)$ & $B(7, 9)$ .

Find the area of rhombus if its vertices are $(3, 0), (4, 5), (-1, 4)$ and $(-2, -1)$ taken in order.
[Hint: Area of rhombus $=\dfrac{1}{2}\times$ product of its diagonals.]

If the point $A(-2,1),\ B(a,b)$ and $C(4,-1)$ are collinear of $a-b=1$ . Find the value of $a$ of $b$ .

The length of line PQ is 10 units and the coordinates of P are $\left( {2, - 3} \right)$ ; calculate the coordinates of point Q, if its abscissa is 10.

Find the value of a , if the distance between points $P(a,-8,4)$ and $Q(-3,-5,4)$ is 5.

Find the equation of the line passing through the point $(2, 3)$ and making intercept of length $3$ unit between the lines $y + 2x = 2$ and $y + 2x = 5$ .

If the point P(x,y) is equidistant from the points A(5,1) and B(-1,5), prove that $3x=2y$ .

Using slopes, find the value of $x$ so that the points $(6,\ -1),\ (5,\ 0)$ and $(x,\ 3)$ are collinear.

The co-ordinate of the mid-points of the sides of a triangle ABC are $D(2, 1), E(5, 3)$ and $F(3, 7)$ . Find the lengths and equations of its sides.

Find the angle between the lines joining the points $(0,\ 0),\ (2,\ 3)$ and $(2,\ -2),\ (3,\ 5)$ .

Find the area of $\triangle PQR$ whose vertices are $P(-1, 1) Q(0, 5) R(3, 2)$

Prove that area if triangle $ABC$ with vertices $A(3,2), B(11,8), C(8,12)$ is $25$ square units.

Prove that the distance between the origin and the point $(-6,\ -8)$ is twice the distance between the points $(4,\ 0)$ and $(0,\ 3)$ .

The medians AD and BE of a triangle with vertices $A(0,b),B(0,0)$ and $C(a,0)$ are perpendicular to each other, if

Find the area of the triangle formed by the point $(0, 0) (4, 0) (0, 3)$ .

Find the equation of line equally inclined to coordinate axes and passes through $(-5, 1, -2)$ .

Find the area (sq.units) of the triangle whose coordinates are
$\left(2,3\right),\left(-1,0\right),\left(2,-4\right)$ .

Find a point on $y-axis$ which is equidistant from $( - 5, - 2)$ and $( 3, 2)$ .

Find the inclination of a line whose slope is
(i) $1$
(ii) $-1$
(iii) $\sqrt 3$
(iv) $-\sqrt 3$
(v) $\frac{1}{{\sqrt 3 }}$

Find the coordinates of a point on $x+y+3=0$ , whose distance from $x+2y+2=0$ is $\sqrt{5}$ .

The line joining two points $A ( 2,0 ) , B ( 3,1 )$ is rotated about $A$ in anticlockwise direction through an angle of $15 ^ { \circ } .$ Find the equation of the line in the new position. If $B$ goes to $C$ in the new position, what will be the coordinates of $C$ ?

Find the slope of the line $\dfrac{x}{3}+\dfrac{y}{2}=1$ .

Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points $P ( 0 , - 4 )$ and $B ( 8,0 )$

Find the slope of the line whose inclination is
$5\pi/6$

Find the area of a triangle with sides $3\ cm,4\ cm,5\ cm.$

Check whether $(5, -2)$ , $(6, 4)$ and $(7, -2)$ are the vertices of an isosceles triangles.

Find the slope of the line passing the two given points
$( a , 0 )$ and $(0, b )$

A point moves such that its distance from the point (4,0) is half that of its distance from the line $x=16$ , find its locus.

The distance of a point $P$ on x-axis from $A(11, 12)$ is $13$ units. Find the co-ordinates of $P.$

Find the distance between the following pairs of points:-
$(-5,7),(-1,3)$

The distance between the points $\left(0,0\right)$ and $\left(-4,3\right)$ is

Find the area of the triangle formed by the mid-points of the sides of $\triangle$ ABC, where A(3,2), B(-5,6) and C(8,3).