Class 11 Commerce Applied Mathematics - Straight Lines

How many points are there on the x-axis which are at a distance of $2 \sqrt{5}$ from the point $(7, 4)$ ?

Given that the point

$P$ is on the

$X$ -axis.

$\therefore$ its co-ordinates are

$P(x,0)$ .

Also

$P$ 's distances

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

$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 7,4 \right) =2\sqrt { 5 }$ units.

$\therefore { d }=\sqrt { { \left( { x }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }-{ y }_{ 1 } \right) }^{ 2 } } =\sqrt { { \left( x-7 \right) }^{ 2 }+{ \left( { 0 }-4 \right) }^{ 2 } } =2\sqrt { 5 }$

$\Rightarrow { x }^{ 2 }-14x+45=0$

$\Rightarrow x=9,5$

Therefore, the co-ordinates of the points on the

$X$ -axis are

$P(9,0)$ and

$P(5,0)$ .

Determine whether the points are vertices of a right triangle $A(-3, -4)$ , $B(2, 6)$ and $C(-6, 10)$

Using the distance formula

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ , we get

$AB^2 = (2 + 3)^2 + (6 + 4)^2 = 5^2 + 10^2 = 25 + 100 = 125$

$BC^2 = (-6, -2)^2 + (10 - 6)^2 = (-8)^2 + 4^2 = 64 +16 = 80$

$CA^2 = (-6 + 3)^2 + (10 + 4)^2 = (-3)^2 + (14)^2 = 9 + 196 = 205$
i.e.,

$AB^2 +BC^2 = 125 + 80 = 205 = CA^2$
Hence ABC is a right angled triangle since the square of one side is equal to sum of the squares of the other two sides.

The point on the y-axis equidistant from $(-5, 2)$ and $(9, -2)$ is $(0,-m)$ , them $m$

Point on y axis has x-coordinates as

$O$

$\therefore$ Let that point be

$P(O,y)$

Let

$A\equiv (-5,2)$ &

$B\equiv (9,-2)$

A.T.Q

$PA=PB$

By distance formula

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

$PA=PB$

$\therefore PA^{2}=PB^{2}$

$\therefore 25+(y-2)^{2}=81+(y+2)^{2}$

$\therefore 25+y^{2}-4y+4=81+y^{2}+4y+4$

$-8y=56$

$y=-7$

According to question

$(0,y)\equiv (0,-m)$

$(0,-7)\equiv (0,-m)$

$\therefore m=7$

Find the perimeter of a triangle whose vertices have the coordinates $(3,10),(5,2)$ and $(14,12)$ .

Solution:

Let

$A(3,10), B(5,2)$ and

$C(14,12)$ be the coordinates of the vertices of a triangle.

$AB=\sqrt{(3-5)^2+(10-2)^2}$

$=\sqrt{4+64}$

$=\sqrt{68}$

$=2\sqrt{17}$ units

$=8.25$ units

$BC=\sqrt{(14-5)^2+(12-2)^2}$

$=\sqrt{81+100}$

$=\sqrt{181}$

$=13.45$ units

$AC=\sqrt{(3-14)^2+(10-12)^2}$

$=\sqrt{121+4}$

$=\sqrt{125}$

$=5\sqrt{5}$ units

$=11.180$ units

So, perimeter of the required triangle

$=AB+BC+AC$

$=8.25+13.45+11.180$ units

$=32.880$ units

Show that the two parabolas $x^2 + 4a (y - 2b - a ) = 0$ and $y^2 = 4b (x - 2a + b)$ intersect at right angles at a common end of the latus rectum of each.

$\dfrac{dy}{dx}=y'$

$x^{2}+4a(y-2b-a)=0$ ...............(1)

diffrentiate the eqn.

$2x+4ay'=0$

$y'=m=\dfrac{-x}{2a}$

$y^{2}=4b(x-2a+b)$ ..........(2)

diffrentiate the eqn.

$2yy'=4b$

$y'=n=\dfrac{2b}{y}$

solving eqn (1) & (2)

$(2a ,2b)$

let

$\theta$ be the angle at intersection point

$m\times n=\dfrac{-x.2b}{y.2a}=\dfrac{-2a.2b}{2b.2a}=-1$

$\Rightarrow \theta = 90^{\circ}$

Which point on x-axis is equidistant from $(5,9)$ and $(-4,6)$ ?

Let the point be

$(h,0)$

let,

$P_1(5,9) \ and \ P_2(-4,6)$

Distance between two points =

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

So,

$D_1=\sqrt { { (5-h) }^{ 2 }+{ (9-0) }^{ 2 } }$

$D_2=\sqrt { { (-4-h) }^{ 2 }+{ (6-0) }^{ 2 } }$

But, it is given that both the points are equidistant

$\therefore \sqrt { { (5-h) }^{ 2 }+{ (9) }^{ 2 } } =\sqrt { { (-4-h) }^{ 2 }+{ (6) }^{ 2 } }$

Squaring both sides

$(5-h)^{2}+81 = {(-4-h)^{2}}+36$

$25+h^2-10h+81 = 16+8h+h^{2}+36$

$-18h = -81-25+16+36$

$18h=54$

$h=3$

So point is (3,0)

The vertices of a triangle are $A\left( { - 2,\,2} \right),\,B\left( {4,\,4} \right)\,and\,C\left( {8,\,2} \right).$ Find the length of the medians to the sides

${\text{let}}{\mkern 1mu} AE,BF,CG{\mkern 1mu} \;{\text{are}}{\mkern 1mu} \;{\text{Medians}}{\mkern 1mu} \;E{\mkern 1mu} \;{\text{is}}{\mkern 1mu} \;{\text{mid}}{\mkern 1mu} \;{\text{point}}{\mkern 1mu} \;{\text{of}}{\mkern 1mu} BC$

$\therefore E = \left( {6,3} \right)$

Similarly, $F = \left( {3,2} \right)$

$G = \left( {1,3} \right)$

$AE = \sqrt {{{\left( {6 + 2} \right)}^2} + {{\left( {3 - 2} \right)}^2}}$

$\sqrt {64 + 1} = \sqrt {65} \,unit$

$BF = \sqrt {{{\left( 1 \right)}^2} + {2^2}} = \sqrt {1 + 4} = \sqrt {5\,} unit$

$CG= \sqrt {{7^2} + {{\left( { - 1} \right)}^2}} = \sqrt {50} unit$

1014430_1054106_ans_ea404d17393a4ea884f26ab8a81c47a6.png

Find the slope of the line passing through the points $(3, -2)$ & $(7, -2)$

$(x1,y1)=(3,-2)$

$(x2,y2)=(7,-2)$

$\text {slope}=\dfrac{y2-y1}{x2-x1}$

$\text {slope}=\dfrac{-2-(-2)}{7-3}$

$\text {slope}=\dfrac{-2+2}{7-3}$

$\text {slope}=\dfrac{0}{7-3}$

$\text {slope}=0$

Hence the

$\text {slope }=0$

Find the value of $x$ for which the distance between $p(2, -3)$ and $q(x, 5)$ is $10$ units.

The distance between

$p(2,-3)$ and

$q(x,5)$ is

$\sqrt{(x-2)^2+(5+3)^2}$ units.

Given,

$\sqrt{(x-2)^2+(5+3)^2}=10$

Squaring we have,

$(x-2)^2+64=100$

or,

$(x-2)^2=36$

or,

$x-2=\pm 6$

or,

$x=8,-4$ .

Find a point on $x-axis$ which is equidistant from $A(2, -5)$ and $(-2, 9)$ .

Any point on

$x$ -axis is taken as

$P(a,0),$

$A(2,-5)$ and

$B(-2, 9)$

So, we must have

$PA = PB \Rightarrow PA^2 = PB^2$

$(a-2)^2 + 5^2 = (a + 2)^2 + 9^2$

$a^2 - 4a + 4 + 25 = a^2 + 4a + 4 + 81$

$-4a + 25 = 4a + 81$

$-8a = 81 - 25$

$-8a = 56$

$a = -7$

Required point is

$(-7, 0)$ .

Find the slope of line which passes through the point (7,11) and (9,15) .

As we know that slope of line

$\left( m \right)$ passing through

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

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

$m = \cfrac{\left( {y}_{2} - {y}_{1} \right)}{\left( {x}_{2} - {x}_{1} \right)}$

As the line passes through the points

$\left( 7, 11 \right)$ and

$\left( 9, 15 \right)$ .

Therefore,

Slope of line

$= \cfrac{15 - 11}{9 - 7} = \cfrac{4}{2} = 2$

Thus the slope of given line is

$2$ .

Hence the correct answer is

$2$ .

Show that the points $(3, -2), (1, 0), (-1, -2)$ and $(1, -4)$ are concyclic

REF.Image

$A(3,-2), B(1,0), C(-1,-2), D(1,-4)$

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

$BD= \sqrt{(1-1)^{2}+(-4)^{2}}= \sqrt{4^{2}}=4$

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

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

$CD= \sqrt{2^{2}+2^{2}}= 2\sqrt{2}$

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

For concyclic :

$AC\times BD= (AD\times BC)+ (AB \times CD)$

$4 \times 4= (2\sqrt{2}\times 2\sqrt{2})+(2\sqrt{2}\times 2\sqrt{2})$

$16= 8+8$

$16=16$

$\therefore$ Hence, the points are concyclic

1125104_1137708_ans_98713cf7fda4412c9d7fdd97625bb277.jpg

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

Let

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

Let

$A(2,-5)$ and

$B(-2,9)$ . Then,

$PA=PB$

$PA^2=PB^2$

$(2-x)^2+(-5-0)^2=(-2-x)^2+(9-0)^2$

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

$-8x=56$

$x=-7$

Therefore, the required point is

$(-7,0)$ .

If the points $A(1,2), B(-4,3), C(x,2)$ and $D(2,3)$ form parallelogram find the value of $x$ .

In a parallelogram

$AB=CD$

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

$\quad 2-x=-5\quad \Rightarrow x=7$

$\quad 2-x=5\quad \Rightarrow x=-3$

1145628_1204854_ans_ec68e4c3a1444df18418afbdde6e08d3.png

If the point $A(a,2)$ is equidistant from the points $B(8,-2)$ and $C(2,-2)$ , the value of $a$ .

Using distance formula

$AB^2=AC^2$

$\Rightarrow(a-8)^2+(2+2)^2=(a-2)^2+(2+2)^2$

$\Rightarrow a^2-16a+64+16=a^2-4a+4+16$

$\Rightarrow-16a+64=-4a+4$

$\Rightarrow -16a+4a=4-64$

$\Rightarrow-12a=-60$

$\Rightarrow a=5$

If $(a,b)$ is the mid-point of the line segment joining the points $A\left( {10, - 6} \right),B\left( {k,4} \right)$ and $a-2b=18$ ,find the value of $k$ and the distance AB.

Solution -

$a=\frac{10+K}{2}$

$b=\frac{-6+4}{2}=-1$

$a= 18-2=16$

$[\because a-26=18]$

$16=\frac{10+k}{2}$

$k=22$

Distance

$AB = \sqrt{(22-10)^{2}+(4+6)^{2}}$

$=\sqrt{144+100}$

$=\sqrt{244}$

$=2\sqrt{61}$

1100650_1187659_ans_a9ea11b35a7e46a6baad53288e1cce04.jpg

A point P on X-axis is at distance 13 from A(11,12). Find the coordinates of P.

A point on X axis can be represented as (a,0)

Square of Distance between (a,0) and (11,12) is

(a-11)^2 + (0-12)^2 = 13^2(given)

(a-11)^2 = 169-144 = 25 = 5^2

a-11 = 5

a = 16

If $P ( 2 , P )$ is the mid point of the line segment joining the the points. $A ( 6 , - 5 )$ and $B ( - 2,11 )$ find the value of $P .$

$P=(2,P)$

mid point of

$A(6,-5) \& B(-2,11)$

$=(2,3)$

$P=3$

Find the slope of the line whose inclination is $30^{o}$

$Slope = tan\theta$

$tan 30^o = 1 / \sqrt3$

21 - 3(b - 7) = b + 20

equation

$\Rightarrow$

$21-3(b-7) = b+20$

$21-3b-3(-7) = b+20$

$21 -3b+21=b+20$

$42-3b=b+20$

$22 = 4b$

$b = \frac{22}{4}$

$b = \frac{11}{2}$

Given line

$5x-y = 1$

slope of line

$= \frac{-5}{-1} = 5$

we have to take a line perpendicular

to this line

1173012_1271548_ans_9c8e76546bd34748aa846b6bdbb873e3.jpeg

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

(i)

$tan\theta =1 \Rightarrow \theta 45^{\circ}$ ACW from

$tan^{x}- axis$ .

(ii)

$-1 =tan\theta \Rightarrow \theta = 135^{\circ}$ ACW.

(iii)

$tan \theta =\sqrt{3}\Rightarrow \theta =60^{\circ}$ ACW.

(iv)

$tan\theta =-\sqrt{3}\Rightarrow \theta = 120^{\circ}$ ACW.

(v)

$tan\theta =\frac{1}{\sqrt{3}}\Rightarrow \theta =30^{\circ}$ ACW.

1213393_1306814_ans_7459c9db553a4da48144f39a0c7d2a60.JPG

Find the slope of the line perpendicular to $AB$ if $A=(0,-5)$ and $B=(-2,4)$ .

Slope of

$AB=\dfrac{4-(-5)}{-2-0}=-\dfrac { 9 }{ 2 }$

$\therefore$ slope of

$\bot$ to

$AB$

$=-1/9/-2=\dfrac { 2 }{ 9 }$

Hence, this is the answer.

Find the distance between the following pairs of points:-
$(2,3),(4,1)$

Let

$A\left(2,3\right)$ and

$B\left(4,1\right)$ be the points of a straight line.

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

$AB=\sqrt{{\left(4-2\right)}^{2}+{\left(1-3\right)}^{2}}$

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

$=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$ units.

Find the angle between the line $\sqrt {3}x+y+1=0$ and $x+1=0$ .

When written in form

$y=mx+c$ ,

$m=$ slope of line

For

$L_1:$

$y=-\sqrt{3}x-1=m_1=-\sqrt{3}$

For

$L_2: o.y=x+1\Rightarrow y=\dfrac{x}{0}+\dfrac{1}{0}\Rightarrow m_2\rightarrow \infty$

Slope between lines

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

$\theta =$ Angle

$\Rightarrow \theta =\tan^{-1}\left(\left|\dfrac{1-\dfrac{m_1}{m_2}}{m_1+\dfrac{1}{m_2}}\right|\right)=\tan^{-1}\left(\left|\dfrac{1}{m_1}\right|\right)=\tan^{-1}\left(\left|\dfrac{1}{\sqrt{3}}\right|\right)$

$\Rightarrow -\theta =150^o$ or

$30^o$ .

1214405_1395691_ans_da0cd1d33aab479bacc3306803b1414e.jpg

If point $(x,y)$ is equidistant from point $(3,k)$ and $(k,3)$ then find the value of $k$ when $x\neq y$

if (x,y)is equidistant form point (3,k)& (k,3).

then

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

on both sides=

$(x-3)^{2}+(y-k)^{2}=(x-k)^{2}+(y-3)^{2}$

$x^{2}+9-6x+y^{2}+k^{2}-2ky=y^{2}+x^{2}-2kx+9+6y+k^{2}$

(2k-6)x=(2k-6)y

(2k-6)(x-y)=0[x+y]then

then 2k-6=0

[k=3]

1211595_1362760_ans_8dfd6b7c00b04f3fa96be1b4b079ccb3.JPG

Show that the point A(-2,3,5), B(1,2,3) and C(7,0,-1) are collinear .

$AB=\sqrt{(1+2)^{2}+(2+3)^{2}+(3-5)^{2}}=\sqrt{14}$

$BC=\sqrt{56}=2\sqrt{14}$

$CA=\sqrt{126}=3\sqrt{14}$

AB+BC=CA
A, B, C are collinear.

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

The distance of the point

$P(-6,8)$ from the origin is

$\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$ units.

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

Given

$A(6,4),B(2,12)$

Let

$m$ be the slope of AB. Then,

we know that slope of line passing through

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

$m=\dfrac{12-4}{2-6}=\dfrac{8}{-4}=-2$

slope of two perpendicular lines

$m_1m_2=-1\Rightarrow m_2=\dfrac{-1}{m_1}$

So, the slope of a line perpendicular to AB is

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

$A ( 0,2 ) , B ( 0,5 )$ be any two points, then the distance between them is _________.

Given

$A(0,2),B(0,5)$

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

$=\sqrt{0+9}$

$=3$ units

Distance between them is

$3$ units

The distance between the two points is $5.$ One of them is $(3, 2)$ and the ordinate of the second is $-1$ , then find sum of its x-coordinate.

Using distance formula:

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

$\Rightarrow (3-x)^{2}+9=25$

$\Rightarrow 9+x^{2}-6x+9-25=0$

$\Rightarrow x^{2}-6x-7=0$

$\Rightarrow x=7, -1$

Mid-points of the sides AB and AC of $\Delta ABC$ are $(3, 5)$ and $(-3, -3)$ , respectively, then find the length of BC.

Let

$D=(3, 5), E=(-3. -3),$ then

$DE=\sqrt{(3+3)^{2}+(5+3)^{2}}$

$=\sqrt{36+64}$

$=\sqrt{100}=10$

$\Rightarrow BC=2 DE=2\times 10=20$ units

The vertices of a triangle are A $(2, 2)$ , B $(-4, -4)$ , C $(5, -8)$ . Then, if the length of the median through C is $\sqrt m$ .Find $m$

Mid point of AB

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

$=(-1, -1)=D$
Now

$CD=\sqrt{(5+1)^{2}+(-8+1)^{2}}$

$=\sqrt{36+49}\sqrt{85} units$

If P $(x, y)$ is a point in the coordinate plane such that locus of P is

(a)

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

$\Rightarrow x=y$

(b)

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

$\Rightarrow x^{2}+y^{2}-2ax-2by-2ab=0$

(c)

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

$\Rightarrow y^2=4x^2\Rightarrow y\pm 2x=0$

(d)

$\sqrt{x^{2}+y^{2}}=\frac{1}{2}(x+y)\Rightarrow 4(x^{2}+y^{2})=(x+y)^{2}$

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

Find the angles between the lines $\displaystyle \sqrt{3}x+y= 1$ and $x+\sqrt{3}y= 1$

The given lines are

$\displaystyle \sqrt{3}x+y= 1$ and

$x+\sqrt{3}y= 1$

$\Rightarrow \displaystyle y=- \sqrt{3}x+1$ and

$\displaystyle y= -\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}$
Thus slope of line (1) is

$\displaystyle m_{1}= -\sqrt{3}$ while the slope of line (2) is

$\displaystyle m_{2}= -\frac{1}{\sqrt{3}}$
The acute angle i.e;

$\displaystyle \theta$ between the two lines is given by

$\displaystyle \tan \theta = \left | \frac{m_{1}-m_{2}}{1+m_{1}m_{2}} \right |$

$\displaystyle \quad = \left | -\frac{\sqrt{3}+\frac{1}{\sqrt{3}}}{1+(-\sqrt{3})(-\frac{1}{\sqrt{3}})} \right |$

$\displaystyle \quad = \left | \frac{\frac{-3+1}{\sqrt{3}}}{1+1} \right |= \left | \frac{-2}{2\sqrt{3}} \right |$

$\displaystyle\Rightarrow \tan \theta = \frac{1}{\sqrt{3}} , \theta = 30^{\circ}$
Thus, the angle between the given lines is either

$\displaystyle 30^{\circ}$ or

$180^{\circ}-30^{\circ}= 150^{\circ}$

Find the slope of a line passing through the points with coordinates $(-1, 0)$ and $(1, 3)$ .

The slope of the line is

$m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}=\dfrac{3-0}{1-(-1)}=\dfrac{3}{2}$ .

Find the equation of the line that passes through the points $(-1,0)$ and $(-2,4)$

Equation line from

$(-1,0)$ and

$(-2,4)$

$y-y_{1}=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$

$y=\dfrac{4-0}{-2+1}(x+1)$

$y=\dfrac{4}{-1}(x+1)$

$y=-4(x+1)$

$4x+y+4=0$

The distance between A(1, 3) and B(x, 7) isCalculate the possible values of x

$A(1,3)$ ,

$B(x,7)$

$AB=5$ units

Distance between two points

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

$(x_1,y_1)$ and

$(x_2,y_2)$

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

Squaring Both sides

$(AB)^2=(x-1)^2+(4)^2$

$(5)^2=x^2-2x+1+16$

$25=x^2-2x+17$

$x^2-2x-8=0$

$x^2-4x+2x-8=0$

$(x-4)(x+2)$

$x=-2$ or

$4$

A chair is $5$ feet from one wall of a room and $7$ feet from the wall at a right angle to it. How far is the chair from the intersection of the two walls?

From intersection point , the distance of chair is $\sqrt{{7}^{2}+{5}^{2}} = \sqrt{74}$

Find the slope of the line with inclination $60^\circ$

Inclination of the line is

$60^\circ$ .

$\therefore \theta = 60^\circ$
Now, the slope of the line

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

Prove that the equations to the straight lines passing through the origin which make an angle $\alpha$ with the straight line $y + x = 0$ are given by the equation
$x^2 + 2xy \sec 2 \alpha + y^2 = 0 .$

The lines makes an angle $\alpha$ with the line $y+x=0$

Slope of given line that is ${ m }_{ 1 }$ $=-1$

Let the slope of the other line be ${ m }_{ 2 }$

The angle between straight lines thatt is $\tan { \theta } =\left| \dfrac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right|$

$\tan { \alpha } =\left| \dfrac { -1-{ m }_{ 2 } }{ 1-1.{ m }_{ 2 } } \right| =\left| \dfrac { { m }_{ 2 }+1 }{ { m }_{ 2 }-1 } \right| \\ \dfrac { { m }_{ 2 }+1 }{ { m }_{ 2 }-1 } =\tan { \alpha } ,\dfrac { { m }_{ 2 }+1 }{ { m }_{ 2 }-1 } =-\tan { \alpha } \\ \Rightarrow { m }_{ 2 }=\dfrac { \tan { \alpha } +1 }{ \tan { \alpha } -1 } ,\dfrac { \tan { \alpha } -1 }{ \tan { \alpha } +1 }$

Equation of straght line with given slope and a point is $y=mx+c$

Lines passes through origin $0=0.m+c$

$\Rightarrow c=0$

So the equation of lines are $y=\dfrac { \tan { \alpha } +1 }{ \tan { \alpha } -1 } x$ and $y=\dfrac { \tan { \alpha } -1 }{ \tan { \alpha } +1 } x$

$y=\dfrac { \sin { \alpha +\cos { \alpha } } }{ \sin\alpha -\cos\alpha } x,y=\dfrac { \sin\alpha -\cos\alpha }{ \sin { \alpha +\cos { \alpha } } } x$

Combined equation of straight lines

$\left( y-\dfrac { \sin { \alpha +\cos { \alpha } } }{ sin\alpha -cos\alpha } x \right) \left( y-\dfrac { sin\alpha -cos\alpha }{ \sin { \alpha +\cos { \alpha } } } x \right) =0\\ { y }^{ 2 }-xy\left( \dfrac { \sin\alpha -\cos\alpha }{ \sin { \alpha +\cos { \alpha } } } \right) -xy\left( \dfrac { \sin { \alpha +\cos { \alpha } } }{ \sin\alpha -\cos\alpha } \right) +\left( \dfrac { \sin { \alpha +\cos { \alpha } } }{ \sin\alpha -\cos\alpha } \right) \left( \dfrac { \sin\alpha -\cos\alpha }{ \sin { \alpha +\cos { \alpha } } } \right) { x }^{ 2 }=0\\ { y }^{ 2 }-xy\left( \dfrac { { (\sin\alpha -\cos\alpha ) }^{ 2 }+{ (\sin\alpha +\cos\alpha ) }^{ 2 } }{ (\sin { \alpha +\cos { \alpha } } )(\sin\alpha -\cos\alpha ) } \right) +{ x }^{ 2 }=0\\ { y }^{ 2 }-xy\left( \dfrac { 2 }{ \sin ^{ 2 }{ \alpha } -\cos ^{ 2 }{ \alpha } } \right) +{ x }^{ 2 }=0\\ { y }^{ 2 }-xy\left( \dfrac { 2 }{ -\cos2\alpha } \right) +{ x }^{ 2 }=0\\ { y }^{ 2 }+2xy\sec { 2\alpha } +{ x }^{ 2 }=0$

Hence proved

Prove that the angle between the straight lines joining the origin to the intersection of the straight line $y = 3x + 2$ with the curve $x^2 + 2xy + 3y^2 + 4x + 8y - 11 = 0$ is $\tan^{-1} \dfrac {2 \sqrt2}{3}$

${ x }^{ 2 }+2xy+3{ y }^{ 2 }+4x+8y-11=0\\ y=3x+2$

The equation of pair straight line joining the origin is find by method of homogenisation. We make the coefficient of constant term of straight line $1$ and put it the equation of the curve so that degree of each term of the curve become $2$

$\dfrac { y-3x }{ 2 } =1...(i)\\ x^{ 2 }+2xy+3{ y }^{ 2 }+4x(1)+8y(1)-11{ (1) }^{ 2 }=0\\ x^{ 2 }+2xy+3{ y }^{ 2 }+4x\left( \dfrac { y-3x }{ 2 } \right) +8y\left( \dfrac { y-3x }{ 2 } \right) -11{ \left( \dfrac { y-3x }{ 2 } \right) }^{ 2 }=0\\ x^{ 2 }+2xy+3{ y }^{ 2 }+2xy-6x^{ 2 }+4{ y }^{ 2 }-12xy-\dfrac { 11{ y }^{ 2 } }{ 4 } -\dfrac { 99{ x }^{ 2 } }{ 4 } +\dfrac { 33xy }{ 2 } =0\\ -119x^{ 2 }+17{ y }^{ 2 }+34xy=0\\ 7x^{ 2 }-{ y }^{ 2 }-2xy=0$

is the desired equation of straight lines.

Angle between pair of straight lines that is $\tan { \theta } =\left| \dfrac { 2\sqrt { { h }^{ 2 }-ab } }{ a+b } \right|$

$\tan { \theta } =\left| \dfrac { 2\sqrt { 1+7 } }{ 7-1 } \right| =\dfrac { 2\sqrt { 2 } }{ 3 } \\ \Rightarrow \theta =\tan ^{ -1 }{ \left( \dfrac { 2\sqrt { 2 } }{ 3 } \right) }$

Hence proved.

Prove that the points $(3, 0) (6, 4)$ and $(-1, 3)$ are the vertices of a right angled isosceles triangle.

R.E.F image

We can safely assume that AB and BC are the isosceles sides.

By distance formula,

$AB=\sqrt{(-1-3)^{2}+(3-0)^{2}}=\sqrt{16+9}=5$

$BC=\sqrt{(3-6)^{2}+(0-4)^{2}}=\sqrt{9+16}=5$

Since AB=BC, the

$\Delta ABC$ is isosceles.

For proving

$\angle ABC$ is right, we use the relation

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

where

$m_{1}$ is slope of line AB and

$m_{2}$ of BC :

$m_{1}=\dfrac{3-0}{-1-4}=\dfrac{3}{-4}$

$m_{2}=\dfrac{4-0}{6-3}=\dfrac{4}{3}$

$m_{1}\times m_{2}=\dfrac{-3}{4}\times \dfrac{4}{3}=-1$

Hence proved,

$\Delta ABC$ is right angled isosceles.

1083007_785838_ans_2794fe16987743199413ac4efbce3f79.PNG

Find the value of $k$ if the angle between St. lines $4x - y + 7 = 0, kx - 5y - 9 = 0$ is $45^{\circ}$ .

Line

$1$ is

$4x-y+7=0$ or

$y=4x+7$ is of the form

$y={m}_{1}x+c$

Line

$2$ is

$kx-5y-9=0$ or

$5y=kx-9$ or

$y=\dfrac{kx}{5}-\dfrac{9}{5}$ is of the form

$y={m}_{2}x+c$

Slope of line

$1$ is

${m}_{1}=4$ and that of line

$2$ is

$\dfrac{k}{5}$

We have

$\tan{\theta}=\dfrac{{m}_{1}-{m}_{2}}{1+{m}_{1}{m}_{2}}$

$\Rightarrow \tan{{45}^{\circ}}=\dfrac{4-\dfrac{k}{5}}{1+\dfrac{4k}{5}}$

$\Rightarrow 1=\dfrac{20-k}{5+4k}$

$\Rightarrow 5+4k=20-k$

$\Rightarrow 5k=20-5=15$

$\therefore k=\dfrac{15}{5}=3$

Hence

$k=3$

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

Perpendicular Length from Point to line

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

$ax+by+c = 0$

$d = \dfrac{|ax_{1}+by_{1}+c|}{\sqrt{a^{2}+b^{2}}}$

given line

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

given Point

$(-\sqrt{a^{2}-b^{2}},0)$ and

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

Product of Perpendicular distance from

both Point to line

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

$= \dfrac{\dfrac{-cos^{2}\theta }{a^{2}}(a^{2}-b^{2})+1}{\dfrac{cos^{2}\theta }{a^{2}}+\dfrac{sin^{2}\theta }{b^{2}}} = \dfrac{(-cos^{2}\theta +\dfrac{b^{2}}{a^{2}}cos^{2}\theta+1)a^{2}b^{2}}{(b^{2}cos^{2}\theta +a^{2}sin^{2}\theta )}$

$= \dfrac{\dfrac{(a^{2}sin^{2}\theta +b^{2}cos^{2}\theta )a^{2}b^{2}}{a^{2}}}{(a^{2}sin^{2}\theta +b^{2}cos^{2}\theta )} = b^{2}$

1169919_875587_ans_5fe32498b8b943a0b84b6fdb94abce2c.jpg

Show that the quadrilateral whose vertices are $(2,-1),(3,4),(-2,3)$ and $(-3,-2)$ is a rhombus.

In rhombus, all sides are equal and diagonals are not equal.

Distance between two points

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

$AB=\sqrt { { (3-2) }^{ 2 }+{ (4+1) }^{ 2 } } =\sqrt { 26 } \\ BC=\sqrt { { (3+2) }^{ 2 }+{ (3-4) }^{ 2 } } =\sqrt { 26 } \\ CD=\sqrt { { (-3+2) }^{ 2 }+{ (-2-3) }^{ 2 } } =\sqrt { 26 } \\ DA=\sqrt { { (-3-2) }^{ 2 }+{ (-2+1) }^{ 2 } } =\sqrt { 26 } \\ AC=\sqrt { { (2+2) }^{ 2 }+{ (4) }^{ 2 } } =4\sqrt { 2 } \\ BD=\sqrt { { (-3-3) }^{ 2 }+{ (4+2) }^{ 2 } } =6\sqrt { 2 }$

$AB=BC=CD=DA$ All sides are equal

$AC \ne BD$ Diagonals are not equal

Hence, It is a rhombus.

Which point on y-axis is equidistant from $(2,3)$ and $(-4,1)$ ?

Let the point on the y axis be (0,k)

Given that,

$\sqrt { { (x_2-x_1) }^{ 2 }+{ (y_2-y_1) }^{ 2 } }=\sqrt { { (x'_2-x'_1) }^{ 2 }+{ (y'_2-y'_1) }^{ 2 } }$

$\therefore$

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

Squaring both sides

$4+9+k^{2}+6k = 16+1+k^{2}-2k$

$8k=17-13$

$8k=4$

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

So point =

$(0,\cfrac{1}{2})$

Two vertices of an isosceles triangle are $(2,0)$ and $(2,5)$ . Find the third vertex if the length of the equal sides is $3$ .

Let

$\Delta$ ABC be the isosceles triangle, the third vertex be C(a,b).

and A(2,0) and B(2,5)

Let AC and BC be equal sides of the triangle

By distance formula we have,

$D=\sqrt{(x_2-x_1)+(y_2-y_1)}$

$\therefore AB=\sqrt{(2-2)^2+(5-0)^2}=\sqrt{25}=5$

Now,

$AC=3$ [length of equal sides=3 ]

$AC^2=3^2=9$ ..........(a)

but by distance formula,

$AC^2=\sqrt{(a-2)^2+b^2}$ ...........(b)

Combining (a) and (b)

$\Rightarrow (a-2)^2+b^2=9$

$\Rightarrow a^2+4-4a+b^2=9$

$\Rightarrow a^2+b^2-4a=5$ ---------- (1)

Consider BC

$BC^2=\sqrt{(a-2)^2+(b-5)^2}$

but BC=3

$\therefore (a-2)^2+(b-5)^2=9$

$\Rightarrow a^2+4-4a+b^2+25-10b=9$

$\Rightarrow a^2+b^2-4a-10b+20=0$ ------- (2)

$\Rightarrow -10b+20+5=0$ ---------(from 1 )

$\Rightarrow 10b=25$

$\Rightarrow b=\dfrac{25}{10}=\dfrac{5}{2}$

Now,

$a^2+\dfrac{25}{4}-4a=5$

$\Rightarrow a^2-4a=\dfrac{-5}{4}$

$\Rightarrow a^2-4a+\dfrac{5}{4}=0$

$\Rightarrow a=\dfrac{4\pm\sqrt{16-4*1*\dfrac{5}{4}}}{2}$

$=\dfrac{4\pm\sqrt{11}}{2}$

$=\dfrac{2+\sqrt{11}}{2},\dfrac{2-\sqrt{11}}{2}$

$\therefore$ Coordinate of third vertex (a,b) =

$(\dfrac{2+\sqrt{11}}{2},\dfrac{5}{2})$ and

$(\dfrac{2-\sqrt{11}}{2},\dfrac{5}{2})$

Find the centre of the circle passing through $(6,-6),(3,-7)$ and $(3,3)$

Let the given points be

$A(6,-6)$ ,

$B(3,-7)$ and

$C(3,3)$ and let the centre be

$O(x,y)$

$OA = OB = OC$

So by distance formula we have,

Distance between two points = $\sqrt { { (x_2-x_1) }^{ 2 }+{ (y_2-y_1) }^{ 2 } }$

consider,

$OA=OB$

$\Rightarrow{\left( {OA} \right)^2} = {\left( {OB} \right)^2}$

$\Rightarrow {\left( {x - 6} \right)^2} + {\left( {y + 6} \right)^2} = {\left( {x - 3} \right)^2} + {\left( {y + 7} \right)^2}$

$\Rightarrow {x^2} + 36 - 12x + {y^2} + 36 + 12y = {x^2} + 9 - 6x + {y^2} + 49 + 14y$

$\Rightarrow 3x + y = 7$ ---(i)

Similarly,

$OB=OC$

$\Rightarrow{(OB)^2} = {\left( {OC} \right)^2}$

$\Rightarrow {\left( {x - 3} \right)^2} + {\left( {y + 7} \right)^2} = {\left( {x - 3} \right)^2} + {\left( {y - 3} \right)^2}$

$\Rightarrow 14y + 6y = 18 - 58$

$\Rightarrow 20y = - 40$

$\Rightarrow$

$y=-2$

putting

$y=-2$ in (i)

$3x+(-2)=7$

$\Rightarrow3x-2=7$

$\Rightarrow$

$x=3$

So, The centre of the circle is

$(3,-2)$

Name the quadrilateral formed, if any, by the following points and give reasons for your answers:
$A(-1,-2),B(1,0),C(-1,2),D(-3,0)$

Let

$A(-3,5),\ B(3,1),\ C(0,3)$

$D(-1, -4)$ be the given points

[distance

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

$AB=\sqrt {(3+3)^2 + (1-5)^2}= \sqrt {6^2 + (-4)^2}$

$AB=\sqrt {36+16}=\sqrt {52}=2\sqrt {13}$

$\boxed {AB=2\sqrt {13}}$

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

$BC=\sqrt {(-3)^2 +(2)^2 }= \sqrt {9+4}=\sqrt {13}$

$\boxed {BC=\sqrt {13}}$

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

$CD= \sqrt {(-1)^2 +(-7)^2}=\sqrt {1+49}$

$\boxed {CD=\sqrt {50}=5\sqrt {2}}$

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

$DA=\sqrt {(2)^2 +(-9)^2 }= \sqrt {4+81}=\sqrt {85}$

$\boxed {DA=\sqrt {85}}$

Now distances

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

$AC=\sqrt {9+4}=\sqrt {13}$

$\boxed {AC=\sqrt {13}}$

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

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

$\boxed {BD=\sqrt {41}}$

Hence we see that from above sides

$AC+BC=AB$

$\sqrt {13}+\sqrt {13}=2\sqrt {13}$

$A, B, C$ are collinear

Hence, points

$A, B, C$ &

$D$ does not from a quadrilateral.

Prove that the points $(3,0),(4,5),(-1,4)$ and $(-2,-1)$ , form a rhombus. Also, find its area.

Let the points of rhombus are A(3,0), B(4,5), C(-1,4) and D(-2,-1).

We know that all the sides of a rhombus are Equal.

So by distance formula we have,

Distance between two points = $\sqrt { { (x_2-x_1) }^{ 2 }+{ (y_2-y_1) }^{ 2 } }$

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

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

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

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

$\therefore AB=BC=CD=AD=\sqrt{26}$

$\therefore ABCD$ is a rhombus (Proved)

AC and BD are the diagonals of the rhombus ABCD

$AC=\sqrt{(3-(-1))^2+(0-4)^2}=\sqrt{16+16}=\sqrt{32}$

$BD=\sqrt{(4-(-2))^2+(5-(-1))^2}=\sqrt{36+36}=\sqrt{72}$

Area=

$0.5\times (AC)\times(BD)$

Area =

$0.5\times\sqrt32 \times \sqrt72=24$ Sq. units

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

Given,

$P(x,y), A(5,1), B(1,5)$

Given,

$PA=PB$ (equidistant)

So by distance formula we have,

Distance between two points = $\sqrt { { (x_2-x_1) }^{ 2 }+{ (y_2-y_1) }^{ 2 } }$

$PA^2=PB^2$

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

$\Rightarrow x^2+25-10x+y^2+1-2y+1-2y=x^2+1-2x+y+25-10y$

$\Rightarrow -10x-2y+26=-2x-10y+26$

$\Rightarrow -5x-y=-x-5y$

$\Rightarrow 4x=4y$

$\Rightarrow x=y$ (proved)

Prove that the points $(3,-2), (4,0), (6,-3)$ and $(5,-5)$ are the vertices of a parallelogram.

Let the vertices of a parallelogram be

$A(-3,2)$ ,

$B(4,0)$ ,

$C(6,-3)$ and

$D(5,-5)$

So by distance formula we have,

Distance between two points = $\sqrt { { (x_2-x_1) }^{ 2 }+{ (y_2-y_1) }^{ 2 } }$

$AB=\sqrt { { (4+3) }^{ 2 }+{ (0+2) }^{ 2 } } =\sqrt { 5 } \\ CD=\sqrt { { (5-6) }^{ 2 }+{ (-5+3) }^{ 2 } } =\sqrt { 5 } \\ BC=\sqrt { { (6-4) }^{ 2 }+{ (-3) }^{ 2 } } =\sqrt { 13 } \\ AD=\sqrt { { (5-3) }^{ 2 }+{ (-5+2) }^{ 2 } } =\sqrt { 13 }$

Opposite site of a parallelogram are equal.

Therefore

$AB=CD=\sqrt{5}$ and

$BC=AD=\sqrt{13}$

$ABCD$ is a parallelogram (PROVED)

If $A=(2,3)$ , $B=(-2,3)$ and $P$ moves such that $|PA-PB|=4$ , then find the locus of $P$ .

1351328_1000655_ans_87532b58d0fa4e4b969c6891f11ae55e.png

If $(-4,0)$ and $(4,0)$ are two vertices of an equilateral triangle, find the coordinates of its third vertex.

Let C(x,y) be the third vertex of triangle ABC having two vertices at A(-4,0) and B(4,0).Since

$\triangle$ ABC is equilateral. Therefore,

AC=BC=AB

Now,AC=BC

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

$\Rightarrow \quad \quad { \left( x+4 \right) }^{ 2 }+{ y }^{ 2 }={ \left( x-4 \right) }^{ 2 }+{ y }^{ 2 }$

$\Rightarrow \quad \quad 16x=0$

$\Rightarrow \quad \quad x=0$
Again

,AC=BC=AB

$\Rightarrow \quad \quad AC=AB$

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

$\Rightarrow \quad \quad { \left( x+4 \right) }^{ 2 }+{ y }^{ 2 }=64$

$\Rightarrow \quad \quad { \left( 0+4 \right) }^{ 2 }+{ y }^{ 2 }=64$

$\Rightarrow \quad \quad { y }^{ 2 }=48$

$\Rightarrow \quad \quad y=\pm 4\sqrt { 3 }$

Hence, the coordinates of the third vertex are

$C\left( 0,4\sqrt { 3 } \right) and\quad D\left( 0,-4\sqrt { 3 } \right) .$

If $P(2,-1), Q(3,4), R(-2,3)$ and $S(-3,-2)$ be four points in a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus.

The given points are P(2,-1), Q(3,4), R(-2,3) and S(-3,-2).We have

$PQ=\sqrt { { \left( 3-2 \right) }^{ 2 }+{ \left( 4+1 \right) }^{ 2 } } =\sqrt { { 1 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 26 } units$

$QR=\sqrt { { \left( -2-3 \right) }^{ 2 }+{ \left( 3-4 \right) }^{ 2 } } =\sqrt { 25+1 } =\sqrt { 26 } units$

$RS=\sqrt { { \left( -3+2 \right) }^{ 2 }+{ \left( -2-3 \right) }^{ 2 } } =\sqrt { 1+25 } =\sqrt { 26 } units$

$SP=\sqrt { { \left( -3-2 \right) }^{ 2 }+{ \left( -2-3 \right) }^{ 2 } } =\sqrt { 26 } units$

$PR=\sqrt { { \left( -2-2 \right) }^{ 2 }+{ \left( 3+1 \right) }^{ 2 } } =\sqrt { 16+16 } =4\sqrt { 2 } units$
and,

$QS=\sqrt { { \left( -3-3 \right) }^{ 2 }+{ \left( -2-4 \right) }^{ 2 } } =\sqrt { 36+36 } =6\sqrt { 2 } units$

$\therefore \quad \quad PQ=QR=RS=SP=\sqrt { 26 } units$
and,

$PR\neq QS$
This means that PQRS is quadrilateral whose sides are equal but diagonals are not equal.

Thus, PQRS is a rhombus but not a square.

.Now, Area of rhombus PQRS=

$\frac { 1 }{ 2 } \times \left( Product\quad of\quad lengths\quad of\quad diagonals \right)$

$\Rightarrow \quad \quad Area\quad of\quad rhombus\quad PQRS=\frac { 1 }{ 2 } \times \left( PR\times QS \right)$

$\Rightarrow \quad \quad Area\quad of\quad rhombus\quad PQRS=\left( \frac { 1 }{ 2 } \times 4\sqrt { 2 } \times 6\sqrt { 2 } \right) sq.\quad units=24sq.units$

Prove that the points $(-2, -1), (1, 0), (4,3)\ and\ (1,2)$ are the vertices of a parallelo-gram. Is it a rectangle?

Let the given point be A,B,C and D respectively. Then,

Coordinates of the mid-point of AC are

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

Coordinates of the mid-point of BD are

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

Thus, AC and BD have the same mid-point. Hence, ABCD is a parallelogram. Now, we shall see whether ABCD is a rectangle or not.

We have

$AC=\sqrt { { \left( 4-\left( -2 \right) \right) }^{ 2 }+{ \left( 3-\left( -1 \right) \right) }^{ 2 } } =2\sqrt { 13 }$

$and,\quad \quad \quad BD=\sqrt { { \left( 1-1 \right) }^{ 2 }+{ \left( 0-2 \right) }^{ 2 } } =2$
Clearly,

$AC\neq BD.$ So, ABCD is not a rectangle.

If coordinate of two points $A$ and $B$ are $(3,4)$ and $(5,-2)$ find coordinates of $P$ if $PA$ = $PB$ and area of $\triangle PAB=10$

Let the coordinate

$P$ be

$(x,y)$

Since it is given that

$PA = PB$

So, firstly we will calculate the distance

$PA$ .

$PA = (x,y) (3,4)$

Distance

$PA = \sqrt{(3-x)^{2}+(4-y)^{2}}$

$PB=(x,y) (5,-2)$

Distance

$PB = \sqrt{(5-x)^{2}+(-2-y)^{2}}$

So,

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

Squaring both the sides in the above equation,

${(3-x)^{2}+(4-y)^{2}}={(5-x)^{2}+(-2-y)^{2}}$

$9+x^{2}-6x+16+y^{2}-8y=25+x^{2}-10x+4+y^{2}+4y$

$-6x-8y=-10x+4+4y$

$4x-12y=4$

$x-3y=1$ (Equation 1)

Now,it is given that Area of

$\triangle PAB = 10$

Area of triangle of

$(3,4)\ (5,-2)$ and

$(x,y)$

Area of triangle is given by the formula

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

Area of

$\triangle PAB = \frac{1}{2}[3(-2-y)+5(y-4)+x(4+2)]=10$

$-6+2y-20+6x=20$

$46=2y+6x$

$3x+y=23$ (Equation 2)

Now, solving equations 1 and 2.

Since

$x-3y=1$

therefore,

$x = 3y+1$

Equation 2 implies,

$3(3y+1)+y=23$

$9y+3+y=23$

$10y=20$

$y= 2$

$x=3y+1$

$x=(3 \times 2)+1$

$x= 7$

Therefore, the coordinates of

$P$ are

$(7,2)$

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

1320616_1078937_ans_3e53e02c932e461482bae38aa958b30d.jpg

Find the points on the X-axis which are at a distance of $2\sqrt{5}$ from the point $(7,-4)$ . How many such points are there?

Let the pt. be

$(x, 0)$ then

$\sqrt{(7 - x^2) + (-4 - 0)^2} = 2 \sqrt{5}$

$49 + x^2 - 14 x + 16 = (2 \sqrt{5} )^2 = 20$

$x^2 = 14 x + 45 = 0$

$x^2 - 9x - 5x + 45 = 0$

$x (x - 9) -5 (x - 9) =0$

$(x - 5) (x - 9) = 0$

$x = 5$ or

$x = 9$

$\therefore$ pt. is

$(5, 0)$ or

$(9, 0)$

$\therefore 2$ pts. are there

If $R(x, y)$ is a point on the line segment joining the points $P(a, b)$ and $Q(b, a)$ , the prove that $x + y = a + b$ .

$R(x,y),P(a,b) and Q(b,a)$

Line

$PQ=>(y-b)=\cfrac { a-b }{ b-a } (x-a)\\ =>y+x=b+a$

Now

$R(x,y)$ lies on line PQ , it must satisfy its equation , which we can see it does

$=>x+y=b+a$

Show that $O(0,0),C(a,0),B(a,b),C(0,b)$ are the vertices of a rectangle. If $P(x,y)$ is a point in the coordinates plane, then prove using formula that $PQ^{2}+PB^{2}=PA^{2}+PC^{2}$ .

$O(o, o), A(o,o), B(a,b), c(a,b)$

$OA=BC, OC=AB$

$OA=2, BC=a$

$OC=b, AB=b$

$\therefore OA=BC$ &

$OC=AB$

eqn of

$OC, x=0$

$OA\Rightarrow y=0$

$AB\Rightarrow x=a$

$BC\Rightarrow y=b$

Let

$P(x,y)$ be in co-ordinates plane

To prove

$PO^{2}+PB^{2}=PA^{2}+PC^{2}$

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

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

$(1)$

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

$(2)$

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

$=\sqrt{(x-a)^{2}+y^{2}}$ ...........

$(3)$

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

$=\sqrt{x^{2}+(y-b)^{2}}$ ..........

$(4)$

$PO^{2}+PB^{2}=x^{2}+y^{2}+(x-a)^{2}+(y-b)^{2}$

$PA^{2}+PC^{2}=(x-a)^{2}+y^{2}+x^{2}+(y-b)^{2}$

$\therefore PO^{2}+PB^{2}=PA^{2}+PC^{2}$

Hence proved

1379355_1140876_ans_6847f589f41549919106ad5909fe26de.png

In each of the following find the value of $'k'$ , for which the points are collinear.
$(7, -2), (5, 1), (3, k)$ .

Given points $\left(7,-2\right) \left(5,1\right) \left(3,k\right)$

$A\left(7,-2\right)\ B\left(5,1\right) \ C\left(3,k\right)$ we have to find $k$

$x_{1} y_{1}$ $x_{2} y_{2}$ $x_{3} y_{3}$

The point are collinear if they lie on same line

$\dfrac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right) +x_{3}\left(y_{1}-y_{2}\right)\right]=0$

$\dfrac{1}{2}\left[7\left(1-k\right)+5\left(k+2\right)+3\left(-2-1\right)\right]=0$

$7-7k+5k+10-6-3=0$

$-2k+17-9=0$

$-2k=-8$

$k=\dfrac{8}{2}$

$\boxed{k=4}$

If the distance between points $(x,0)$ and $(0,3)$ is $5$ , what are the values of $x$ ?

Distance between

$(x,0)$ &

$(0,3)$

$=5$

${ x }^{ 2 }+{ 3 }^{ 2 }={ (5) }^{ 2 }\\ =>{ x }^{ 2 }=25-9=16\\ =>x=\pm 4$

Find the point on Y-axis which is equidistant from $(-6,4)$ and $(2,-8)$ .

Given the point is on Y axis is

$x=0$

the point is

$(0,y)$

$(0,y)$ is equidistant from

$(-6,4)$ and

$(2,-8)$

$\Rightarrow$

$\sqrt { { \left( 0+6 \right) }^{ 2 }+{ \left( y-4 \right) }^{ 2 } } =\sqrt { { \left( 2 \right) }^{ 2 }+{ \left( y+8 \right) }^{ 2 } }$

Squaring both sides

$\Rightarrow$

$36+{y}^{2}+16-8y=4+{y}^{2}+64+16y$

$\Rightarrow$

$52-68=24y$

$\Rightarrow$

$-\cfrac{16}{24}=y$

$\Rightarrow$

$y=\cfrac{-2}{3}$

The point on y axis equidistant from

$(-6,4)$ and

$(2,-8)$ is

$(0,\cfrac{-2}{3})$

Prove that the points $A(-5,4),B(-1,-2)$ and $(5,2)$ the vertices of an isosceles right-angled triangle. Find the coordinates of D, so that ABCD is a square.

$A(-5, 4)\ B(-1, -2)\ C(5, 2)$ .

$AB=\sqrt{4^2+6^2}=\sqrt{52}$

$BC=\sqrt{6^2+4^2}=\sqrt{52}$

$AC=\sqrt{10^2+2^2}=\sqrt{104}$

It is isosceles triangle.

with

$AB=BC$

For making square

$BD=AC=\sqrt{104}$

$\sqrt{(x+1)^2+(y+2)^2}=\sqrt{104}$ ------(1)

Also, slope of

$AC$ is

$\dfrac{-1}{slope\ of\ BD}$

Slope of

$BD=\dfrac{y+2}{x+1}=\dfrac{-1}{\dfrac{-2}{10}}$

$\dfrac{y+2}{x+1}=5$ ------(2)

from (1) & (2)

$x=1\ y=8$

$D(1, 8)$ is new vertex of square.

1234602_1187527_ans_3ddd981886c048ce887a12933f0f39cf.png

If the image of the point $(2, 1)$ with respect to a line mirror be $(5, 2)$ , find the equation of the mirror.

We have

$y-\dfrac{3}{2}=-3\left(x-\dfrac{7}{2}\right)$

$2y-3=-3(2x-7)$

$2y-3=-6x+21$

$6x+2y=24$

$3x+y=12$

Hence this is the centre

1368733_1177773_ans_66e8450ceae249c598b6e3b538e12e9e.png

If the points $(1,1), (-1,-1)$ and $(-\sqrt{3},k)$ are vertices of an equilateral triangle find the value of $k$ .

$A (1, 1) , B(-1, -1), C(-\sqrt{3}, k)$

$AB = AC$

$\Rightarrow \sqrt{4 + 4} = \sqrt{(1 + \sqrt{3})^2 + (1 - k)^2}$

$\Rightarrow 8 = 1 + 3 + 2 \sqrt{3} + (1 - k)^2$

$\Rightarrow 4 - 2 \sqrt{3} = (1 - k)^2$

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

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

$\therefore 1 - k = \sqrt{3} - 1,$ or

$1 - k = 1 - \sqrt{3}$

$\Rightarrow k = 2 -\sqrt{3}$ or

$-k = \sqrt{3}$

$\therefore k = 2 - \sqrt{3}, k = \sqrt{3}$

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

Let the given point be P(2,-5) and Q(-2,9)

And the point required be R(a,0)

As per question,

Point R is equidistant from P & Q

Hence,

$PR=QR$

Finding PR

$x_1=2,\ y_1=-5$

$x_2=a,\ y_2=0$

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

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

$=\sqrt{a^2+4-4a+25}$

$=\sqrt{a^2-4a+29}$

Finding QR

$x_1=-2,\ y_1=9$

$x_2=a,\ y_2=0$

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

$=\sqrt{(a+2)^2+(0-9)^2}$

$=\sqrt{a^2+4+4a+81}$

$=\sqrt{a^2+4a+85}$

As per question

$PR=QR$

$\sqrt{a^2-4a+29}=\sqrt{a^2+4a+85}$

Squaring both sides

$a^2-4a+29=a^2+4a+85$

$-4a-4a=85-29$

$-8a=56$

$a=\dfrac{56}{-8}$

$a=-7$

Hence the required point is R(-7,0)

Find the ordinate of point whose abscissa is $10$ and which is at a distance of $10$ units from the point $P(2,-3)$ .

Given

$x_1=10\quad y_1=y$

$x_2=2\quad y_2=-3$

Distance

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

$d^2=(x_1-x_2)^2+(y_1-y_2)^2$

$100=(10-2)^2+(y+3)^2$

$(y+3)^2=100-64$

$(y+3)^2=36$

$y+3=6\Rightarrow y=3$

$y+3=-6\Rightarrow y=-9$

Find all possible values of $a$ for which the distance between the points $A(a,-1)$ and $B(5,3)$ is $5$ units.

Given,

$A(a, -1), B(5, 3)$

Given,

$AB=5$

$\therefore AB^2=25$

$(a-5)^2+(-1-3)^2=25$

$\Rightarrow (a-5)^2+16=25$

$\Rightarrow (a-5)^2=9$

$\therefore a-5=-3, a-5=3$

$\therefore a=2, a=8$

Prove that the points $A(-2, -1), B(1, 0), C(4, 3)$ and $D(1, 2)$ are the vertices of a parallelogram $ABCD$ .

Given points are

$A(-2,-1), B(1,0), C(4,3)$ and

$D(1,2)$

A quadrilateral is a parallelogram if the opposite sides are equal.

$\therefore AB^2 = (-2-1)^2+(-1-0)^2$

$=9+1$

$=10$

$BC^2 = (4-1)^2+(3-0)^2$

$= 9+9$

$= 18$

$CD^2 = (1-4)^2+(2-3)^2$

$9+1$

$=10$

$AD^2 = (-2-1)^2+(-1-2)^2$

$=9+9$

$=18$

$\therefore AB=CD=\sqrt{10}$ and

$BC = AD =\sqrt{18}$

$\therefore$ The opposite sides of the quadrilateral ABCD are equal, the four points from a parallelogram.

Find a point on the x-axis which is equidistant from the points $(7,6)$ and $(3,4)$ .

Let

$P(\alpha, 0)$ be the point.

From distance formula,

$\sqrt{(7-\alpha)^2+(6)^2}=\sqrt{(3-\alpha)^2+4^2}$

$\Rightarrow (7-\alpha)^2-(3-\alpha)^2=16-36$

$\Rightarrow 49-14\alpha -9+6\alpha =-20$

$\Rightarrow 8\alpha =60$

$\Rightarrow \alpha =\dfrac{15}{2}$

$\Rightarrow P\left(\dfrac{15}{2}, 0\right)$ .

Find that point on y axis which as equidistant from point $(6,5)$ and $(-4,3)$

$(0.4)\,\,(6,5)\,\,(4,3)$

$d=\sqrt{(0-6)^2+(y-5)^2}$ …….

$(1)$

$d=\sqrt{(0+4)^2+(y-3)^2}$ ……

$(2)$

$(1)$

$=$

$(2)$

squaring both sides

$6^2+(y-5)^2=4^2+(4-3)^2$

$36-24+y^2+25-10y=y^2+9-64$

$28=4y$

$y=7$ .

Find the co-ordinates of the point equidistant from the points $A( - 2, - 3),B( - 1,0)$ and $C(7, - 6)$

Given,

the point is equidistant from

$A(-2, -3) B(-1, 0)$ and

$C(7, -6)$

Let the point be

$(x, y)=D$

$AD=BD=CD$

$\sqrt{(x+2)^2+(y+3)^2}=\sqrt{(x+1)^2+(y-6)^2}=\sqrt{(x-7)^2+(y+6)^2}$

Squaring on all sides

$\Rightarrow (x+2)^2+(y+3)^2=(x+1)^2+y^2=(x-7)^2+(y+6)^2$

$\Rightarrow x^2+4x+4+y^2+6y+9y=x^2+2x+1+y^2=x^2-14x+49+y^2+12y+36$

$\Rightarrow 4x-6y+13=2x+1=-14x+12y+85$

Taking first and second equation

$\Rightarrow 4x+6y+13=2x+1$

$\Rightarrow 2x+6y=-12$ ,

$x+3y=-6$ ……(i)

Taking first and third equation

$\Rightarrow 4x+6y+13=-14x+12y+85$

$\Rightarrow 18x-6y=72$ ,

$3x-y=12$ ……(ii)

By equation (i) and (ii)

$3x+9y=-18$

$3x-y=12$

We get

$8y=-6$

$\Rightarrow y=-\dfrac{3}{4}$ and

$x\Rightarrow \dfrac{15}{4}$

$\therefore (x, y)=\left(\dfrac{15}{4}, \dfrac{-3}{4}\right)$ .

Find the cosine of the angle $A$ of the triangle with vertices $A(1, -1), B(6, 11)$ and $C(1, 2)$ .

Slope of side

$AB$ is

$\dfrac{11+1}{6-1}=2.4$

Angle made with

$X- axis$ is

$\tan^{-1}2.4$

Since

$AC$ is parallel to

$Y axis$ angle made with Y-axis is

$\dfrac{\pi}{2}-\tan^{-1}2.4=\text{cot}^{-1}(2.4)=\text{cot}^{-1}\bigg(\dfrac{12}{13}\bigg)$

$\implies \cot A=\dfrac{12}{13}$

Find the distance between the points A (a,b) and B(b,a) if a-b=4.

Distance b/w 2pts

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

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

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

Distance between points (a,b) and (b,a) is

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

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

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

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

$d^{2} = 2\times 4^{2}$

$d^{2} = 32$

$d = 4\sqrt{2}$ units

If point (x,y) is equidistant from points (7,1) and (3,5) show that y =x-2

$\rightarrow$ using distance formula

$\rightarrow \sqrt{(x-7)^{2}+(y-1)^{2}} = \sqrt{(x-3)^{2}+(y-5)^{2}}$

$\Rightarrow -14x+49-2y+1 = -6x+9-10y+25$

$\Rightarrow 8x-8y = 16$

$\Rightarrow x - y = 2 \Rightarrow y = x-2$

1165307_1229431_ans_f0b6d9ab027a469fb73e9138a5ad46ce.jpg

Find a point on $Y$ -axis which is equidistant from $P(-6,4)$ and $Q(2,-8)$ .

$P(-6,4)Q(2,-8)$

Point on Y-axis (0,y) (0,y) equidistant from two points.

$\sqrt{(0+6)^{2}+(y-4)^{2}}=\sqrt{(0-2)^{2}+(y+8)^{2}}$

squaring on both sides

$36+y^{2}-8y+16=4+y^{2}+64+16y$

$52-8y= 68+16y$

$68+16y - 52+8y=0$

$16+24y=0$

$y=\dfrac{-16}{24}=\dfrac{-8}{12}=\dfrac{-2}{3}$

$\therefore$ The point on Y axis is

$(0,-2/3)$

Find the equation of a straight line:
with slope $-1/3$ and $y-intercept$ $-4$ .

Slope(m)

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

y-intercept(i)

$=-4$

$\therefore$ Straight line

$\Rightarrow y=\cfrac{-1}{3}x-4$

$\Rightarrow 3y+x+12=0$

Prove that the points $(0,\,-1).(2,\,1).(0,\,3)$ and $(-2,\,1).$ are the vertices of square?

1334555_1263304_ans_3ddf4bb675984c1f95bfae9ab1d56030.PNG

Show that the points (12 , 8),(-2 , 6) and (6 , 0), are the vertices of a right angled triangle.

1334060_1263451_ans_0ec0163ce3ee4727a931805fac966d19.PNG

Find the coordinates of point P which divides the line segment joining $A(-2,-7)$ and $B(6,1)$ in the ratio $5:3$

Let

$P=\left(x,y\right)$

$\Rightarrow x=\dfrac{5.6+3.\left(-2\right)}{5+3}=\dfrac{30-6}{8}=3$

$y=\dfrac{1.5+3.\left(-7\right)}{8}=\dfrac{5-21}{8}=-2$

$\therefore P=\left( 3,-2\right)$

Hence, the answer is

$\left( 3,-2\right).$

If $A$ is $( - 1,2,0 ) , B$ is $( 1,2,3 )$ and $C$ is $( 4,2,1 ) ,$ then prove that $\Delta A B C$ is an isosceles right triangle.

Distance formula

$\sqrt{\left( x_2-x_1\right)^2+\left( y_2-y_1\right)^2+\left(z_2-z_1\right)^2}$

$\Rightarrow AB=\sqrt{\left( 1+1\right)^2+\left( 2-2\right)^2+\left( 3-0\right)^2}$

$=\sqrt{4+9}$

$=\sqrt{13}$

$\Rightarrow BC=\sqrt{\left( 4-1\right)^2+\left(2-2\right)^2+\left( 1-3\right)^2}$

$=\sqrt{9+4}$

$=\sqrt{13}$

$\Rightarrow AC=\sqrt{\left( 4+1\right)^2+\left( 2-2\right)^2+\left(1-0\right)^2}$

$=\sqrt{25+1}$

$=\sqrt{26}$

Since

$AB=BC$

$\therefore ABC$ is an isosceles triangle

$\Rightarrow AB^2+BC^2=\left( \sqrt{13}\right)^2+\left( \sqrt{13}\right)^2$

$=13+13$

$=26$

$=\left(AC\right)^2$

$\therefore AB^2+BC^2=AC^2$

$\therefore \angle B=90$

$\Rightarrow \triangle ABC$ is an isosceles right angled triangle.

Hence, proved.

1216562_1284710_ans_45194bf136f24e559567f08efe71e8b8.png

How many points are there on the $x-$ axis whose distance from the point $(2,3)$ is less than $3$ units ?

Given point (2, 3)

Let point be (x, 0)

$\sqrt{(x-2)^{2}+(3-0)^{2}} < 3$

$=\sqrt{(x-2)^{2}+9} < 3$

$=(x-2)^{2}+9 < 9$

$=x^{2}+4-4x+9 <9$

$=x^{2}-4x+4 < 0$

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

Thus there are infinite points.

1186710_1291917_ans_2878879dad664669b11d6b735035c6e6.jpg

If the distances of ${ P }({ x },{ y })$ from ${ A }(5,1)$ and ${ B }(-1,5)$ are equal then, prove that $3{ x }=2{ y }.$

Let

$\begin{array}{l} P\left( { x,y } \right) , A\left( { 5,1 } \right) \, B\left( { -1,5 } \right) \\ Given, \\ PA=PB \\ P{ A^{ 2 } }=P{ B^{ 2 } } \\ { \left( { x-5 } \right) ^{ 2 } }+{ \left( { y-1 } \right) ^{ 2 } }={ \left( { x+1 } \right) ^{ 2 } }+{ \left( { y-5 } \right) ^{ 2 } } \\ { x^{ 2 } }+{ y^{ 2 } }-10x-2y+26={ x^{ 2 } }+{ y^{ 2 } }+2x-10y+26 \\ 12x=8y \\ 3x=2y \\ Hence,proved \end{array}$

If the point $P(1,2)$ lies on line segment joining the points $A(2,4)$ and $B(4,8)$ then find $\dfrac{{AP}}{{AB}}$ .

we have:

$P(1, 2)$ ,

$A(2, 4)$ ,

$B(4, 8)$

Distance formula we have,

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

$AP=\sqrt{(2-1)^2+(4-2)^2}=\sqrt{5}$

$AB=\sqrt{(2-4)^2+(4-8)^2}=2\sqrt{5}$

$\Rightarrow \left(\dfrac{AP}{AB}\right)=\dfrac{1}{2}$ .

Find the angle between the curves given below :
$y^{2}=4x,\ x^{2}+y^{2}=5$

We have

$y^2=5$ and

$x^2+y^2=5$

Let's find the intersection point of the two curves

Substituting the value of

$y^2=4x$

We get

$x^2+4x-5=0$

$\implies (x+5)(x-1)=0$

So,

$x=-5$ or

$x=1$

For

$x=5, y=\pm2$

$x=-5$ is neglected as

$y$ have real value.

Slope of tangent of curve

$y^2=4x$ is

$m_1=\cfrac{dy}{dx}=\cfrac{4}{2y}|_{(1,2)}=1$

Slope of tangent of curve

$x^2+y^2=5$ is

$m_2=\cfrac{dy}{dx}=\cfrac{-x}{y}|_{(1,2)}=\cfrac{-1}{2}$

Let angle between them be

$\theta$

Then

$\tan \theta=|\cfrac{m_1-m_2}{1+m_1m_2}|=|\cfrac{1+)(1/2)}{1-(1/2)}|=3$

$\implies \theta=\tan^{-1}3$

evaluate:
$3x-5y=16;x-3y=8$

$3x-5y=16\quad \longrightarrow \left( 1 \right)$

$x-3y=8\quad \longrightarrow \left( 2 \right)$

From eqn

$(2)$ ,

$x=8+3y$

In eqn

$(1)$

$3(8+3y)-5y=16$

$\Rightarrow 24+9y-5y=16$

$4y=-8$

$y=-2$

So,

$x=8+3(-2)$

$=8-6=2$

Hence, this is the answer.

If the slope of a line joining $A(2,x)B(-3,7)$ is $1$ then find $x$ .

As we know that the slope of a line joining the points

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

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

$m = \cfrac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

Therefore,

Slope of line joining the points

$A \left( 2, x \right)$ and

$B \left( -3, 7 \right)$ is-

${m}_{AB} = \cfrac{7 - x}{-3 - 2} = 1 \quad \left( \text{Given} \right)$

$\Rightarrow \cfrac{7 - x}{-5} = 1$

$\Rightarrow 7 - x = -5 \Rightarrow x = 7 + 5 = 12$

Find the equation of the set of points such that the sum of the square of its distance from the points (3,4,5) and (-1, 3, -7) is a constant.

Let (a,b,c) be a set of points whose square of sum of distance from (3,4,5) and (-1,3,-7) is a constant (k)

$(a-3)^{2}+(b-4)^{2}+(c-5)^{2}+(a+1)^{2}+(b-3)^{2}+(c+7)^{2}=k$

$a^{2}+b^{2}+c^{2}-6a-8b-10c+9+16+25+a^{2}+b^{2}+c^{2}+2a-6b+140+1+9+44=k$

$2[a^{2}+b^{2}+c^{2}]-4a-14b+2c+109=k$

$a^{2}+b^{2}+c^{2}-2a-7b+c=\frac{k-109}{2}$

$a^{2}+b^{2}+c^{2}-2a-7b+c=\frac{k-109}{2}$ _______ (1)

Hence equation (1) is the required equation.

1194386_1372632_ans_ee522a1ab74746d386cfc039a8679bc6.JPG

Are the points given below collinear? If, state which point lies between the other two $d(P,Q)=11, d(Q,R)=6.5, d(P,R)=4.5$

$d\left(P,Q\right)>d\left(Q,R\right)>d\left(P,R\right)$

$PQ=11$ units.

$QR=6.5$ units and

$PR=4.5$ units.

We observe that

$6.5+4.5=11.0$ units

$\therefore QR+PR=PQ$

Thus, the points

$P,Q,R$ are collinear.

And

$R$ lies between

$P$ and

$Q$

If $\left(1,x\right)$ is at $\sqrt{10}$ units from $\left(0,0\right)$ then find the values of $x$

Let

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

$B\left({x}_{2},{y}_{2}\right)=\left(0,0\right)$

$\therefore$ Distance

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

$\Rightarrow \sqrt{10}=\sqrt{{\left(0-1\right)}^{2}+{\left(0-x\right)}^{2}}$

$\Rightarrow \sqrt{10}=\sqrt{1+{x}^{2}}$

$\Rightarrow 1+{x}^{2}=10$ by squaring on both sides

$\Rightarrow {x}^{2}=10-1=9$

$\therefore x=\pm 3$

If the points $A\left(2,0,0\right),\,B\left(4,p,q\right)$ and $C\left(1,5,2\right)$ are collinear,find the values of $p$ and $q$

Given that the points

$A\left(2,0,0\right),\,B\left(4,p,q\right)$ and

$C\left(1,5,2\right)$ are collinear.Let the point

$B$ divide the line segment

$AC$ in the ratio

$k:1$ , then by section formula the coordinates of the point

$B$ are

$\left(\dfrac{k+2}{k+1},\,\dfrac{5k+0}{k+1},\,\dfrac{2k+0}{k+1}\right)$

But the coordinates of the point

$B$ are

$\left(4,p,q\right)$ , so we have,

$\dfrac{k+2}{k+1}=4,\,\dfrac{5k}{k+1}=p,\,\dfrac{2k}{k+1}=q$

$\Rightarrow\, k+2=4k+4,\,\dfrac{5k}{k+1}=p,\,\dfrac{2k}{k+1}=q$

$\Rightarrow\, k-4k=4-2,\,\dfrac{5k}{k+1}=p,\,\dfrac{2k}{k+1}=q$

$\Rightarrow\, -3k=2,\,\dfrac{5k}{k+1}=p,\,\dfrac{2k}{k+1}=q$

$\Rightarrow\, k=\dfrac{-2}{3},\,\dfrac{5k}{k+1}=p,\,\dfrac{2k}{k+1}=q$

$\Rightarrow\, k=\dfrac{-2}{3},\,\dfrac{5\times \dfrac{-2}{3}}{\dfrac{-2}{3}+1}=p,\,\dfrac{2\times \dfrac{-2}{3}}{\dfrac{-2}{3}+1}=q$

$\Rightarrow\, k=\dfrac{-2}{3},\,p=\dfrac{\dfrac{-10}{3}}{\dfrac{-2+3}{3}},\,q=\dfrac{\dfrac{-4}{3}}{\dfrac{-2+3}{3}}$

$\Rightarrow\, k=\dfrac{-2}{3},\,p=\dfrac{\dfrac{-10}{3}}{\dfrac{1}{3}},\,q=\dfrac{\dfrac{-4}{3}}{\dfrac{-1}{3}}$

$\therefore\, k=\dfrac{-2}{3},\,p=-10,\,q=4$

If the points $A\left(1,0,-2\right),\,B\left(2,p,q\right)$ and $C\left(-1,5,2\right)$ are collinear,find the values of $p$ and $q$

Given that the points

$A\left(1,0,-2\right),\,B\left(2,p,q\right)$ and

$C\left(-1,5,2\right)$ are collinear.Let the point

$B$ divide the line segment

$AC$ in the ratio

$k:1$ , then by section formula the coordinates of the point

$B$ are

$\left(\dfrac{-k+1}{k+1},\,\dfrac{5k+0}{k+1},\,\dfrac{2k-2}{k+1}\right)$

But the coordinates of the point

$B$ are

$\left(2,p,q\right)$ , so we have,

$\dfrac{-k+1}{k+1}=2,\,\dfrac{5k}{k+1}=p,\,\dfrac{2k-2}{k+1}=q$

$\Rightarrow\,-k+1=2k+2,\,\dfrac{5k}{k+1}=p,\,\dfrac{2k-2}{k+1}=q$

$\Rightarrow\,-k-2k=2-1,\,\dfrac{5k}{k+1}=p,\,\dfrac{2k-2}{k+1}=q$

$\Rightarrow\,-3k=1,\,\dfrac{5k}{k+1}=p,\,\dfrac{2k-2}{k+1}=q$

$\Rightarrow\,k=\dfrac{-1}{3},\,\dfrac{5k}{k+1}=p,\,\dfrac{2k-2}{k+1}=q$

$\Rightarrow\,k=\dfrac{-1}{3},\,\dfrac{\dfrac{-5}{3}}{\dfrac{-1+3}{3}}=p,\,\dfrac{\dfrac{-2}{3}-2}{\dfrac{-1+3}{3}}=q$

$\Rightarrow\,k=\dfrac{-1}{3},\,\dfrac{\dfrac{-5}{3}}{\dfrac{2}{3}}=p,\,\dfrac{\dfrac{-2-6}{3}}{\dfrac{2}{3}}=q$

$\Rightarrow\,k=\dfrac{-1}{3},\,p=\dfrac{\dfrac{-5}{3}}{\dfrac{2}{3}},\,q=\dfrac{\dfrac{-8}{3}}{\dfrac{2}{3}}$

$\Rightarrow\,k=\dfrac{-1}{3},\,p=\dfrac{-5}{2},\,q=\dfrac{\dfrac{-8}{3}}{\dfrac{2}{3}}$

$\Rightarrow\,k=\dfrac{-1}{3},\,p=\dfrac{-5}{2},\,q=\dfrac{-8}{2}$

$\therefore\,k=\dfrac{-1}{3},\,p=\dfrac{-5}{2},\,q=-4$

If the points $A\left(-1,-2,1\right),\,B\left(-2,p,q\right)$ and $C\left(1,-2,-1\right)$ are collinear,find the values of $p$ and $q$ .

Given that the points

$A\left(-1,-2,1\right),\,B\left(-2,p,q\right)$ and

$C\left(1,-2,-1\right)$ are collinear.Let the point

$B$ divide the line segment

$AC$ in the ratio

$k:1$ , then by section formula the coordinates of the point

$B$ are

$\left(\dfrac{k-1}{k+1},\,\dfrac{-2k-2}{k+1},\,\dfrac{-k+1}{k+1}\right)$

But the coordinates of the point

$B$ are

$\left(-2,p,q\right)$ , so we have,

$\dfrac{k-1}{k+1}=-2,\,\dfrac{-2k-2}{k+1}=p,\,\dfrac{-k+1}{k+1}=q$

$\Rightarrow\, k-1=-2k-2,\,\dfrac{-2\left(k+1\right)}{k+1}=p,\,\dfrac{-k+1}{k+1}=q$

$\Rightarrow\, k+2k=-2+1,\,p=-2,\,\dfrac{-k+1}{k+1}=q$

$\Rightarrow\, 3k=-1,\,p=-2,\,\dfrac{-k+1}{k+1}=q$

$\Rightarrow\, k=\dfrac{-1}{3},\,p=-2,\,q=\dfrac{-k+1}{k+1}$

$\Rightarrow\, k=\dfrac{-1}{3},\,p=-2,\,q=\dfrac{-\dfrac{-1}{3}+1}{\dfrac{-1}{3}+1}$

$\Rightarrow\, k=\dfrac{-1}{3},\,p=-2,\,q=\dfrac{\dfrac{1+3}{3}}{\dfrac{-1+3}{3}}$

$\Rightarrow\, k=\dfrac{-1}{3},\,p=-2,\,q=\dfrac{\dfrac{4}{3}}{\dfrac{2}{3}}$

$\Rightarrow\, k=\dfrac{-1}{3},\,p=-2,\,q=\dfrac{4}{2}$

$\therefore\,k=\dfrac{-1}{3},\,p=-2,\,q=2$

If the points $A\left(4,2,2\right),\,B\left(4,p,q\right)$ and $C\left(2,3,3\right)$ are collinear,find the values of $p$ and $q$

Given that the points

$A\left(4,2,2\right),\,B\left(4,p,q\right)$ and

$C\left(2,3,3\right)$ are collinear.Let the point

$B$ divide the line segment

$AC$ in the ratio

$k:1$ , then by section formula the coordinates of the point

$B$ are

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

But the coordinates of the point

$B$ are

$\left(4,p,q\right)$ , so we have,

$\dfrac{2k+4}{k+1}=4,\,\dfrac{3k+2}{k+1}=p,\,\dfrac{3k+2}{k+1}=q$

$\Rightarrow\, 2k+4=4k+4,\,\dfrac{3k+2}{k+1}=p,\,\dfrac{3k+2}{k+1}=q$

$\Rightarrow\, 2k-4k=4-4,\,\dfrac{3k+2}{k+1}=p,\,\dfrac{3k+2}{k+1}=q$

$\Rightarrow\, -2k=0,\,\dfrac{3k+2}{k+1}=p,\,\dfrac{3k+2}{k+1}=q$

$\Rightarrow\, k=0,\,\dfrac{3\times 0+2}{0+1}=p,\,\dfrac{3\times 0+2}{0+1}=q$

$\therefore\,k=0,\,p=2,\,q=2$

Find the relation between x and y such that the point (x,y) is equidistnat from the points (7,1) and (3,5)

1321852_1586384_ans_5ec1785b8e84493ebdc8d0046227f152.jpg

Given $A=(3,1)$ and $B(0,y-1)$ . Find $y$ if $AB=5$ .

As we know that

Distance between two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

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

Given

$A=(3,1),B=(0,y-1)$

$AB=5$

$\implies \sqrt{(3-0)^2+(1-(y-1))^2}=5$

$\implies \sqrt{9+(y-2)^2}=5$

$\implies 9+(y-2)^2=25$

$\implies (y-2)^2=4^2$

$\implies y-2=\pm 4$

$\implies y=2\pm 4$

$\implies y=6,-2$

find the distance between A(cos $\theta$ ,Sin $\theta$ ) and B (sin $\theta$ ,-Cos $\theta$ ) using distance formula.find the co-ordinates of the mid point of the line joining the points (3,-10 ) and (6,-8).

1298573_1607113_ans_5c1de996b99745eba52b3013a53de709.jpg

If $A(1, 2), B(-3, 2)$ and $C(3, -2)$ be the vertices of a $\triangle ABC$ , show that
$\tan A=2$

1847586_1710209_ans_0940552756b44366a615f966e6c20808.jpeg

If $A(1, 2), B(-3, 2)$ and $C(3, -2)$ be the vertices of a $\triangle ABC$ , show that
$\tan B=\dfrac{2}{3}$

Given vertices of

$\Delta$ ABC are A(1, 2), B(-3, 2), C(3, -2)

B is the angle between BC and AB

Slope of a line joining two points

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

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

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

Let

$m_1$ = Slope of line BC =

$\dfrac{-2-2}{3-(-3)}$

$\Rightarrow m_1$ =

$\dfrac{-4}{6}$

$\Rightarrow m_1$ =

$\dfrac{-2}{3}$

Let

$m_2$ = Slope of line AB =

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

$\Rightarrow m_2$ =

$\dfrac{0}{-4}$

$\Rightarrow m_2$ = 0

Now,

$tanB=\mid \dfrac {m_1-m_2}{1+m_1m_2}\mid$

$tanB=\mid \dfrac {(\dfrac{-2}{3})-(0)}{1+(0)(\dfrac{-2}{3})}\mid$

$tanB=\mid \dfrac {(\dfrac{-2}{3})}{1}\mid$

Hence, tanB =

$\dfrac{2}{3}$

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 $-9$ .

1563944_1740445_ans_7b8e69a40cdf4f0785c1b1e86a3052ff.jpg

$A$ is a point on the parabola $y^{2}=4 a x .$ The normal at $A$ cuts the parabola again at point $B$ .If $A B$ subtends a right angle at the vertex of the parabola. find the slope of $A B$

Normal at point

$A\left(a t_{1}^{2}, 2 a t_{1}\right)$ meets the parabola at point

$B\left(a t_{2}^{2}, 2 a t_{2}\right)$

$\Rightarrow t_{2}=-t_{1}-\dfrac{2}{t_{1}} \qquad\qquad\qquad(1)$

Also

$A B$ subtends right angle at vertex,

$\Rightarrow t_{1} t_{2}=-4 \qquad\qquad\qquad(2)$

Eliminating

$t_{2}$ from Eqs. (1) and (2),

$\Rightarrow -\dfrac{4}{t_{1}}=-t_{1}-\dfrac{2}{t_{1}}$

$\Rightarrow \dfrac{2}{t_{1}}=t_{1}$

Then slope of

$A B=$ slope of normal at

$A=\pm \sqrt{2}$

Find the slope of the line passing through the following pairs of points:
$(a, -b)$ and $(b, -a)$

We know that,

$slope = \dfrac{(y_2 - y_1) }{ (x_2 - x_1)}$

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

Find the slope of the line whose inclination is:
$72^\circ 30'$

We know that, the slope of a line is given by the tan of its inclination.

Slope

$= tan 72^\circ 30' = 3.1716$ (Refer table)

1757293_1832785_ans_c9f94008c38544af9c9d410048faff40.jpg

Find the slope of the line passing through the following pairs of points:
$(-2, -3)$ and $(1, 2)$

We know that,

$slope = \dfrac{(y_2 - y_1) }{ (x_2 - x_1)}$

Slope

$=\dfrac{ (2 + 3) }{ (1 + 2)} = \dfrac5 3$

Find the slope of the line whose inclination is:
$46^\circ$

We know that, the slope of a line is given by the tan of its inclination.

Slope

$= tan 46^\circ = 1.0355$ (Refer table)

1757297_1832788_ans_c91829b9b6164574bc120da39d21f97c.jpg

Find the slope of the line passing through the following pairs of points:
$(-4, 0)$ and origin

We know that,

$slope = \dfrac{(y_2 - y_1) }{ (x_2 - x_1)}$

$Slope = \dfrac{(0 - 0) }{ (0 + 4)} = 0$

Find the inclination of the line whose slope is:
$1.0875$

Slope

$= tan\theta = 1.0875$

$\Rightarrow \theta = 47^\circ 24'$ (Refer table)

1757314_1832792_ans_1377059245af4419b2b0e8bf19d67b50.jpg

Find the inclination of the line whose slope is:
$0.7646$

Slope

$= tan \theta = 0.7646$

$\Rightarrow \theta = 37^\circ 24'$ (Refer table)

1757300_1832790_ans_900655735b1a458496fbb990eec3f33c.jpg

Without using the distance formula, show that the points $A (4, 5), B (1, 2), C (4, 3)$ and $D (7, 6)$ are the vertices of a parallelogram.

$\textbf{Step-1: Assume the given points to be equal to some variables & find slopes.}\\$

$\text{Given points are}$

$A (4, 5), B (1, 2), C (4, 3)$

$\text{and}$

$D (7, 6).$

$\text{Slope of}$

$AB = \dfrac{(2 – 5)}{ (1 – 4)} = -\dfrac3{ -3} = 1$

$\text{Slope of}$

$CD = \dfrac{(6 – 3)}{ (7 – 4)} = \dfrac33 = 1$

$\text{As the slope of}$

$AB$

$\text{= slope of}$

$CD$

$\text{So, we can conclude}$

$AB \parallel CD$

$\text{Now,}$

$\text{Slope of}$

$BC = \dfrac{(3 – 2)}{ (4 – 1)} = \dfrac13$

$\text{Slope of}$

$DA = \dfrac{(5 – 6)}{ (4 – 7)} = -\dfrac1{-3} = \dfrac 1 3$

$\text{As, slope of}$

$BC$

$\text{= slope of}$

$DA$

$\text{So, we can say}$

$BC \parallel DA$

$\text{Therefore, both pairs of opposite side of quadrilateral ABCD are parallel.}$

$\text{Thus, ABCD is a parallelogram.}$

$\textbf{Hence, ABCD is a parallelogram.}$

Find the slope of the line parallel to AB if:
$A = (0, -3)$ and $B = (-2, 5)$

We know that,

$slope = \dfrac{(y_2 - y_1) }{ (x_2 - x_1)}$

Slope of

$AB = \dfrac{(5 + 3)}{ (-2 – 0)} =\dfrac 8{-2} = -4$

Hence, the slope of the line parallel to

$AB$ should be = Slope of

$AB$ =

$-4$

Without using the distance formula, show that the points $A (4, -2), B (-4, 4)$ and $C (10, 6)$ are the vertices of a right-angled triangle.

Given points are

$A (4, -2), B (-4, 4)$ and

$C (10, 6).$

Calculating the slopes, we have

Slope of

$AB =\dfrac{ (4 + 2)}{ (-4 – 4)} = \dfrac6{ -8} = -\dfrac34$

Slope of

$BC = \dfrac{(6 – 4)}{ (10 + 4)} = \dfrac2{ 14} = \dfrac17$

Slope of

$AC = \dfrac{(6 + 2)}{ (10 – 4)} = \dfrac8 6 =\dfrac 43$

It’s clearly seen that,

Slope of

$AB = \dfrac{-1}{ Slope\ of\ AC}$

Thus,

$AB \perp AC.$

Therefore, the given points are the vertices of a right-angled triangle.

Find the slope of the line perpendicular to $AB$ if:
$A = (0, -5)$ and $B = (-2, 4)$

We know that,

$slope = \dfrac{(y_2 - y_1) }{ (x_2 - x_1)}$

Slope of

$AB =\dfrac{ (4 + 5)}{ (-2 – 0)} = \dfrac{-9}2$

Slope of the line perpendicular to

$AB$

$\dfrac {-1} {\text{Slope of AB}}$

$\dfrac{-1}{(-9/2 )} = \dfrac 2 9$

Find the slope of the line parallel to AB if:
$A = (-2, 4)$ and $B = (0, 6)$

We know,

$slope = \dfrac{(y_2 - y_1) }{ (x_2 - x_1)}$

Slope of

$AB = \dfrac{(6 - 4)}{ (0 + 2)} = \dfrac22 = 1$

Hence, the slope of the line parallel to

$AB$ should be = Slope of

$AB$ =

$1$

The line passing through $(0, 2)$ and $(-3, -1)$ is parallel to the line passing through $(-1, 5)$ and $(4, a)$ . Find $a.$

We know that,

$slope = \dfrac{(y_2 - y_1) }{ (x_2 - x_1)}$

Slope of the line passing through

$(0, 2)$ and

$(-3, -1) =\dfrac{ (-1 – 2)}{ (-3 – 0)} = \dfrac{-3}{-3} = 1$

Slope of the line passing through

$(-1, 5)$ and

$(4, a) = \dfrac{(a – 5)}{ (4 + 1)} = \dfrac{(a -5)} 5$

As, the lines are parallel.

The slopes must be equal.

$1 = \dfrac{(a – 5)} 5$

$a – 5 = 5$

$a = 10$

Find the slope of the line perpendicular to AB if:
$A = (3, -2)$ and $B = (-1, 2)$

We know that,

$slope = \dfrac{(y_2 - y_1) }{ (x_2 - x_1)}$

Slope of

$AB =\dfrac{ (2+2)}{ (-1 – 3)} = \dfrac{-4}4=-1$

Slope of the line perpendicular to

$AB$

$\dfrac {-1} {\text{Slope of AB}}$

$\dfrac{-1}{(-1 )} = 1$

$(-2, 4), (4, 8), (10, 7)$ and $(11, -5)$ are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.

Let the given points be

$A (-2, 4), B (4, 8), C (10, 7)$ and

$D (11, -5)$ .

And, let

$P, Q, R$ and

$S$ be the mid-points of

$AB, BC, CD$ and

$DA$ respectively.

So,

Co-ordinates of P are

$\left ( \dfrac{-2 + 4}{2}, \dfrac{4 + 8}{2} \right ) = (1, 6)$

Co-ordinates of Q are

$\left ( \dfrac{4 + 10}{2}, \dfrac{8 + 7}{2} \right ) = \left (7, \dfrac{15}{2} \right )$

Co-ordinates of R are

$\left ( \dfrac{10 + 11}{2}, \dfrac{7 - 5}{2} \right ) = \left ( \dfrac{21}{2}, 1 \right )$

Co-ordinates of S are

$\left ( \dfrac{11 - 2}{2}, \dfrac{-5 + 4}{2} \right ) = \left ( \dfrac{9}{2}, \dfrac{-1}{2} \right )$

Slope of

$PQ = \dfrac{\dfrac{15}{2} - 6}{7 - 1} = \dfrac{\dfrac{15 - 12}{2}}{6} = \dfrac{3}{12} = \dfrac{1}{4}$

Slope of

$RS = \dfrac{\dfrac{-1}{2} - 1}{\dfrac{P9}{2} - \dfrac{21}{2}} = \dfrac{\dfrac{-1 - 2}{2}}{\dfrac{9 - 21}{2}} = \dfrac{-3}{-12} = \dfrac{1}{4}$

As, slope of

$PQ =$ Slope of

$RS$ , we can say

$PQ \parallel RS.$

Now,

Slope of

$QR = \dfrac{1 - \dfrac{15}{2}}{\dfrac{21}{2} - 7} = \dfrac{\dfrac{2 - 15}{2}}{\dfrac{21 - 14}{2}} = \dfrac{-13}{7}$

Slope of

$SP = \dfrac{6 + \dfrac{1}{2}}{1 - \dfrac{9}{2}} = \dfrac{\dfrac{12 + 1}{2}}{\dfrac{2 - 9}{2}} = \dfrac{13}{-7} = \dfrac{-13}{7}$

As, slope of

$QR =$ Slope of

$SP$ , we can say

$QR \parallel SP.$

Therefore,

$PQRS$ is a parallelogram.

The line passing through $(-4, -2)$ and $(2, -3)$ is perpendicular to the line passing through $(a, 5)$ and $(2, -1)$ . Find $a.$

We know that,

$slope = \dfrac{(y_2 - y_1) }{ (x_2 - x_1)}$

Slope of the line passing through

$(-4, -2)$ and

$(2, -3) = \dfrac{(-3 + 2)}{ (2 + 4)} = -\dfrac16$

Slope of the line passing through

$(a, 5)$ and

$(2, -1) = \dfrac{(-1 – 5)}{ (2 – a)} = \dfrac{-6}{ (2 –a)}$

As, the lines are perpendicular we have

$-\dfrac16 = \dfrac{-1}{(-6/ 2 – a)}\\$

$-\dfrac16 =\dfrac{ (2 – a)}6$

$2 – a = -1$

$a = 3$

Determine $x$ so that the slope of the line through $(1, 4)$ and $(x, 2)$ is $2.$

We know that,

$slope = \dfrac{(y_2 - y_1) }{ (x_2 - x_1)}$

Given, the slope of the line through

$(1, 4)$ and

$(x, 2)$ is

$2.$

So,

$\dfrac{2 - 4}{x - 1} = 2\\$

$\dfrac{-2}{x - 1} = 2\\$

$\dfrac{-1}{x - 1} = 1\\$

$-1 = x - 1$

$x = 0$

Coordinates of $A$ are $(1,1)$ and of $A_{n}$ are $(n, 2n + 1)$ , $O$ is the origin, then match the columns :

For

$P\left( { x }_{ 1 },{ y }_{ 1 } \right) ,Q\left( { x }_{ 2 },{ y }_{ 2 } \right)$ distance formula is

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

${ \left( O{ A }_{ n } \right) }^{ 2 }={ \left( n-0 \right) }^{ 2 }+{ \left( 2n+1-0 \right) }^{ 2 }={ n }^{ 2 }+4{ n }^{ 2 }+1+4n=5{ n }^{ 2 }+4n+1$

B)

${ \left( A{ A }_{ n } \right) }^{ 2 }={ \left( n-1 \right) }^{ 2 }+{ \left( 2n+1-1 \right) }^{ 2 }={ n }^{ 2 }+1-2n+4{ n }^{ 2 }=5{ n }^{ 2 }-2n+1$

C)

${ \left( A_{ 1 }{ A }_{ n } \right) }^{ 2 }={ \left( n-1 \right) }^{ 2 }+{ \left( 2n+1-3 \right) }^{ 2 }={ n }^{ 2 }+1-2n+4{ n }^{ 2 }+4-8n=5{ n }^{ 2 }-10n+5$

D)

${ \left( A_{ n-1 }{ A }_{ n } \right) }^{ 2 }={ \left( n-\left( n-1 \right) \right) }^{ 2 }+{ \left( 2n+1-\left( 2\left( n-1 \right) +1 \right) \right) }^{ 2 }={ \left( 1 \right) }^{ 2 }+{ \left( 2 \right) }^{ 2 }=1+4=5$

$A_{1}, A_{2},....A_{n}$ are points on the line $y=x$ lying in the positive quadrant such that $OA_{n}=n O \ A_{n-1}, O$ being the origin . If $OA_{1}=1$ then the coordinates of $A_{8}$ are $(3a \sqrt 2, 3a \sqrt 2)$ where $a$ is equal to

We can see that for

$n=8$ we have

$OA_8= 8OA_7 = 8(7OA_6) = 8\times 7(6A_5)..........= 8\times 7\times 6 \times 5 \times 4 \times 3 \times 2 \times 1\ OA_1$
Hence we get the expression written below after substituting the value of

$OA_1 =1$

$OA_8=8!OA_1=40320$

$(3a \sqrt{2},3a \sqrt{2}) = (OA_8 \cos 45^o, OA_8 \sin 45^o)$

$a= \dfrac{1}{3 \sqrt{2}} \times 40320 \times \dfrac{1}{ \sqrt{2}} = 6720$

Find the value of $x$ for which the distance between the points $P(2x,-5)$ and $Q(3x,-2)$ is 5 units.

1415824_1000865_ans_3439e7b7c18044f7a92490ecde176db3.jpg

The x-coordinate of point P is twice its y-coordinate. If P is equidistant from $Q(2, -3)$ and $R(-3, 6)$ then find the coordinate of P.

1415822_1000481_ans_bf67bac1f3784755a05dc46a07bb6503.png

Let $ABC$ ,be a triangle such that the coordinates of the vertex $A$ are $(-3,1)$ .Equation of the median through $B$ is $2x+y-3=0$ and equation of the angular bisector of $C$ is $7x-4y-1=0$ .Find the slope of line $BC$ .

Consider the coordinates of C as $\left( {a,b} \right)$ , and since the point C lies on the angular bisector of C, therefore the equation becomes,

$7a - 4b = 1$ (1)

The midpoint of AC will be $\left( {\dfrac{{a - 3}}{2},\dfrac{{b + 1}}{2}} \right)$ and as the median lies on the midpoint of B, therefore the equation becomes,

$2\left( {\dfrac{{a - 3}}{2}} \right) + \left( {\dfrac{{b + 1}}{2}} \right) = 3$

$2a - 6 + b + 1 = 6$

$2a + b = 11$ (2)

From equation (1) and (2),

$a = 3,b = 5$

Slope of AC is,

${m_1} = \dfrac{{5 - 1}}{{3 - \left( { - 3} \right)}}$

$= \dfrac{2}{3}$

The slope of bisector is,

${m_2} = - \dfrac{{{\rm{Coeff}}{\rm{.}}\;{\rm{of}}\;x}}{{{\rm{Coeff}}{\rm{.}}\;{\rm{of}}\;y}}$

$= \dfrac{7}{4}$

The angle between AC and the bisector will be,

$\tan \alpha = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$

$\ = \left| {\dfrac{{\dfrac{2}{3} - \dfrac{7}{4}}}{{1 + \left( {\dfrac{2}{3}} \right)\left( {\frac{7}{4}} \right)}}} \right|$

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

Let $m$ be the slope of BC.

The angle between BC and bisector will be equal to the angle between AC and bisector.

$\left| {\dfrac{{m - {m_2}}}{{1 + m{m_2}}}} \right| = \frac{1}{2}$

$\left| {\dfrac{{m - \frac{7}{4}}}{{1 + m\left( {\frac{7}{4}} \right)}}} \right| = \frac{1}{2}$

$\dfrac{{4m - 7}}{{7m + 4}} = \pm \dfrac{1}{2}$

$m = \dfrac{2}{3}\;;\;m = 18$

When the slope of BC is $\dfrac{2}{3}$ , then A,B,C will be collinear and it will form no triangle.

Therefore, the slope of BC is 18.

Prove that the points $(at^2_1, 2at_1)$ , $(at^2_2, 2at_2)$ and $(0, a)$ will be collinear, if $t_1t_2=-1$ .

$A(a{ t }_{ 1 }^{ 2 },{ 2at }_{ 1 })$ and

$B(a{ t }_{ 2 }^{ 2 },{ 2at }_{ 2 })$ and

$C(0,a)$

=>Area of

$\triangle ABC=0$ for collinearity

=>Area of

$\triangle ABC=\cfrac { 1 }{ 2 } \begin{vmatrix} a{ t }_{ 1 }^{ 2 } & { 2at }_{ 1 } \\ a{ t }_{ 2 }^{ 2 } & { 2at }_{ 2 } \\ 0 & a \\ a{ t }_{ 1 }^{ 2 } & { 2at }_{ 1 } \end{vmatrix}\\ 0=\cfrac { 1 }{ 2 } \left| 2{ a^{ 2 } }{ t }_{ 1 }^{ 2 }{ t }_{ 2 }-2a^{ 2 }{ t }_{ 1 }{ t }_{ 2 }^{ 2 }+a^{ 2 }{ t }_{ 2 }^{ 2 }-a^{ 2 }{ t }_{ 1 }^{ 2 } \right| \\ 0=\cfrac { 1 }{ 2 } \left| a^{ 2 }({ t }_{ 1 }{ t }_{ 2 }({ 2t }_{ 1 }-{ 2t }_{ 2 }))+a^{ 2 }({ t }_{ 2 }-{ t }_{ 1 })({ t }_{ 2 }+{ t }_{ 1 }) \right| \\ 0=\cfrac { 1 }{ 2 } \left| a^{ 2 }({ t }_{ 1 }{ t }_{ 2 }+1)({ t }_{ 1 }+{ t }_{ 2 }) \right| \\ =>({ t }_{ 1 }{ t }_{ 2 }+1)=0\quad or \quad { t }_{ 1 }-{ t }_{ 2 }=0\\ \\$

${ t }_{ 1 }-{ t }_{ 2 }$ cannot be zero,if so they will coincide.

${ t }_{ 1 }{ t }_{ 2 }=-1$

Show that $A(2, 3), B(4, 5)$ and $C(3, 2)$ can be the vertices of a rectangle. Find the coordinates of the fourth vertex.

Given:

$A=\left( 2,3 \right) ;B=\left( 4,5 \right) ;C=\left( 3,2 \right)$

Slope of

$AB=\cfrac { 5-3 }{ 4-2 } =1$

Slope of

$BC=\cfrac { 2-5 }{ 3-4 } =\cfrac { -3 }{ -1 } =3$

Slope of

$AC=\cfrac { 2-3 }{ 3-2 } =-1$

Slope of

$AB\times$ Slope of

$AC=-1$

$\therefore AB\bot AC$

Let

$D=(x,y)$

Slope of AC=Slope of BD

$\therefore \cfrac { 3-2 }{ 2-3 } =\cfrac { 5-y }{ 4-x }$

$\Rightarrow 5-y=x-4\Rightarrow x+y-9=0$

Also, Slope of

$BD\times$ Slope of

$DC=-1$

$\Rightarrow \cfrac { 5-y }{ 4-x } \times \cfrac { y-2 }{ x-3 } =-1$

$\Rightarrow \left( 5-y \right) \left( y-2 \right) =\left( 3-x \right) \left( 4-x \right)$

$\Rightarrow 5y-10-{ y }^{ 2 }+2y=12-3x-4x+{ x }^{ 2 }$

$\Rightarrow { y }^{ 2 }-7y+10=-{ x }^{ 2 }+7x-12$

Puttig

$y=9-x$

${ \left( 9-x \right) }^{ 2 }-7\left( 9-x \right) +10+{ x }^{ 2 }-7x+12=0$

$\Rightarrow 81+{ x }^{ 2 }-18x-63+7x+10+{ x }^{ 2 }-7x+10=0$

$\Rightarrow 2{ x }^{ 2 }-18x+40=0$

${ x }^{ 2 }-9x+20=0\Rightarrow x=5$ or

$x=4$

$x=5,y=4$

$x=4,\quad y=5$

$\left( 4,5 \right)$ is already a vertex (B)

$\therefore D=(5,4)$

$\therefore$ Four vertices are

$A(2,3);B(4,5);C(3,2);D(5,4)$

1166684_1184568_ans_311a46e3ee2d47aabcf125019089f06a.JPG

Find the points on the x-axis, whose distance from the line $\dfrac{x}{3}+\dfrac{y}{4}=1$ are $4$ units.

$Po{ { int } }\, \, on\, x-axis\, \, is\, (x,0) \\ Dis\tan ce\, from\, \, \, line=\frac { { \left| { \frac { { x1 } }{ 3 } +\frac { { y1 } }{ 4 } -1 } \right| } }{ { \sqrt { \frac { 1 }{ { { 3^{ 3 } } } } +\frac { 1 }{ { { 4^{ 2 } } } } } } } \\ 4=\frac { { \left| { \frac { x }{ 3 } +0-1 } \right| } }{ { \sqrt { \frac { 1 }{ 9 } +\frac { 1 }{ 6 } } } } \\ 4=\frac { { \left| { \frac { x }{ 3 } -1 } \right| } }{ { \sqrt { \frac { { 25 } }{ { 9\times 16 } } } } } \\ 4\times \sqrt { \frac { { 25 } }{ { 9\times 16 } } } =\frac { x }{ 3 } -1 \\ 4\times \frac { 5 }{ { 3\times 4 } } =\frac { x }{ 3 } -1 \\ \frac { 5 }{ 3 } +1=\frac { x }{ 3 } =\frac { 8 }{ 3 } ,\, \therefore \, x=8 \\ Po{ { int } }\, \, is\, \, (8,0)$

Find the angle between the lines whose slopes are $(2-\sqrt 3)$ and $(2+\sqrt 3)$ .

1972515_1710206_ans_567532a632f24b1d84f84af0669bf787.jpg

If $\theta$ us the angle between the diagonals of a parallelogram $ABCD$ whose vertices are $A(0, 2), B(2, -1), C(4, 0)$ and $D(2, 3)$ . Show that $\tan \theta=2$ .

1855011_1710222_ans_c7fe49edaef441c583223d51a0833d89.jpeg

If $\theta$ is the angle between the lines joining the points $A(0, 0)$ and $B(2, 3)$ , and the points $C(2, -2)$ and $D(3, 5)$ , show that $\tan \theta=\dfrac{11}{23}$ .

1948993_1710217_ans_283cfa7493854ad3be0dd5ecf9ce3f62.jpg

If $A(1, 2), B(-3, 2)$ and $C(3, -2)$ be the vertices of a $\triangle ABC$ , show that
$\tan C=\dfrac{4}{7}$

Given vertices of

$\Delta$ ABC are A(1, 2), B(-3, 2), C(3, -2)

C is the angle between BC and AC

Slope of a line joining two points

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

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

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

Let

$m_1$ = Slope of line BC =

$\dfrac{-2-2}{3-(-3)}$

$\Rightarrow m_1$ =

$\dfrac{-4}{6}$

$\Rightarrow m_1$ =

$\dfrac{-2}{3}$

Let

$m_2$ = Slope of line AC =

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

$\Rightarrow m_2$ =

$\dfrac{-4}{2}$

$\Rightarrow m_2$ = -2

Now,

$tanC=\mid \dfrac {m_1-m_2}{1+m_1m_2}\mid$

$tanC=\mid \dfrac {(\dfrac{-2}{3})-(-2)}{1+(-2)(\dfrac{-2}{3})}\mid$

$tanC=\mid \dfrac {(\dfrac{4}{3})}{(\dfrac{7}{3})}\mid$

Hence, tanC =

$\dfrac{4}{7}$

Find the angle between the lines whose slopes are $\sqrt 3$ and $\dfrac{1}{\sqrt 3}$ .

1972510_1710205_ans_1d49e6d4d1e14a8d94b07b91cc43c049.jpg

Straight Lines - Class 11 Commerce Applied Mathematics - Extra Questions

How many points are there on the x-axis which are at a distance of 2√5 2 \sqrt{5} from the point (7,4)(7, 4)?

Determine whether the points are vertices of a right triangle A(−3,−4)A(-3, -4), B(2,6)B(2, 6) and C(−6,10)C(-6, 10)

The point on the y-axis equidistant from (−5,2)(-5, 2) and (9,−2)(9, -2) is (0,−m)(0,-m), them mm

Find the perimeter of a triangle whose vertices have the coordinates (3,10),(5,2)(3,10),(5,2) and (14,12)(14,12).

Show that the two parabolas x2+4a(y−2b−a)=0 x^2 + 4a (y - 2b - a ) = 0 and y2=4b(x−2a+b) y^2 = 4b (x - 2a + b) intersect at right angles at a common end of the latus rectum of each.

Which point on x-axis is equidistant from (5,9)(5,9) and (−4,6)(-4,6)?

The vertices of a triangle are A(−2,2),B(4,4)andC(8,2).A\left( { - 2,\,2} \right),\,B\left( {4,\,4} \right)\,and\,C\left( {8,\,2} \right). Find the length of the medians to the sides

Find the slope of the line passing through the points (3,−2)(3, -2) & (7,−2)(7, -2)

Find the value of xx for which the distance between p(2,−3)p(2, -3) and q(x,5)q(x, 5) is 1010 units.

Find a point on x−axisx-axis which is equidistant from A(2,−5)A(2, -5) and (−2,9)(-2, 9).

Find the slope of line which passes through the point (7,11) and (9,15) .

Show that the points (3,−2),(1,0),(−1,−2)(3, -2), (1, 0), (-1, -2) and (1,−4)(1, -4) are concyclic

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

If the points A(1,2),B(−4,3),C(x,2)A(1,2), B(-4,3), C(x,2) and D(2,3)D(2,3) form parallelogram find the value of xx.

If the point A(a,2)A(a,2) is equidistant from the points B(8,−2)B(8,-2) and C(2,−2)C(2,-2), the value of aa.

If (a,b)(a,b) is the mid-point of the line segment joining the points A(10,−6),B(k,4)A\left( {10, - 6} \right),B\left( {k,4} \right) and a−2b=18a-2b=18,find the value of kk and the distance AB.

A point P on X-axis is at distance 13 from A(11,12). Find the coordinates of P.

If P(2,P)P ( 2 , P ) is the mid point of the line segment joining the the points. A(6,−5)A ( 6 , - 5 ) and B(−2,11)B ( - 2,11 ) find the value of P.P .

Find the slope of the line whose inclination is 30o30^{o}

21 - 3(b - 7) = b + 20

Find the inclination of a line whose slope is(i) 11(ii) −1-1(iii) √3\sqrt 3 (iv) −√3-\sqrt 3 (v) 1√3\frac{1}{{\sqrt 3 }}

Find the slope of the line perpendicular to ABAB if A=(0,−5)A=(0,-5) and B=(−2,4)B=(-2,4).

Find the distance between the following pairs of points:-(2,3),(4,1)(2,3),(4,1)

Find the angle between the line √3x+y+1=0\sqrt {3}x+y+1=0 and x+1=0x+1=0.

If point (x,y)(x,y) is equidistant from point (3,k)(3,k) and (k,3)(k,3) then find the value of kk when x≠yx\neq y

Show that the point A(-2,3,5), B(1,2,3) and C(7,0,-1) are collinear .

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

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

A(0,2),B(0,5)A ( 0,2 ) , B ( 0,5 ) be any two points, then the distance between them is _________.

The distance between the two points is 5.5. One of them is (3,2)(3, 2) and the ordinate of the second is −1-1 , then find sum of its x-coordinate.

Mid-points of the sides AB and AC of ΔABC\Delta ABC are (3,5)(3, 5) and (−3,−3)(-3, -3), respectively, then find the length of BC.

The vertices of a triangle are A (2,2)(2, 2), B (−4,−4)(-4, -4), C (5,−8)(5, -8). Then, if the length of the median through C is √m\sqrt m.Find mm

If P(x,y)(x, y) is a point in the coordinate plane such that locus of P is

Find the angles between the lines √3x+y=1\displaystyle \sqrt{3}x+y= 1 and x+√3y=1x+\sqrt{3}y= 1

Find the slope of a line passing through the points with coordinates (−1,0)(-1, 0) and (1,3)(1, 3).

Find the equation of the line that passes through the points (−1,0)(-1,0) and (−2,4)(-2,4)

The distance between A(1, 3) and B(x, 7) isCalculate the possible values of x

A chair is 55 feet from one wall of a room and 77 feet from the wall at a right angle to it. How far is the chair from the intersection of the two walls?

Find the slope of the line with inclination 60∘60^\circ

Prove that the equations to the straight lines passing through the origin which make an angle α\alpha with the straight line y+x=0 y + x = 0 are given by the equationx2+2xysec2α+y2=0. x^2 + 2xy \sec 2 \alpha + y^2 = 0 .

Prove that the angle between the straight lines joining the origin to the intersection of the straight line y=3x+2y = 3x + 2 with the curve x2+2xy+3y2+4x+8y−11=0 x^2 + 2xy + 3y^2 + 4x + 8y - 11 = 0 is tan−12√23 \tan^{-1} \dfrac {2 \sqrt2}{3}

Prove that the points (3,0)(6,4)(3, 0) (6, 4) and (−1,3)(-1, 3) are the vertices of a right angled isosceles triangle.

Find the value of kk if the angle between St. lines 4x−y+7=0,kx−5y−9=04x - y + 7 = 0, kx - 5y - 9 = 0 is 45∘45^{\circ}.

Prove that the product of the lengths of the perpendiculars drawn from the points (√a2−b2,0)(\sqrt{a^2-b^2}, 0) and(−√a2−b2,0) (-\sqrt{a^2-b^2}, 0) to the line xacosθ+ybsinθ=1\dfrac{x}{a} cos\theta+\dfrac{y}{b}sin\theta= 1 is b2b^2.

Show that the quadrilateral whose vertices are (2,−1),(3,4),(−2,3)(2,-1),(3,4),(-2,3) and (−3,−2)(-3,-2) is a rhombus.

Which point on y-axis is equidistant from (2,3)(2,3) and (−4,1)(-4,1)?

Two vertices of an isosceles triangle are (2,0)(2,0) and (2,5)(2,5). Find the third vertex if the length of the equal sides is 33.

Find the centre of the circle passing through (6,−6),(3,−7)(6,-6),(3,-7) and (3,3)(3,3)

Name the quadrilateral formed, if any, by the following points and give reasons for your answers:A(−1,−2),B(1,0),C(−1,2),D(−3,0)A(-1,-2),B(1,0),C(-1,2),D(-3,0)

Prove that the points (3,0),(4,5),(−1,4)(3,0),(4,5),(-1,4) and (−2,−1)(-2,-1), form a rhombus. Also, find its area.

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

Prove that the points (3,−2),(4,0),(6,−3)(3,-2), (4,0), (6,-3) and (5,−5)(5,-5) are the vertices of a parallelogram.

If A=(2,3)A=(2,3), B=(−2,3)B=(-2,3) and PP moves such that |PA−PB|=4|PA-PB|=4, then find the locus of PP.

If (−4,0)(-4,0) and (4,0)(4,0) are two vertices of an equilateral triangle, find the coordinates of its third vertex.

If P(2,−1),Q(3,4),R(−2,3)P(2,-1), Q(3,4), R(-2,3) and S(−3,−2)S(-3,-2) be four points in a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus.

Prove that the points (−2,−1),(1,0),(4,3) and (1,2)(-2, -1), (1, 0), (4,3)\ and\ (1,2) are the vertices of a parallelo-gram. Is it a rectangle?

If coordinate of two points AA and BB are (3,4)(3,4) and (5,−2)(5,-2) find coordinates of PP if PAPA=PBPB and area of △PAB=10\triangle PAB=10

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

Find the points on the X-axis which are at a distance of 2√52\sqrt{5} from the point (7,−4)(7,-4). How many such points are there?

If R(x,y)R(x, y) is a point on the line segment joining the points P(a,b)P(a, b) and Q(b,a)Q(b, a), the prove that x+y=a+bx + y = a + b.

Show that O(0,0),C(a,0),B(a,b),C(0,b)O(0,0),C(a,0),B(a,b),C(0,b) are the vertices of a rectangle. If P(x,y)P(x,y) is a point in the coordinates plane, then prove using formula that PQ2+PB2=PA2+PC2PQ^{2}+PB^{2}=PA^{2}+PC^{2}.

In each of the following find the value of ′k′'k', for which the points are collinear.(7,−2),(5,1),(3,k)(7, -2), (5, 1), (3, k).

If the distance between points (x,0)(x,0) and (0,3)(0,3) is 55, what are the values of xx?

Find the point on Y-axis which is equidistant from (−6,4)(-6,4) and (2,−8)(2,-8).

Prove that the points A(-5,4),B(-1,-2)A(-5,4),B(-1,-2) and (5,2)(5,2) the vertices of an isosceles right-angled triangle. Find the coordinates of D, so that ABCD is a square.

If the image of the point (2, 1)(2, 1) with respect to a line mirror be (5, 2)(5, 2), find the equation of the mirror.

If the points (1,1), (-1,-1)(1,1), (-1,-1) and (-\sqrt{3},k)(-\sqrt{3},k) are vertices of an equilateral triangle find the value of kk.

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

Find the ordinate of point whose abscissa is 1010 and which is at a distance of 1010 units from the point P(2,-3)P(2,-3).

Find all possible values of aa for which the distance between the points A(a,-1)A(a,-1) and B(5,3)B(5,3) is 55 units.

Prove that the points A(-2, -1), B(1, 0), C(4, 3)A(-2, -1), B(1, 0), C(4, 3) and D(1, 2)D(1, 2) are the vertices of a parallelogram ABCDABCD.

Find a point on the x-axis which is equidistant from the points (7,6)(7,6) and (3,4)(3,4).

Find that point on y axis which as equidistant from point (6,5)(6,5) and (-4,3)(-4,3)

Find the co-ordinates of the point equidistant from the points A( - 2, - 3),B( - 1,0)A( - 2, - 3),B( - 1,0) and C(7, - 6)C(7, - 6)

Find the cosine of the angle AA of the triangle with vertices A(1, -1), B(6, 11)A(1, -1), B(6, 11) and C(1, 2)C(1, 2).

Find the distance between the points A (a,b) and B(b,a) if a-b=4.

If point (x,y) is equidistant from points (7,1) and (3,5) show that y =x-2

Find a point on YY-axis which is equidistant from P(-6,4)P(-6,4) and Q(2,-8)Q(2,-8).

Find the equation of a straight line:with slope -1/3-1/3 and y-intercepty-intercept -4-4.

How many points are there on the x-axis which are at a distance of $2 \sqrt{5}$ from the point $(7, 4)$ ?

Determine whether the points are vertices of a right triangle $A(-3, -4)$ , $B(2, 6)$ and $C(-6, 10)$

The point on the y-axis equidistant from $(-5, 2)$ and $(9, -2)$ is $(0,-m)$ , them $m$

Find the perimeter of a triangle whose vertices have the coordinates $(3,10),(5,2)$ and $(14,12)$ .

Show that the two parabolas $x^2 + 4a (y - 2b - a ) = 0$ and $y^2 = 4b (x - 2a + b)$ intersect at right angles at a common end of the latus rectum of each.

Which point on x-axis is equidistant from $(5,9)$ and $(-4,6)$ ?

The vertices of a triangle are $A\left( { - 2,\,2} \right),\,B\left( {4,\,4} \right)\,and\,C\left( {8,\,2} \right).$ Find the length of the medians to the sides

Find the slope of the line passing through the points $(3, -2)$ & $(7, -2)$

Find the value of $x$ for which the distance between $p(2, -3)$ and $q(x, 5)$ is $10$ units.

Find a point on $x-axis$ which is equidistant from $A(2, -5)$ and $(-2, 9)$ .

Show that the points $(3, -2), (1, 0), (-1, -2)$ and $(1, -4)$ are concyclic

If the points $A(1,2), B(-4,3), C(x,2)$ and $D(2,3)$ form parallelogram find the value of $x$ .

If the point $A(a,2)$ is equidistant from the points $B(8,-2)$ and $C(2,-2)$ , the value of $a$ .

If $(a,b)$ is the mid-point of the line segment joining the points $A\left( {10, - 6} \right),B\left( {k,4} \right)$ and $a-2b=18$ ,find the value of $k$ and the distance AB.

If $P ( 2 , P )$ is the mid point of the line segment joining the the points. $A ( 6 , - 5 )$ and $B ( - 2,11 )$ find the value of $P .$

Find the slope of the line whose inclination is $30^{o}$

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 slope of the line perpendicular to $AB$ if $A=(0,-5)$ and $B=(-2,4)$ .

Find the distance between the following pairs of points:-
$(2,3),(4,1)$

Find the angle between the line $\sqrt {3}x+y+1=0$ and $x+1=0$ .

If point $(x,y)$ is equidistant from point $(3,k)$ and $(k,3)$ then find the value of $k$ when $x\neq y$

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

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

$A ( 0,2 ) , B ( 0,5 )$ be any two points, then the distance between them is _________.

The distance between the two points is $5.$ One of them is $(3, 2)$ and the ordinate of the second is $-1$ , then find sum of its x-coordinate.

Mid-points of the sides AB and AC of $\Delta ABC$ are $(3, 5)$ and $(-3, -3)$ , respectively, then find the length of BC.

The vertices of a triangle are A $(2, 2)$ , B $(-4, -4)$ , C $(5, -8)$ . Then, if the length of the median through C is $\sqrt m$ .Find $m$

If P $(x, y)$ is a point in the coordinate plane such that locus of P is

Find the angles between the lines $\displaystyle \sqrt{3}x+y= 1$ and $x+\sqrt{3}y= 1$

Find the slope of a line passing through the points with coordinates $(-1, 0)$ and $(1, 3)$ .

Find the equation of the line that passes through the points $(-1,0)$ and $(-2,4)$

A chair is $5$ feet from one wall of a room and $7$ feet from the wall at a right angle to it. How far is the chair from the intersection of the two walls?

Find the slope of the line with inclination $60^\circ$

Prove that the equations to the straight lines passing through the origin which make an angle $\alpha$ with the straight line $y + x = 0$ are given by the equation
$x^2 + 2xy \sec 2 \alpha + y^2 = 0 .$

Prove that the angle between the straight lines joining the origin to the intersection of the straight line $y = 3x + 2$ with the curve $x^2 + 2xy + 3y^2 + 4x + 8y - 11 = 0$ is $\tan^{-1} \dfrac {2 \sqrt2}{3}$

Prove that the points $(3, 0) (6, 4)$ and $(-1, 3)$ are the vertices of a right angled isosceles triangle.

Find the value of $k$ if the angle between St. lines $4x - y + 7 = 0, kx - 5y - 9 = 0$ is $45^{\circ}$ .

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

Show that the quadrilateral whose vertices are $(2,-1),(3,4),(-2,3)$ and $(-3,-2)$ is a rhombus.

Which point on y-axis is equidistant from $(2,3)$ and $(-4,1)$ ?

Two vertices of an isosceles triangle are $(2,0)$ and $(2,5)$ . Find the third vertex if the length of the equal sides is $3$ .

Find the centre of the circle passing through $(6,-6),(3,-7)$ and $(3,3)$

Name the quadrilateral formed, if any, by the following points and give reasons for your answers:
$A(-1,-2),B(1,0),C(-1,2),D(-3,0)$

Prove that the points $(3,0),(4,5),(-1,4)$ and $(-2,-1)$ , form a rhombus. Also, find its area.

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

Prove that the points $(3,-2), (4,0), (6,-3)$ and $(5,-5)$ are the vertices of a parallelogram.

If $A=(2,3)$ , $B=(-2,3)$ and $P$ moves such that $|PA-PB|=4$ , then find the locus of $P$ .

If $(-4,0)$ and $(4,0)$ are two vertices of an equilateral triangle, find the coordinates of its third vertex.

If $P(2,-1), Q(3,4), R(-2,3)$ and $S(-3,-2)$ be four points in a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus.

Prove that the points $(-2, -1), (1, 0), (4,3)\ and\ (1,2)$ are the vertices of a parallelo-gram. Is it a rectangle?

If coordinate of two points $A$ and $B$ are $(3,4)$ and $(5,-2)$ find coordinates of $P$ if $PA$ = $PB$ and area of $\triangle PAB=10$

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

Find the points on the X-axis which are at a distance of $2\sqrt{5}$ from the point $(7,-4)$ . How many such points are there?

If $R(x, y)$ is a point on the line segment joining the points $P(a, b)$ and $Q(b, a)$ , the prove that $x + y = a + b$ .

Show that $O(0,0),C(a,0),B(a,b),C(0,b)$ are the vertices of a rectangle. If $P(x,y)$ is a point in the coordinates plane, then prove using formula that $PQ^{2}+PB^{2}=PA^{2}+PC^{2}$ .

In each of the following find the value of $'k'$ , for which the points are collinear.
$(7, -2), (5, 1), (3, k)$ .

If the distance between points $(x,0)$ and $(0,3)$ is $5$ , what are the values of $x$ ?

Find the point on Y-axis which is equidistant from $(-6,4)$ and $(2,-8)$ .

Prove that the points $A(-5,4),B(-1,-2)$ and $(5,2)$ the vertices of an isosceles right-angled triangle. Find the coordinates of D, so that ABCD is a square.

If the image of the point $(2, 1)$ with respect to a line mirror be $(5, 2)$ , find the equation of the mirror.

If the points $(1,1), (-1,-1)$ and $(-\sqrt{3},k)$ are vertices of an equilateral triangle find the value of $k$ .

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

Find the ordinate of point whose abscissa is $10$ and which is at a distance of $10$ units from the point $P(2,-3)$ .

Find all possible values of $a$ for which the distance between the points $A(a,-1)$ and $B(5,3)$ is $5$ units.

Prove that the points $A(-2, -1), B(1, 0), C(4, 3)$ and $D(1, 2)$ are the vertices of a parallelogram $ABCD$ .

Find a point on the x-axis which is equidistant from the points $(7,6)$ and $(3,4)$ .

Find that point on y axis which as equidistant from point $(6,5)$ and $(-4,3)$

Find the co-ordinates of the point equidistant from the points $A( - 2, - 3),B( - 1,0)$ and $C(7, - 6)$

Find the cosine of the angle $A$ of the triangle with vertices $A(1, -1), B(6, 11)$ and $C(1, 2)$ .

Find a point on $Y$ -axis which is equidistant from $P(-6,4)$ and $Q(2,-8)$ .

Find the equation of a straight line:
with slope $-1/3$ and $y-intercept$ $-4$ .

Prove that the points $(0,\,-1).(2,\,1).(0,\,3)$ and $(-2,\,1).$ are the vertices of square?

Find the coordinates of point P which divides the line segment joining $A(-2,-7)$ and $B(6,1)$ in the ratio $5:3$

If $A$ is $( - 1,2,0 ) , B$ is $( 1,2,3 )$ and $C$ is $( 4,2,1 ) ,$ then prove that $\Delta A B C$ is an isosceles right triangle.

How many points are there on the $x-$ axis whose distance from the point $(2,3)$ is less than $3$ units ?

If the distances of ${ P }({ x },{ y })$ from ${ A }(5,1)$ and ${ B }(-1,5)$ are equal then, prove that $3{ x }=2{ y }.$

If the point $P(1,2)$ lies on line segment joining the points $A(2,4)$ and $B(4,8)$ then find $\dfrac{{AP}}{{AB}}$ .

Find the angle between the curves given below :
$y^{2}=4x,\ x^{2}+y^{2}=5$

evaluate:
$3x-5y=16;x-3y=8$

If the slope of a line joining $A(2,x)B(-3,7)$ is $1$ then find $x$ .