Class 10 Maths - Coordinate Geometry

If the points(-3, 6), (-9, a) and (0, 15) are colliner then find a

Let the given points be A(-3, 6), B(-9, a) and C(0, 15)
Since the points A, B and C are collinear the slope of

$\displaystyle \overleftrightarrow{AB}$ = the slope of

$\displaystyle \overleftrightarrow{BC}$

$\displaystyle \Rightarrow \frac{6-a}{6}= \frac{15-a}{9}$

$\displaystyle \Rightarrow$

$\displaystyle 18-3a= 30-2a\Rightarrow -12$

Find the coordinates of the point which divides the line segment joining the points $(5,-2)$ and $(9,6)$ in the ratio $3:1$ .

Using the section formula, if a point $(x,y)$ divides the line joining the points $({ x }_{ 1 },{ y }_{ 1 })$ and $({ x }_{ 2 },{ y }_{ 2 })$ in the ratio $m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

Let the points be

$A(5,-2)$ and

$B(9,6)$ . Let a point

$P(x,y)$ divides

$AB$ in the ratio

$3:1$ .

Therefore, we have

$P(x,y)=\left(\dfrac{3\times 9+1\times 5}{3+1}, \dfrac{3\times 6+1\times (-2)}{3+1} \right)$

$P(x,y)=(8,4)$

Hence, the required point is

$(8,4)$ .

Find the midpoint of the segment connecting the points (a,b) and (3a, c)

$x = (a+3a)/2 = 4a/2 = 2a$

$y = (b+c)/2$

$(2a,(b+c)/2)$

Find the coordinates of the mid-point of the line segment joining the points $(2, 3)$ and $(4, 7)$ .

Given:

$(x_{1}, y_{1}) = (2, 3)$ and

$(x_{2}, y_{2}) = (4, 7)$

Mid-point

$= \left (\dfrac {x_{1} + x_{2}}{2}, \dfrac {y_{1} + y_{2}}{2}\right )$

$= \left (\dfrac {2 + 4}{2}, \dfrac {3 + 7}{2}\right )$

$= \left (\dfrac {6}{2}, \dfrac {10}{2}\right )$

$= (3, 5)$ .

Find the distance between the points $(3,0)$ and $(0,4)$ without using the distance formula.

1714573_622123_ans_2efe21761dbc41d7ab954f93afcf53c8.jpg

Find the co-ordinates of a point $A$ , where $AB$ is the diameter of a circle, whose centre is $(2, - 3)$ and $B$ is $(1, 4)$ .

Given,

$AB$ is a diameter of circle, point

$B$ is

$(1,4)$ and centre of circle is

$(2,-3)$

Let point

$A$ is

$(x,y)$

As we know mid point of diameter is centre of circle

$\therefore \cfrac {x+1}2=2$ and

$\cfrac {y+4}2=-3$

$\Rightarrow x+1=4$ and

$y+4=-6$

$\Rightarrow x=3 , y=-10$

point

$A$ is

$(3,-10)$

Find the midpoint of the line segment joining the points $(3,0)$ and $(-1,4)$ .

Midpoint

$M(x,y)$ of the line segment joining the points

$({x}_{1},{y}_{1})$ and

$({x}_{1},{y}_{1})$ us

$M(x,y)=M\left( \cfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\cfrac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right)$

$\therefore$ Midpoint of the line segment joining the points

$(3,0)$ and

$(-1,4)$ is

$M(x,y)=\left( \cfrac { 3-1 }{ 2 } ,\cfrac { 0+4 }{ 2 } \right) =M(1,2)$

Prove that the coordinates, x and y, of the middle point of the line joining the point (2,3) to the point (3, 4) satisfy the equation
$x-y + 1=0$ .

Let

$(2,3)$ and

$(3,4)$ Let us have point

$R (x , y)$ be the mid-point of line joining

$P$ and

$Q$

$\Rightarrow$ Ratio

$= 1 : 1$
Applying section formula

$x = \dfrac{2\times 1+3\times1}{1+1}$

$\Rightarrow x = \dfrac{5}{2}$ and

$y = \dfrac{3\times 1+4\times1}{1+1}$

$\Rightarrow y=\dfrac{7}{2}$

To check,

$R\left(\dfrac{5}{2} , \dfrac{7}{2}\right)$ satisfies the given equation i.e

$x-y+1= 0$

$\Rightarrow LHS= \dfrac{5}{2} -\dfrac{7}{2} +1$

$= -1+1=0$

$\Rightarrow LHS= RHS = 0$

Hence,

$R\left(\dfrac{5}{2} , \dfrac{7}{2}\right)$ satisfy the given equation.

Find the coordinates of the point which divides the line joining the points $\left(1, 3\right)$ and $\left(2, 7\right)$ in the ratio $3 : 4$ .

If a points divides line joining two points

$(a,b)$ and

$(c,d)$ in ratio

$m:n$ then its coordinates are

$\left( \dfrac { mc+na }{ m+n } ,\dfrac { md+nb }{ m+n } \right)$

Let the point be

$P(h,k)$

$(1,3)$ and

$(2,7)$ is divided by

$(h,k)$ in

$(3:4)$ then

$h=\dfrac { 3(2)+4(1) }{ 3+4 } =\dfrac { 10 }{ 7 } \\ k=\dfrac { 3(7)+4(3) }{ 7 } =\dfrac { 33 }{ 7 }$

So, the coordinates of the point are

$\left( \dfrac { 10 }{ 7 } ,\dfrac { 33 }{ 7 } \right)$

If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.

We know that the diagonals of a parallelogram bisect each other. So, coordinates of the mid-point of diagonal AC are same as the coordinates of the mid-point of diagonal BD.

$\therefore \quad \quad \left( \dfrac { 6+9 }{ 2 } ,\dfrac { 1+4 }{ 2 } \right) =\left( \dfrac { 8+p }{ 2 } ,\dfrac { 2+3 }{ 2 } \right)$

$\Rightarrow \quad \quad \left( \dfrac { 15 }{ 2 } ,\dfrac { 5 }{ 2 } \right) =\left( \dfrac { 8+p }{ 2 } ,\dfrac { 5 }{ 2 } \right)$

$\Rightarrow \quad \quad \dfrac { 15 }{ 2 } =\dfrac { 8+p }{ 2 } \quad \Rightarrow \quad 15=8+p\quad \Rightarrow \quad p=7$

Find the co ordinates of the midpoint line joining of $A\left(3,2\right)$ and $B\left(5,4\right)$ .

$A(3,2)\,\,\,\,\,\, B(5,4)$

Coordinates of Midpoint of line joining

$A$ and

$B$ are

$(\dfrac{3+5}{2}\,, \dfrac{2+4}{2})$

I.e

$(4,3)$ Answer

Find the sum of 'x' co-ordinates of the vertices of a triangle if the mid-points of its sides be the points $(6, -1), (-1,-2)and(1,-4)$

$\dfrac{x_2 \, + \, x_3}{2} \, = \, 6,\dfrac{y_2 \, + \, y_3}{2} \, = \, -1.$

$\therefore \, x_2 \, + \, x_3 \, = \, 12, \, x_3 \, + \, x_1 \, = \, -2, \, x_1 \, + \, x_2 \, = \, 2.$

$\therefore \, 2(x_1 \, + \, x_2 \, + \, x_3) \, = \, 12.$
or

$x_1 \, + \, x_2 \, + \, x_3 \, = \, 6$

1106653_1006845_ans_fa2c04772146497a9e9cb0e02e9e8b0f.png

Find the co-ordinates of the mid-point of the line joining the points $(2, 3)$ and $(4, 7)$ .

Suppose

$(x,y)$ the mid-point of line joining the points

$(x_1,y_1)$ And

$(x_2,y_2)$

Then,

$x= \dfrac{x_1+x_2}{2} , y= \dfrac{y_1+y_2}{2}$

By comparing from the above points

$x_1=2, y_1=3,x_2=4,y_2=7$

And we are asked to find

$(x,y)$

$x=\dfrac{2+4}{2}=\dfrac{6}{2}=3$

and

$y=\dfrac{3+7}{2}=\dfrac{10}{2}=5$

So the mid-point is

$(3,5)$ .

Find the coordinates of the point which divides the line segment joining the points(6,3) and (-4,5) in the ratio 3:2 internally.

Given: A point divides the line segment joining

$(6,\ 3)$ and

$(-4,\ 5)$ in the ratio

$3:2$ internally

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n,$

Then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

Let

$P(x,\ y)$ be the required point.

Then,

$x=\dfrac { 3\times -4+2\times 6 }{ 3+2 }$ and

$y=\dfrac { 3\times 5+2\times 3 }{ 3+2 }$

$\Rightarrow$

$x=0$ and

$y=\dfrac { 21 }{ 5 }$

So, the coordinates of

$P$ are

$\bigg( 0,\ \dfrac{21}{5}\bigg)$

Find the coordinates of the point which divides internally the join of the points
a) $(8,9)$ and $(-7,4)$ in the ratio $2:3$
b) $(1,-2)$ and $(4,7)$ in the ratio $1:2$ .

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

(a) Let the point be

$P(x,y)$

$(x,y)=(\dfrac{2 \times -7+3 \times 8 }{2+3},\dfrac{2 \times 4+3 \times 9}{2+3})$

$X=2,\;Y==7$

So the point is

$(2,7)$

(b) Let the required point be

$P(x,y)$

Then,

$(x,y)=(\dfrac{1 \times 4+2 \times 1}{1+2},\dfrac{1 \times 7+2 \times -2}{1+2})$

$x=2,\;y=1$

therefore, required point

$(2,1)$

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

$(a,b)$ is the mid-point of Line segment joining the points

$A(10,-6)$ and

$B(k,4)$

So,

$a=\cfrac{10+k}{2}$ and

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

It is given that,

$a-2b=18$

Put

$b=-1$

$a-2(-1)=18$

$a=18-2=16$

Now,

$a=\cfrac{10+k}{2}$

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

$k+10=32$

$k=22$

Distance between the points

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

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

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

Distance between the points

$(10,-6)$ and

$(22,4)$ is

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

$=\sqrt{244}$

$=2\sqrt{61}$

If the points $A(-2, -1), B(1, 0), C(a, 3)$ and $D(1, b)$ form a parallelogram $ABCD$ , find the value of $a$ and $b$ .

In the parallelogram

$ABCD$ ,

$AC$ and

$BD$ are the diagonals.

Co-ordinates of mid-point of

$AC=(\dfrac{-2+a}{2}, \dfrac{1+1}{2})$

Co-ordinate of mid-point of

$BD=(\dfrac{1+1}{2}, \dfrac{0+b}{2})$

It is known that diagonals of the parallelogram bisect each other. Thus mid point of

$AC$ and

$BD$ coincide each other.

Thus,

$\dfrac{-2+a}{2}=1\implies a=4$

$\dfrac{1+1}{2}=\dfrac{b}{2}\implies b=2$

Given $A\left( {4, - 3} \right)$ , $B\left( {8,5} \right)$ . Find the coordinates of the point that divides segment $\text AB$ in two equal parts.

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

& in case of mid point we have

$m:n = 1:1$ .

Given points:

$A(4,-3)$ and

$B(8,5)$ ,

Coordinates of mid point

$=\left(\dfrac{4+8}{2}, \dfrac{-3+5}{2}\right)=(6,1)$

A(2,3,7), B (-1,3,2) and C (p,5,r) are vertices of a triangle. If the median through A is equally inclined to the co-ordinate axes then find the co-ordinates of the midpoint of BC in terms of p and r.

Here

$A \equiv(2,3,7) B\equiv(-1,3,2)C\equiv (p,5,r)$
Let the mid point of BC is D.

$\therefore$ D has coordinates =

$(\dfrac{p-1}{2},\dfrac{5+3}{2},\dfrac{r+2}{2})$
=

$(\dfrac{p-1}{2},4,\dfrac{r+2}{2})$

Find the mid point of line joining $(2,3)$ and $(-6,5)$ .

Using the midpoint formula, if a point $(x,y)$ is midpoint of line the joining the points $({ x }_{ 1 },{ y }_{ 1 })$ and $({ x }_{ 2 },{ y }_{ 2 }),$ then

$(x,y) = \left( \dfrac { { x }_{ 2 } + { x }_{ 1 } }{ 2 },\dfrac { { y }_{ 2 } + { y }_{ 1 } }{ 2 } \right)$

The given points are $(2,3)$ and $(-6,5).$ Let the midpoint of line joining the given points is $(h,k).$

Now, by using the above midpoint formula,

$\begin{aligned}{}\left( {h,k} \right) &= \left( {\frac{{2 - 6}}{2},\frac{{3 + 5}}{2}} \right)\\ &= \left( { - \frac{4}{2},\frac{8}{2}} \right)\\ &= \left( { - 2,4} \right)\end{aligned}$

Find the value of $a$ , for which point $P( a , 2 )$ is the midpoint of the lines segment joining the points $Q ( - 5,4 )$ and $R ( - 1,0 )$ .

$(x_1,y_1)$ and

$(x_2,y_2)$ are the coordinates of two points, then the coordinates of the mid point of the line joining these points is given by:

$\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

So, if the mid point of

$(-5, 4)$ and

$(-1, 0)$ is

$(9, 2)$ , then using the mid-point formula:

$\left(\dfrac{-5-1}{2}, \dfrac{4+0}{2}\right)=(a, 2)$

$(-3, 2)=(a, 2)$

On comparing the co-ordinates:

$a=-3$ .

Find the other point if the origin is mid point of them and one vertex is $(4,3)$

Given point

$(4,3)$

The mid point is origin

$\implies 0=\dfrac{x+4}{2};0=\dfrac{y+3}{2}\\x+4=0;y+3=0\\x=-4,y=-3$

Find the co-ordinates of points which divides the line segment joining the points $(-2,5)$ and $(6,8)$ in the ratio $3:2$ internally.

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

Let

$\left( h, k \right)$ be the required point.

Given that the point

$\left( h, k \right)$ divides the line segment joining the points

$\left( -2, 5 \right)$ and

$\left( 6, 8 \right)$ in the ratio

$3 : 2$ internally.

Therefore,

$\left( h, k \right) = \left( \cfrac{-2 \times 2 + 6 \times 3}{3 + 2}, \cfrac{5 \times 2 + 8 \times 3}{3 + 2} \right)$

$\Rightarrow \left( h, k \right) = \left( \cfrac{-4 + 18}{5}, \cfrac{10 + 24}{5} \right)$

$\Rightarrow \left( h, k \right) = \left( \cfrac{14}{5}, \cfrac{34}{5} \right)$

Hence the coordinate of required point are

$\left( \cfrac{14}{5}, \cfrac{34}{5} \right)$ .

The shape of a garden is rectangular in the middle and semicircular at the ends.Find the axes of the garden?

To find the area of the shown figure,it may be considered as the sum of area of a rectangle and

$2$ semicircles.

Length of the rectangle

$=20-\left(3.5+3.5\right)=20-7=13$ m

Width of rectangle

$=7$ cm

Area of semicircle

$A=\dfrac{1}{2}\times\dfrac{22}{7}\times\,3.5\times\,3.5$

Area of semicircle

$B=\dfrac{1}{2}\times\dfrac{22}{7}\times\,3.5\times\,3.5$

Area of

$A$ and

$B=\dfrac{1}{2}\times\dfrac{22}{7}\times\,3.5\times\,3.5+\dfrac{1}{2}\times\dfrac{22}{7}\times\,3.5\times\,3.5$

$=\dfrac{22}{7}\times\,3.5\times\,3.5=38.5$ sq.m

Total area

$=$ Area of

$\left(A+B+C\right)$

$=91+38.5$ sq.m

$=129.5$ sq.m

$A$ and $B$ are the points $(2,0)$ and $(0,2)$ respectively. Find the coordinates of point $P$ if it is the mid point.

Co-ordinates of point

$(x_1,y_1)$ and

$(x_2,y_2)$ is given as

$\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$ .

Given:

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

Thus co-ordinates of mid point

$P$ is given as

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

Find the ratio in which the point $P$ whose abscissa is $3$ divides the line joining $A(6,5)$ and $B(-1,4)$ and hence find the coordinates of $P$ .

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

& for mid points we have,

$m:n = 1:1$ .

Let the ratio in which

$P$ divides

$AB$ be

$r:1$

Given Abscissa of

$P$ is

$3$

$A(6,5),B(-1,4)$

Co-ordinates of

$P$ is

$\bigg(\dfrac{-r+6}{r+1},\dfrac{4 r+5}{r+1}\bigg)$

$\implies \dfrac{-r+6}{r+1}=3\implies -r+6=3 r+3$

$\implies 4 r=3\implies r=\dfrac{3}{4}$

So the ratio is

$3:4$

Coordinates of

$P$ is

$\bigg(\dfrac{4(6)+3(-1)}{4+3},\dfrac{4(5)+3(4)}{4+3}\bigg)=\bigg(3,\dfrac{32}{7}\bigg)$

Find the ratio in which the line joining A(6,5) & B(4, -3) is divided by the y =2, Also find the coordinate of the point of divisions.

1311632_1438178_ans_817c56c1f30d4d2895b09a7d1dadec90.jpeg

Find the mid-points of the line segment joining the points.
$\left( {a,b} \right)$ and $\left( {a + 2b,2a - b} \right)$

Given: Two points

$(a,b)$ and

$(a+2b, 2a-b)$

Let P(x,y) be the mid point of given points.

$\therefore P(x,y) = P(\frac{a+a+2b}{2},\frac{b+2a-b}{2})$

$\Rightarrow p(a+b,a)$ is the required midpoint.

The mid point of $(2,3)$ and $(3,4)$ is

Given points are

$(2,3)$ and

$(3,4)$

The mid point is given as

$=\left(\dfrac{2+3}2,\dfrac {3+4}2\right)\\=\left( \dfrac 52,\dfrac 72\right)$

Find the area of the triangle whose vertices are (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

$:-$

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

$=\dfrac 12|3(2+1)-4(-1-8)+5(8-2) |$

$=\dfrac 12|9+36+30|$

$=\dfrac {75}2$

By given,

$\dfrac{a}{2}$ =

$\dfrac{75}{2}$

$\Rightarrow$

$a= 75$

If A and B are $(-2, -2)$ and $(2, -4)$ , respectively, find the coordinates of P such that $AP = \dfrac{3}{7} AB$ and P lies on the line segment AB.

Given:- A line segment joining the points

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

$B \left( 2, -4 \right)$ .

$P$ is a point on

$AB$ such that

$AP = \cfrac{3}{7} AB$ .

Now,

$AP = \cfrac{3}{7} AB \quad \left( \text{Given} \right)$

$7AP = 3 \left( AP + BP \right)$

$7AP = 3AP + 3 BP$

$\Rightarrow 7AP - 3 AP = 3 BP$

$\Rightarrow \cfrac{AP}{BP} = \cfrac{3}{4}$

Therefore,

Point

$P$ divides

$AB$ internally in the ratio

$3 : 4$ .

As we know that if a point

$\left( h, k \right)$ divides a line joining the point

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

$\left( {x}_{2}, {y}_{2} \right)$ in the ration

$m : n$ , then coordinates of the point is given as-

$\left( h, k \right) = \left( \cfrac{m{x}_{2} + n{x}_{1}}{m+n}, \cfrac{m{y}_{2} + n{y}_{1}}{m+n} \right)$

Therefore,

Coordinates of

$P = \left( \cfrac{3 \times \left( 2 \right) + 4 \times \left( -2 \right)}{3 + 4}, \cfrac{3 \times \left( -4 \right) + 4 \times \left( -2 \right)}{3 + 4} \right) = \left( \cfrac{-2}{7}, \cfrac{-20}{7} \right)$

Hence, the coordinates of

$P$ are

$\left( \cfrac{-2}{7}, \cfrac{-20}{7} \right)$ .

If the midpoints of the line segment joining the points $P ( 6 , b - 2 )$ and $( -2 , 4)$ is $( 2 , - 3 ) ,$ find the value of $b .$

$P(6,b-2), Q=(-2,4)$

midpoint of PQ is

$(2,-3)$

$x=2$

$y=-3$

$y=\frac{y_{1}+y_{2}}{2}$ , Acording to midpoint formula

$-3=\frac{b-2+4}{2}$

$-6=b+2$

$\therefore b=-8$

1236826_1502561_ans_ee808bfb830c483792603acab71fbc8f.jpg

Find the mid point of $(9,0)$ and $(5,4)$

Given points

$(9,0)$ and

$(5,4)$

The mid point is given as

$\left(\dfrac{x_1+x_2}2,\dfrac {y_1+y_2}2\right) \\\left(\dfrac {14}2,\dfrac 42\right)=(7,2)$

Solve the following :
Find the co-ordinates of midpoint of the segment joining the points $(22, 20)$ and $(0,16)$ .

Given

$(22, 20)$ and

$(0,16)$

mid-point of

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$X=\left ( \dfrac{x_1+x_2}{2}\right ) and \ Y=\left ( \dfrac{y_1+y_2}{2}\right )$

Here

$(x_1,y_1)=(22,20)$ and

$(x_2,y_2)=(0,16)$

$\Rightarrow X=\dfrac {22+0}{2}$ and

$Y=\dfrac {20+16}{2}$

$\Rightarrow X=11$ and

$Y=18$

$\therefore$ the mid point of

$(22, 20)$ and

$(0,16)$ is

$(11,18)$

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

$:-$

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

$=\dfrac 12|2(1-8)+1(8-7)+10(7-1) |$

$=\dfrac 12|-14+1+60|$

$=\dfrac {47}2$

By given,

$\dfrac{b}{2}$ =

$\dfrac{47}{2}$

$\Rightarrow$

$b= 47$

If the points $A(a, -11), B(5, b), C(2, 15)$ and $D(1, 1)$ and the vertices of a parallelogram $ABCD$ , find the values of $a$ and $b$ .

We know that diagonals of parallelograms bisect each other. Therefore coordinates of the midpoint of

$AC$ and

$BD$ will be the same.

Hence,

$\begin{aligned}{}\left( {\frac{{a + 2}}{2},\frac{{15 - 11}}{2}} \right)& = \left( {\frac{{5 + 1}}{2},\frac{{b + 1}}{2}} \right)\\\left( {\frac{{a + 2}}{2},\frac{4}{2}} \right) &= \left( {\frac{6}{2},\frac{{b + 1}}{2}} \right)\\\left( {\frac{{a + 2}}{2},2} \right) &= \left( {3,\frac{{b + 1}}{2}} \right)\quad\quad\quad\dots(i)\end{aligned}$

From equation

$(i),$

$\begin{aligned}{}\frac{{a + 2}}{2} = 3&\text{ and }\frac{{b + 1}}{2} = 2\\a + 2 = 6&\text{ and }b + 1 = 4\\a = 4&\text{ and }b = 3\end{aligned}$

Hence,

$a=4$ and

$b=3.$

Draw a quadrilateral in the Cartesian plane, whose vertices are (-4, 5), (0, 7), (5, -5) and (-4, -2). Also, find its area

Let ABCD be the quadrilateral with vertices are (-4, 5), (0, 7), (5, -5) and (-4, -2).
Then, by plotting A, B, C and D on the Cartesian plane and joining AB, BC, CD and DA the given quadrilateral can be drawn as
To find the area of quadrilateral ABCD, we draw one diagonal, say AC.
Accordingly, area (ABCD) = area (

$\triangle$ ABC) + area (

$\triangle$ ACD)
We know that the area of a triangle whose vertices are

$\left ( x_1, y_1 \right ), \left ( x_2, y_2 \right )$ and

$\left ( x_3, y_3 \right )$ is

$\dfrac{1} {2} \left | x_1\left (y_2 - y_1 \right ) + x_2 \left (y_3 - y_1 \right ) + x_3 \left (y_1 - y_2 \right )\right |$
Therefore, area of

$\triangle ABC$

$= \dfrac{1} {2} \left | -4\left (7 + 5 \right ) + 0 \left (-5 - 5 \right ) + 5 \left (5 - 7 \right ) unit^{2}\right |$

$= \dfrac{1} {2} \left | -4\left (12 \right ) + 5 \left (-2 \right ) \right | unit^{2}$

$= \dfrac{1} {2} \left | -48 -10\right | unit^{2}$

$= \dfrac{1} {2} \left | -58\right | unit^{2}$

$= \frac{1} {2} 58\ unit^{2}$

$= 29\ unit^{2}$
Area of

$\triangle ACD$

$= \dfrac{1} {2} \left | -4\left (7 + 5 \right ) + 0 \left (-5 - 5 \right ) + 5 \left (5 - 7 \right ) unit^{2}\right |$

$= \dfrac{1} {2} \left | -4\left (-5 + 2 \right ) + 5 \left (-2 - 5\right ) + (-4) \left (5 - 5\right )\right | unit^{2}$

$= \dfrac{1} {2} \left | -4\left (-3 \right ) + 5 \left (-7 \right ) + (-4) \left (10\right )\right | unit^{2}$

$= \dfrac{1} {2} \left | 12 - 35 - 40\right | unit^{2}$

$= \dfrac{1} {2} \left | -63\right | unit^{2}$

$= \dfrac{63} {2} unit^{2}$
Thus, area (ABCD) =

$\left (29 + \dfrac{63} {2} \right )\ unit^{2} = \dfrac{58 + 63} {2}\ unit^{2} = \dfrac{121} {2}\ unit^{2}$
1763985_1797296_ans_2aa142c25b014966b9aaaad6c432a840.png

What point on the x-axis is equidistant from the points $(7,6)$ and $(-3,4)$ ?

Let the coordinates of the required point on x-axis

$=P(x,0)$
Points are

$A(7,6)$ and

$B(-3,4)$
According to question

$PA=PB$

${ \left( PA \right) }^{ 2 }={ \left( PB \right) }^{ 2 }$

${ \left( x-7 \right) }^{ 2 }+{ \left( 0-6 \right) }^{ 2 }={ \left( x+3 \right) }^{ 2 }+{ \left( 0-4 \right) }^{ 2 }\Rightarrow { x }^{ 2 }+49-14x+36={ x }^{ 2 }+9+6x+!6\Rightarrow 60=20x\Rightarrow x=3$
So, the required point is

$(3,0)$

Determine whether the point is collinear.
$L(-2,3),M(1,-3),N(5,4)$

$L(-2,3),M(1,-3),N(5,4)$

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

$=\sqrt{(-3)^2 6^2}=\sqrt{9+36}$

$=\sqrt{45}=3\sqrt{5}$

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

$\sqrt{16+49}$

$=\sqrt{65}$

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

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

$=\sqrt{49+1}=\sqrt{50}$

$=5\sqrt{2}$
Sum of two sides is not equal to the third side.
Hence, the given points are not collinear.

Given $M$ is the mid-point of $AB$ , find the co-ordinates of :
$B$ ; if $A = (3 , -1)$ and $M = (-1 , 3)$ .

Let's assume the co-ordinates of

$B$ be

$(x , y)$ .
So,

$(-1 , 3) = \dfrac{3 + x}{2} , \dfrac{-1 + y}{2}$

$-1 = \dfrac{3 + x}{2}$ and

$3 = \dfrac{-1 + y}{2}$

$-2 = 3 + x$ and

$6 = -1 + y$

$x = -5$ and

$y = 7$
Thus , the co-ordinates of

$B$ are

$(-5 , 7)$

Find a point on the y-axis which is equidistant from the points $(5,2)$ and $(-4,3)$

${\textbf{Step - 1: Stating the distance formula}}{\text{.}}$

${\text{We can write distance between two points }}({x_1},{y_1}),({x_2},{y_2}){\text{ as }}$ $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$

${\textbf{Step - 2: Taking a general point on y - axis and equating its distance from the two points}}{\text{.}}$

${\text{Let the point be}}{\text{ (0,}}y){\text{, so its distance from the point (5,2) can be written as}}$

${d_1} = \sqrt {{{(5 - 0)}^2} + {{(2 - y)}^2}} = \sqrt {25 + {{(2 - y)}^2}}$

$\Rightarrow {d_1} = \sqrt {25 + 4 + {y^2} - 4y} = \sqrt {{y^2} - 4y + 29}$ $\quad \quad \text{...eqn(i)}$

${\text{Now the distance of this point from ( - 4,3) can be written as}}$

${d_2} = \sqrt {{{( - 4 - 0)}^2} + {{(3 - y)}^2}} = \sqrt {16 + {{(3 - y)}^2}}$

$\Rightarrow {d_2} = \sqrt {16 + 9 + {y^2} - 6y} = \sqrt {{y^2} - 6y + 25}$ $\quad \quad \text{...eqn(ii)}$

${\text{Now, as this point is equidistant from both the points}}$

$\therefore {d_1} = {d_2}$

$\Rightarrow \sqrt {{y^2} - 4y + 29} = \sqrt {{y^2} - 6y + 25}$ $\quad \quad \textbf{[from eqn(i) and eqn(ii)]}$

${\text{Squaring both the sides, we get, }}$

${y^2} - 4y + 29 = {y^2} - 6y + 25$

$\Rightarrow 2y = - 4$

$\Rightarrow y = - 2$

${\textbf{Thus,the point is (0, - 2)}}{\text{.}}$

Verify whether the following points are collinear or not:
$A (1, -3), B (2, -5), C (-4, 7)$

1896551_1911587_ans_e45621ee4b844c8d82cf552d93ebd9cc.jpeg

The line segment joining the points $A (3, 2)$ and $B (5,1)$ is divided at the point P in the ratio $1:2$ and it lies on the line $3x -18y + k = 0.$ Find the value of $k$ .

Given that point divides the line segment joining the points A(3,2) and B(5,1) in the ratio 1:2

The coordinate of P is ,

By section formula

$\left( \dfrac { m{ x }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\dfrac { { m }{ y }_{ 2 }+n{ y }_{ 1 } }{ m+n } \right)$

here m:n =1:2,

$({ x }_{ 1 },{ y }_{ 1 })=(3,2)and ({ x }_{ 2 },{ y }_{ 2 })=(5,1)$

$\therefore$

$\left( \dfrac { 1\times 5+2\times 3 }{ 1+2 } ,\dfrac { 1\times 1+2\times 2 }{ 1+2 } \right) =\left( \dfrac { 11 }{ 3 } ,\dfrac { 5 }{ 3 } \right) \\ 3\left( \dfrac { 11 }{ 3 } \right) -18\left( \dfrac { 5 }{ 3 } \right) +K=0$

$K-19=0$

$\Rightarrow$

$K=19$

Column II gives the coordinates of the point P that divides the line segment joining the points given in column I, match them correctly

We shall apply the section formula

$P(x,y)=\left( \dfrac { n{ x }_{ 1 }+m{ x }_{ 2 } }{ m+n } ,\dfrac { n{ y }_{ 1 }+m{ y }_{ 2 } }{ m+n } . \right)$

When the point

$P(x,y)$ divides a line segment AB joining

$A\left( { x }_{ 1 },{ y }_{ 1 } \right) \& B\left( { x }_{ 2 },{ y }_{ 2 } \right)$ internally in the ratio

$m:n$

A) Let the line segment AB joining

$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( -1,3 \right)$ &

$B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 5,-6 \right)$ internally in the ratio

$m:n=1:2$ .

Then, by the section formula,

$x=\dfrac { n{ x }_{ 1 }+m{ x }_{ 2 } }{ m+n } =\dfrac { 2\times (-1)+1\times (5) }{ 1+2 } =1$ and

$y=\dfrac { n{ y }_{ 1 }+m{ y }_{ 2 } }{ m+n } =\dfrac { 2\times 3+1\times (-6) }{ 1+2 } =0$

$P(x,y)=(1,0)$ , which matches with 4.

B) Let the line segment AB joining

$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( -2,1 \right)$ &

$B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 1,4 \right)$ internally in the ratio

$m:n=2:1$ .

Then, by the section formula,

$x=\dfrac { n{ x }_{ 1 }+m{ x }_{ 2 } }{ m+n } =\dfrac { 1\times (-2)+2\times (1) }{ 1+2 } =0$ and

$y=\dfrac { n{ y }_{ 1 }+m{ y }_{ 2 } }{ m+n } =\dfrac { 1\times 1+2\times (4) }{ 1+2 } =3$

So,

$P(x,y)=(0,3)$ which matches with 2.

C) Let the line segment AB joining

$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( -1,7 \right)$ &

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

$m:n=2:3$

Then, by the section formula,

$x=\dfrac { n{ x }_{ 1 }+m{ x }_{ 2 } }{ m+n } =\dfrac { 3\times (-1)+2\times 4 }{ 2+3 } =1$ and

$y=\dfrac { n{ y }_{ 1 }+m{ y }_{ 2 } }{ m+n } =\dfrac { 3\times 7+2\times (-3) }{ 2+3 } =3$

So,

$P(x,y)=(1,3)$ which matches with 3.

D) Let the line segment AB joining

$A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 4,-3 \right)$ &

$B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 8,5 \right)$ internally in the ratio

$m:n=3:1$

Then, by the section formula,

$x=\dfrac { n{ x }_{ 1 }+m{ x }_{ 2 } }{ m+n } =\dfrac { 1\times (4)+3\times (8) }{ 3+1 } =7$ and

$y=\dfrac { n{ y }_{ 1 }+m{ y }_{ 2 } }{ m+n } =\dfrac { 1\times (-3)+3\times 5 }{ 3+1 } =3$

So,

$P(x,y)=(7,3)$ which matches with 1.

$A\longrightarrow$ matches with

$4$

$B\longrightarrow$ matches with

$2$

$C\longrightarrow$ matches with

$3$

$D\longrightarrow$ matches with

$1$

The co-ordinates of the midpoint P of segment AB with A $\equiv$ (3.5, 9.5), B $\equiv$ $(-1.5, 0.5)$ is $(1,a)$ , then the value of a is _____

$Let\quad the\quad co-ordinates\quad of\quad the\quad point\quad P\quad be\quad P\left( x,y \right) \quad which\quad bisects\\ the\quad line\quad segment\quad AB\quad joining\quad the\quad points\quad A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 3.5,9.5 \right) \quad \& \quad B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( -1.5,0.5 \right) \quad \\ Then,\quad by\quad the\quad section\quad formula,\\ \quad \\ x=\dfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } =\dfrac { 3.5+(-1.5) }{ 2 } =1\quad and\quad \\ \\ y=\dfrac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } =\dfrac { 9.5+0.5 }{ 2 } =5\\$

$\therefore \quad P\left( x,y \right) =P\left( 1,5 \right)$

$\quad P\left( x,y \right) =P\left( 1,5 \right) .\\$

$\therefore P(1,a)= P(1,5)$

$\implies a =5$

$A (-3, 2)$ and B (5, 4) are the end points of a line segment, find the sum of coordinates of the mid points of the line segment.

The coordinates of mid points of AB

$=\left ( \dfrac{x_{1}+x_{2}}{2},\dfrac{y_{1}+y_{2}}{2} \right )=\left ( \dfrac{-3+5}{2},\dfrac{2+4}{2} \right )$

$=(1, 3)$

A line is such that its segment between the axes is bisected at the point $(23, 27),$ the product of the intercepts made by the line on the coordinate axes is

Let intersection points of the line with the

$x$ -axis and

$y$ -axis are

$A(a,0)$ and

$B(0,b)$ respectively.

Let the midpoint of the segment be

$P(23,27)$

Since,

$P$ is the midpoint of

$A$ and

$B$

$P$ divides

$AB$ in the ratio

$1 : 1$

By Midpoint Formula,

$\Rightarrow (23,27)=\left(\dfrac {a+0}{2},\dfrac {0+b}{2}\right)$

Now,

$\dfrac {a+0}{2}=23$

$\Rightarrow\dfrac {a}{2}=23$

$\Rightarrow a=46$

and, also

$\dfrac {0+b}{2}=27$

$\Rightarrow\dfrac {b}{2}=27$

$\Rightarrow b=54$

So, the intercepts in the axes are

$46$ and

$54$ respectively.

Product of intercepts

$=46\times 54$

$=2484$

Hence, the required answer is

$2484$ .

Also, find the value of $m$ if, coordinates of the point of intersection of the diagonals of the square $ABCD$ is $(1,m)$

Plot the points,

$A, C$ and

$D$ . Join them using straight lines to complete the square.

Hence, B

$=(4,4)$ .
Draw the diagonals

$AC$ and

$BD$ .
Since the diagonals of a square bisect each other, the co-ordinates of the point of intersection of the diagonals of the square

$ABCD$ is given as

$(x,y)=\left(\dfrac{-2+4}{2},\dfrac{4+10}{2}\right)$

$\implies (x,y)=(1,7)$

$\therefore$ the value of

$m$ is

$7$ .

The co-ordinates of the mid-point of CD is $(1,m)$ .
Value of $m$ is

Plot the points, A, C and D. Join them using straight lines to complete the square. Hence B

$=(4,4)$ .
Mid point of CD

$=(1,10)$

Find the abscissa of the point which divides the line segment joining the points $(6, 3)$ and $(-4, 5)$ in the ratio $3 : 2$ internally.

Using the section formula, if a point $(x,y)$ divides the line joining the points $({ x }_{ 1 },{ y }_{ 1 })$ and $({ x }_{ 2 },{ y }_{ 2 })$ internally in the ratio $m:n$ , then $(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1} }{ m + n } ,\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

Substituting $({ x }_{ 1 },{ y }_{ 1 }) \equiv (6,3)$ and $({x }_{ 2 },{ y }_{2 }) \equiv (-4,5)$ and $m = 3, n = 2$ in the section formula, we get the point as,

$(x,y)= \left( \dfrac { 3(-4) + 2(6) }{ 3+2} ,\dfrac { 3(5) + 2(3)}{ 3+2 } \right) \\=\left (\dfrac{12-12}{3+2}, \dfrac{15+6}{2+3} \right)\\=\left( 0, \dfrac {21}{5} \right)$

The required abscissa is $0$ .

The area of a triangle is $5$ . Two of its vertices are $(2, 1)$ and $(3, -2)$ . The third vertex lies on $y = x + 3$ . Find the third vertex.

The given vertices are

$\displaystyle \left ( 2,1 \right )$ and

$\displaystyle \left ( 3,-2 \right )$ and the area of triangle is

$5$ sq. unit

Let the third vertex be

$\displaystyle \left ( x_{3},y_{3} \right )$

Area of triangle

$=\displaystyle \frac{1}{2}\left | x_{1}\left ( y_{2}-y_{3} \right )+x_{2}\left ( y_{3}-y_{1} \right )+x_{3}\left ( y_{1}-y_{2} \right ) \right |$
Here,

$\displaystyle x_{1}=2,y_{1}=1;x_{2}=3,y_{2}=-2$ ; area of the triangle

$=5$ sq unit

$\displaystyle \Rightarrow 5=\frac{1}{2}\left | 2\left ( -2-y_{3} \right )+3\left ( y_{3}-1 \right )+x_{3}\left ( 1+2 \right ) \right |$

$\displaystyle \Rightarrow 10=\left | -4-2y_{3} +3y_{3}-3+3x_{3}\right |$

$\displaystyle \Rightarrow 10=\left | 3x_{3}+y_{3}-7 \right |$

$\displaystyle \Rightarrow 3x_{3}+y_{3}-7=\pm 10$

Taking positive sign

$\Rightarrow \displaystyle 3x_{3}+y_{3}-7= 10$

$\Rightarrow 3x_{3}+y_{3}=17$ .....(i)

Taking negative sign

$\displaystyle \Rightarrow 3x_{3}+y_{3}-7=-10$

$\displaystyle \Rightarrow 3x_{3}+y_{3}=-3$ .......(ii)

Given that

$\displaystyle \left ( x_{3},y_{3} \right )$ lies on

$y = x + 3$
So

$\displaystyle y_{3}=x_{3}+3$

$\Rightarrow x_{3}-y_{3}=-3$ ........(iii)

For (i) and (iii)

Adding eq (i) & (iii)

$4x_{3}=14$

$\Rightarrow x_{3}=\dfrac{7}{2}$

Putting in (iii)

$\Rightarrow \dfrac{7}{2}-y_{3}=-3$

$\Rightarrow y_{3}=\dfrac{7}{2}+3$

$\Rightarrow y_{3}=\dfrac{13}{2}$

$\displaystyle x_{3}=\frac{7}{2},y_{3}=\frac{13}{2}$

For eq (ii) and (iii)

Adding eq (ii) & (iii)

$4x_{3}=-6$

$\Rightarrow x_{3}=\dfrac{-6}{4}$

$\Rightarrow x_{3}=\dfrac{-3}{2}$

Putting in (iii)

$\Rightarrow -\dfrac{3}{2}-y_{3}=-3$

$\Rightarrow y_{3}=-\dfrac{3}{2}+3$

$\Rightarrow y_{3}=\dfrac{3}{2}$

$\displaystyle x_{3}=\frac{-3}{2},y_{3}=\frac{3}{2}$

So the coordinates of the third vertex are

$\displaystyle \left ( \frac{7}{2},\frac{13}{2} \right )\: or\: \left ( \frac{-3}{2},\frac{3}{2} \right )$

In the adjoining figure, $D,\,E,\,F$ are the midpoints of the sides $BC,\,CA\;and\;AB$ of $\Delta ABC$ . If $BE\;and\;DF$ intersect at $X$ while $CF\;and\;DE$ intersect at $Y$ , prove that $XY=\displaystyle\frac{1}{4}BC$ .

$Given:\triangle ABC\quad where\quad D,E,F\quad are\quad the\quad midpoints\quad ofthe\quad sides\quad BC,CAandAB\quad of\triangle ABC.\\ To\quad Prove:XY=\dfrac { 1 }{ 4 } BC\\ Proof:\\ In\quad \triangle ABC,F\quad and\quad E\quad are\quad the\quad midpoints\quad ofAB\quad and\quad AC\quad respectively\\ \therefore FE\parallel BC\quad and\quad FE=\dfrac { 1 }{ 2 } BC=BD(The\quad line\quad joining\quad the\quad mid-points\quad oftwo\quad sides\quad ofa\quad triangle\quad is\quad parallel\quad to\quad the\quad third\quad side\quad and\quad equal\quad to\quad half\quad the\quad length\quad of\quad the\quad third\quad side)\\ \therefore FE\parallel BD\quad andFE\quad =BD.\\ Hence\quad BDEF\quad is\quad a\quad parallelogram\quad whose\quad diagonals\quad BE\quad and\quad DF\quad intersect\quad each\quad other\quad atX)\\ \therefore \quad X\quad is\quad the\quad midpoint\quad of\quad DF.\\ Similarly,Y\quad is\quad the\quad midpoint\quad of\quad DE.\\ Thus,\quad in\triangle DEF,\quad X\quad and\quad Y\quad are\quad the\quad midpoints\quad of\quad DFand\quad DE\quad respectively.\\ So,XY\parallel FE\quad and\quad XY=d\frac { 1 }{ 2 } FE(Midpoint\quad Theorm)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\dfrac { 1 }{ 2 } \times \dfrac { 1 }{ 2 } BC=\dfrac { 1 }{ 4 } BC\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad Hence\quad Proved$

Find the midpoint of the segment connecting the points $(6,4)$ and $(3,-4)$ .

1111

1332767_426713_ans_42c223cbdf0a4786ba377365ba91dd29.PNG

The base of an equilateral triangle with side $2a$ lies along the $y$ -axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Let

$ABC$ be the given equilateral triangle with side

$2a$

$\displaystyle \Rightarrow AB=BC=CA=2a$

Since base,

$BC$ lies on the

$y$ -axis with mid-point at origin

$O$ .

$\displaystyle \Rightarrow OB=OC=a$

So, the coordinates of point

$B$ are

$(0,a)$ and the coordinates of

$C$ are

$(0,-a)$

We know that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.

Hence, vertex

$A$ lies on

$x$ -axis.

So, let the coordinates of point

$A$ be

$(x,0)$

On applying Pythagoras theorem

$\triangle ABC$ , we get

$\displaystyle (AC)^{2}= (OA)^{2}+(OC)^{2}$

$\displaystyle \Rightarrow (2a)^{2}= x^{2}+a^{2}$

$\displaystyle \Rightarrow 4a^{2}-a^{2}=x^{2}$

$\displaystyle \Rightarrow x^{2}= 3a^{2}$

$\displaystyle \Rightarrow x = \pm \sqrt{3a}$

$\displaystyle \therefore$ coordinates of point

$A$ are

$\displaystyle (\pm \sqrt{3a},0)$ or

$\displaystyle ( -\sqrt{3a},0)$

Thus, the vertices of the given equilateral triangle are

$(0,a)(0,-a)$ and

$\displaystyle ( \sqrt{3a},0)$ or

$(0,a)(0,-a)$ and

$\displaystyle ( -\sqrt{3a},0)$

The two opposite vertices of a square are $(-1, 2)$ and $(3, 2)$ . Find the coordinates of the other two vertices.

Let

$ABCD$ is a square where two opposite vertices are

$A(-1,2) and C(3,2)$ .

Let B

$(x,y)$ and D

$(x_1.y_1)$ ids the other two vertices.

In Square

$ABCD$

$AB=BC=CD=DA$

Hence

$AB=BC$

$\Rightarrow \sqrt{(x+1)^2+(y-2)^2}=\sqrt{(3-x)^2+(2-y)^2}$ [by distance formula]

Squaring both sides

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

$\Rightarrow x^2+2x+1+y^2+4-4y=9+x^2-6x+4+y^2-4y$

$\Rightarrow 2x+5=13-6x$

$\Rightarrow 2x+6x=13-5$

$\Rightarrow 8x=8$

$\Rightarrow x=1$

$\triangle ABC ,\angle B=90^{\circ}$ [all angles of square are

$90^{\circ}$ ]

Then according to the Pythagorean theorem

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

$AB=BC$

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

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

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

$\Rightarrow 2(x^2+2x+1+y^2+4-4y)=(4)^2$

Put the value of

$x=1$

$\Rightarrow 2(1^1+2\times 1+1+y^2+4-4y)=16$

$\Rightarrow 2(y^2-4y+8)=16$

$\Rightarrow 2y^2-8y+16=16$

$\Rightarrow 2y^2-8y=0$

$\Rightarrow 2y(y-4)=0$

$\Rightarrow (y-4)=0$

Hence

$y=0$ or

$4$ .

As diagonals of a square are equal in length and bisect each other at

$90^{\circ}$

Let

$P$ is the midpoint of

$AC$

$\therefore \,$ CO-ordinates of

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

P is also the midpoint of

$BD$

then co-ordinates of mid-point of

$BD$ =co-ordinates of

$P$

$\Rightarrow (x_1,y_1)=1,2$

$\therefore x_1=1,y_1=2$

Then other two vertices of square

$ABCD$ are

$(1,0)$ or

$(1, 4)$ and

$(1,2)$ .

Prove that the diagonals of a rectangle ABCD, with vertices A $(2, -1)$ , B $(5, -1)$ , C $(5, 6)$ and D $(2, 6)$ , are equal and bisect each other.

$\Delta ADC$ and $\Delta BDC$ are right angled triangle with AD and BC are hypotenuse.

$AC^2=AB^2+DC^2$

$AC^2=(5-2)^2+(6+1)^2=9 + 48 = 58$ sq.unit

$BD^2 = DC^2+CB^2$

$BD^2 = (5-2)^2+(-1-6)^2=9+49=58$ sq.unit

Hence, both the diagonals are equal in length.

In $\Delta ABO$ and $\Delta CDO$

Since, $\angle OAB = \angle OCD$ , $\angle OBA = \angle ODC$ (Both are alternate interior angles of parallel lines)

and $AB = CD$

Therefore $\Delta ABO \cong \Delta CDO$

$\Rightarrow AO = CO$ and $BO = DO$

Therefore, Both diaginals bisects each other.

Find the ratio in which the line segment joining the points $A(3,- 3)$ and $B(- 2, 7)$ is divided by the x-axis. Also find the coordinates of the point of division.

2150506_494192_ans_813fa9a1a6e34389985e8abf8d7da607.jpeg

Let P and Q be the points of trisection of the line segment joining the points A $(2, -2)$ and B $(-7, 4)$ such that P is nearer to A. Find the coordinates of P and Q.

Since $P$ and $Q$ be the points of trisection of the line segment joining the points A $(2, -2)$ and B $(-7, 4)$ such that $P$ is nearer to $A$ .

Therefore, $P$ divides line segment in the ratio $1:2$ and $Q$ divides in $2:1$ as shown in the figure.

Using section formula,

Coordinates of $P$ = $\left[\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}\right]$

= $\left[\dfrac{1(-7)+2(2)}{1+2}, \dfrac{1(4)+2(-2)}{1+2}\right]$

= $\left[\dfrac{-3}{3}, \dfrac{0}{3}\right]$

= $\left[-1, 0\right]$

Coordinates of Q = $\left[\dfrac{2(-7)+1(2)}{2+1}, \dfrac{2(4)-1(-2)}{2+1}\right]$

= $\left[\dfrac{-12}{3}, \dfrac{6}{3}\right]$

= $\left[-4, 2\right]$

Find the ratio in which the $y-$ axis divides the line segment joining the points $(-4, -6)$ and $(10, 12)$ . Also, find the coordinates of the point of division.

Using the section formula, if a point $(x,y)$ divides the line joining the points $({ x }_{ 1 },{ y }_{ 1 })$ and $({ x }_{ 2 },{ y }_{ 2 })$ in the ratio $m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

Let

$y-$ axis divides the line joining points

$A(-4,-6)$ and

$B(10,12)$ in ratio

$y:1$

Then, as per section formula the coordinates of point which divides the line is

$\dfrac{10y-4}{y+1},\dfrac{12y-6}{y+1}$

We know that coordinate at

$y-$ axis of point of

$x$ is zero

Then,

$\dfrac{10y-4}{y+1}=0$

$\Rightarrow 10y-4=0$

$\Rightarrow 10y=4$

$\Rightarrow y=\dfrac{10}{4}=\dfrac{5}{2}$

Then, ratio is

$\dfrac{2}{5}:1 \Rightarrow 2:5$

Substitute the value of

$y$ in

$y-$ coordinates, we get

$\dfrac{12\dfrac{2}{5}-6}{\dfrac{2}{5}+1}= \dfrac{24-30}{2-5}=\dfrac{-6}{-3}=2$

Then, coordinates of point which divides the line joining

$A$ and

$B$ is

$(0,2)$ and ratio

$\dfrac{2}{5}$ .

If the coordinates of points $A$ and $B$ are $(-2, -2)$ and $(2, -4)$ respectively, find the coordinates of $P$ such that $AP = \dfrac{3}{7} AB$ , where $P$ lies on the line segment $AB$ .

Given coordinates of point

$A(-2,-2)$ and

$B(2,-4)$ and point

$P$ divided

$AB$ as

$AP=\dfrac{3}{7}AB$

Then

$BP=\dfrac{4}{7}AB$

So point

$P$ divided

$AB$ in ratio

$3:4$

$m=3$ and

$n=4$

Using Section formula, coordinates of point

$P$ are

$\left [x=\left (\dfrac{mx_1+nx_2} {m+n} \right )\right ]$ and

$\left [y=\left (\dfrac{my_1+ny_2} {m+n} \right )\right ]$

$A(-2,-2)\equiv(x_2,y_2)$ and

$B(2,-4)\equiv(x_1,y_1)$

$P =\left ( \dfrac{3\times 2-2\times 4}{3+4}, \dfrac{-4\times 32\times 4}{3+4} \right )=\left(\dfrac{6-8}{7},\dfrac{-12-8}{7}\right)=\left(-\dfrac{2}{7},-\dfrac{20}{7}\right)$

The coordinates of the midpoint of the line segment joining the points $\left( 2a+2,3 \right)$ and $\left( 4,2b+1 \right)$ are $\left( 2a,2b \right)$ . Find the values of $a$ and $b$ .

Given,

$\left( 2a,2b \right)$ is the midpoint of

$\left( 2a+2,3 \right) ,\left( 4,2b+1 \right)$
Therefore,

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

$\Rightarrow \left( \dfrac { 2a+6 }{ 2 } ,\dfrac { 2b+4 }{ 2 } \right) =\left( 2a,2b \right)$

$\Rightarrow \dfrac { 2a+6 }{ 2 } =2a$ and

$\dfrac { 2b+4 }{ 2 } =2b$

$\Rightarrow 2a+6=4a$ and

$2b+4=4b$

$\Rightarrow 2a=6$ and

$2b=4$

$\Rightarrow { a=3 }$ and

${ b=2 }$

Prove that the points $(-7, -3), (5, 10), (15, 8)$ and $(3, -5)$ taken in order are the vertices of a parallelogram.

Let the given points be

$A(-7,-3), B(5,10), C(15,8), D(3,-5)$ , taken in order.

We know that, the diagonals of a parallelogram bisect each other.

Hence,

$ABCD$ will be a parallelogram if,

Midpoint of

$AC =$ Midpoint of

$BD$

We also know that, the coordinates of the midpoint of the line segment joining

$(x_1,y_1) \ and\ (x_2,y_2)$ are:

$P(x,y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

Hence, coordinates of midpoint of

$AC =\left(\dfrac{-7+15}2,\dfrac{-3+8}2\right)$

$= \left(4,\dfrac52\right)\\$

And, coordinates of midpoint of

$BD =\left(\dfrac{5+3}2,\dfrac{10-5}2\right)$

$= \left(4,\dfrac52\right)$

Since, the coordinates of the midpoint of

$AC$ and

$BD$ are same, the

$ABCD$ must form a parallelogram.

Hence, points

$A,B,C,D$ are the vertices of a parallelogram.

If $M (4, 5)$ is the mid-point of the line segment $AB$ and co-ordinates of $A$ are $(3, 4)$ , then find the co-ordinates of point $B$ .

Let coordinates of B are

$(x,y)$ , Since M is the mid point of AB,

$4=\dfrac{3+x}{2}$

$5=\dfrac{4+y}{2}$

Hence,

$x=8-3=5$

and

$y=10-4=6$

Answer: B=

$(5,6)$

The coordinates of the vertices of a triangle are $\left({x}_{1}, {y}_{1}\right)$ , $\left({x}_{2}, {y}_{2}\right)$ and $\left({x}_{3}, {y}_{3}\right)$ . The line joining the first two is divided in the ratio $l : k$ , and the line joining this point of division to the opposite angular point is then divided in the ratio $m : k+l$ . Find the coordinates of the latter point of section.

Using the section formula, if a point $(x,y)$ divides the line joining the points $({ x }_{ 1 },{ y }_{ 1 })$ and $({ x }_{ 2 },{ y }_{ 2 })$ in the ratio $m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

The vertices of the triangle are given to be

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

$\left( { x }_{ 3 },{ y }_{ 3 } \right)$ . Let these vertices be

$A,B$ and

$C$ respectively.

Then the coordinates of the point

$P$ that divides

$AB$ in

$l:k$ will be

$\left( \dfrac { l{ x }_{ 2 }+k{ x }_{ 1 } }{ l+k } ,\dfrac { l{ y }_{ 2 }+k{ y }_{ 1 } }{ l+k } \right)$

The coordinates of point which divides

$PC$ in

$m:k+l$ will be

$\left\{ \dfrac { m{ x }_{ 3 }+(k+l)\dfrac { l{ x }_{ 2 }+k{ x }_{ 1 } }{ (l+k) } }{ m+k+l } ,\dfrac { m{ y }_{ 3 }+(k+l)\dfrac { l{ y }_{ 2 }+k{ y }_{ 1 } }{ (l+k) } }{ m+k+l } \right\} \\ \Rightarrow \left( \dfrac { k{ x }_{ 1 }+l{ x }_{ 2 }+m{ x }_{ 3 } }{ m+k+l } ,\dfrac { k{ y }_{ 1 }+l{ y }_{ 2 }+m{ y }_{ 3 } }{ m+k+l } \right)$

A line intersects the y-axis and x-axis at points P and Q respectively. If (2, -5) is the mid point of PQ, then find the coordinates of P and Q.

Let the co-ordiantes of

$P$ be

$(x, y)$ and

$Q$ be

$(x_2, y_2)$

midpoint of

$PQ=(2, -5)$

By midpoint formula,

$x=\dfrac{x_1+x_2}{2}$ and

$y=\dfrac{y_1+y_2}{2}$

$2=\dfrac{x_1+x_2}{2}$ and

$-5=\dfrac{y_1+y_2}{2}$

$\therefore x_1+x_2=4$ and

$y_1+y_2=-10$

Since line

$PQ$ intersects the y-axis at

$P$

So,

$x_1=0$ .

Similarly,

$y_2=0$

$\therefore x_2=4$ and

$y_1=-10$

The co-ordinates of

$P$ are

$(0, -10)$ and

$Q$ are

$(4, 0)$ .

A, B, C, D... are n points in a plane whose coordinates are $(x_1, y_1), (x_2, y_2), (x_3, y_3), ...... AB$ is bisected in the point $G_1; G_2C$ is divided at $G_3$ in the ratio $1:2; G_3D$ is divided at $G_4$ in the ratio $1:3; G_4E$ at $G_5$ in the ratio 1 : 4, and so on until all the points are exhausted. Show that the coordinates of the final point so obtained are
$\dfrac{x_1+x_2+x_3+.....+x_n}{n}$ and $\dfrac{y_1+y_2+y_3+...+y_n}{n}$
[This point is called the Centre of Mean Position of the n given points.]

${ G }_{ 1 }\left( \dfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+y_{ 2 } }{ 2 } \right) \\ { G }_{ 2 }\left( \dfrac { \left( \dfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } \right) 2+1({ x }_{ 3 }) }{ 2+1 } ,\dfrac { \left( \dfrac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) 2+1({ y }_{ 3 }) }{ 2+1 } \right) \Rightarrow { G }_{ 2 }\left( \dfrac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\dfrac { { y }_{ 1 }+y_{ 2 }+{ y }_{ 3 } }{ 3 } \right) \\ { G }_{ 3 }\left( \dfrac { \left( \dfrac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } \right) 3+1({ x }_{ 4 }) }{ 3+1 } ,\dfrac { \left( \dfrac { { y }_{ 1 }+y_{ 2 }+{ y }_{ 3 } }{ 3 } \right) 3+1({ y }_{ 4 }) }{ 3+1 } \right) \Rightarrow { G }_{ 3 }\left( \dfrac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+{ x }_{ 4 } }{ 4 } ,\dfrac { { y }_{ 1 }+y_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 } }{ 4 } \right) \\ { G }_{ n }\left( \dfrac { \left( \dfrac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+{ x }_{ 4 }..........{ x }_{ n } }{ n-1 } \right) (n-1)+1({ x }_{ n }) }{ n-1+1 } ,\dfrac { \left( \dfrac { { y }_{ 1 }+y_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 }+............{ y }_{ n } }{ n-1 } \right) (n-1)+1({ y }_{ n }) }{ n-1+1 } \right) \\ \Rightarrow { G }_{ n }\left( \dfrac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+{ x }_{ 4 }..........{ x }_{ n } }{ n } ,\dfrac { { y }_{ 1 }+y_{ 2 }+{ y }_{ 3 }+{ y }_{ 4 }+............{ y }_{ n } }{ n } \right)$

From points on the circle ${ x }^{ 2 }+{ y }^{ 2 }=a^{ 2 }$ , tangents are drawn to the hyperbola ${ x }^{ 2 }-{ y }^{ 2 }=a^{ 2 }$ . Prove that the locus of the mid-points of the chord of the contact is the curve ${ \left( { x }^{ 2 }-{ y }^{ 2 } \right) }^{ 2 }=a^{ 2 }\left( { x }^{ 2 }+{ y }^{ 2 } \right)$

1968202_765349_ans_56ec5bae06bf404fb99ff7163b253523.jpg

If $A$ and $B$ are two points having coordinates $(-2,-2)$ and $(2,-4)$ respectively, find the coordinates of $P$ such that $AP=\cfrac{3}{7}AB$ .

Given:

$AP=\cfrac { 3 }{ 7 } AB$

$\dfrac{AP}{AB}=\dfrac{3}{7}$

$\dfrac{AP}{AP+BP}=\dfrac{3}{3+4}$

$\dfrac{AP}{BP}=\dfrac{3}{4}$

Hence, the line joining the points

$A(-2,-2)$ and

$B(2,-4)$ is divided by

$P$ in ratio

$3:4$ .

$P(x,y)=\left(\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}\right)$

Substituting the values, $x_1=-2,y_1=-2, \ x_2=2,y_2=-4$ in the section formula.

$P=\left( \cfrac { 3(2)+4(-2) }{ 3+4 } ,\cfrac { 3(-4)+4(-2) }{ 3+4} \right)$

$P=\left( \cfrac { -2 }{ 7 } ,\cfrac { -20 }{ 7 } \right)$

$A(4,2),B(6,5)$ and $C(1,4)$ are the vertices of $\triangle {ABC}$ .
(i) The median from $A$ meets $BC$ at $D$ . Find the coordinates of the point $D$ .
(ii) Find the coordinates of point $P$ on $AD$ such that $AP:PD=2:1$

Given,

$A(4,2),B(6,5),C(1,4)$

(i)

Since AD is median, Point D becomes the mid-point of BC.

Coordinates of D

$=\left ( \dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right )$

$=\left ( \dfrac{6+1}{2}, \dfrac{5+4}{2} \right )$

$=\left ( \dfrac{7}{2},\dfrac{9}{2} \right )$

(ii)

Given,

$AP:PD=2:1$

As per section formula, the coordinates of point P

$= \left ( \dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2} \right )$

$= \left ( \dfrac{2\left ( \frac{7}{2} \right )+1(4)}{2+1},\dfrac{2\left ( \frac{9}{2} \right )+1(2)}{2+1} \right )$

$=\left ( \dfrac{11}{3},\dfrac{11}{3} \right )$

If the coordinates of the mid-points of the sides of a triangle are $(1,1), (2,-3)$ and $(3,4)$ , find its vertices.

1568100_973750_ans_ea88a2c56f084025a63bed1ead7de0de.jpg

If the coordinates of the mid-points of the sides of a triangle be $(3,-2),(-3,1)$ and $(4,-3)$ , then find the coordinates of its vertices.

Let's say D,E and F are the midpoint of AB,BC and AC.

We have mid point formula,

Point $P(X,Y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

$\Rightarrow$

$\dfrac{x+a}{2}=3\Rightarrow x+a=6$

$\Rightarrow$

$\dfrac{y+b}{2}=-2\Rightarrow y+b=-4$

$\Rightarrow$

$\dfrac{a+p}{2}=-3\Rightarrow a+p=-6$

$\Rightarrow$

$\dfrac{b+q}{2}=1\Rightarrow b+q=2$

$\Rightarrow$

$\dfrac{x+p}{2}=4\Rightarrow x+p=8$

$\Rightarrow$

$\dfrac{y+q}{2}=-3\Rightarrow y+q=-6$

Consider,

$x+a=6$ and

$a+p=-6$

$\Rightarrow$

$x-6-p=6$

$\Rightarrow$

$x-p=12,x+p=8$

On solving;

$\Rightarrow p=8-10=-2$

$\therefore 2x=20$

$\Rightarrow x=10$

$\therefore$

$a=-10+6= -4$

Now,

$y+q= -6$

$\Rightarrow -b-4+q= -6$

$\Rightarrow -b+q= -6+4$

$\Rightarrow$

$q-b= -2$

$\Rightarrow b-q=2$ and also

$b+q =2$

on solving;

$2b = 4$

$\Rightarrow b= 2$

$\therefore y= -4-2= -6$

$y+q=-6$

$\Rightarrow q=-6+6=0$

$\therefore$ Coordinate of vertices of

$\bigtriangleup$ ABC are

$A(10,-6), B(-4,2), C(-2,0)$

Find the distance of the point (1, 2) from the midpoint of the line segment joining the points (6, 8) and (2, 4).

Let $C$ be the midpoint of the line segment joining $(6, 8)$ and $(2,4)$ .

We know that the coordinates of the midpoint of the line segment joining $(x_1,y_1) \ and\ (x_2,y_2)$ are:

$P(x,y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

Hence, coordinates of the

$C$ will be:

$\left(\cfrac{6+2}{2},\cfrac{8+4}{2}\right)=(4, 6)$

We know that, the distance between two points

$A(x_1,y_1)$ and

$B(x_2,y_2)$ is

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

$\therefore\$ Required distance between

$C(4,6)\text{ and } (1,2)=\sqrt{(4-1)^2+(6-2)^2} =\sqrt{9+16} =5 \; units$

Hence, the distance between

$(1,2)$ from the midpoint of the line segment joining

$(6, 8)$ and

$(2,4)$ is

$5 \ units$ .

If $A$ and $B$ are $(1,4)$ and $(5,2)$ respectively, find the coordinates of $P$ when $\dfrac{AP}{BP}= \dfrac{3}{4}$ .

Here we use the section formula as given below:

$(x,y)$ divides line in ratio

$m:n$ , then

$x$ and

$y$ are given by

$x=\cfrac{mx_2+nx_1}{m+n}, \quad \quad y=\cfrac{my_2+ny_1}{m+n}$

Here, $x_1=1,y_1=4, \ x_2=5,y_2=2$ and $m:n=3:4$

So by putting all the values we have,

$\Rightarrow x=\cfrac{15+4}{7}, \quad \quad y=\cfrac{6+16}{7} \\ \Rightarrow P\left( \cfrac{19}{7},\cfrac{22}{7} \right)$

If 3 $\cos$ $\theta$ = 1, find the value of $\dfrac{6 \sin^2 \theta \, + \, \tan^2 \theta}{4 \cos \theta}$

Given,

$\cos \theta=\dfrac{1}{3}$

$\Rightarrow \sin \theta =\dfrac{2\sqrt 2}{3}$

$\Rightarrow \tan \theta =2\sqrt 2$

$\dfrac{6\sin ^2\theta +\tan ^2\theta }{4\cos \theta }$

$=\dfrac{6\times \tfrac{8}{9}+8}{4\times \tfrac{1}{3}}$

upon solving the above equation, we get,

$=10$

Show that the mid-point of the line segment joining the points $(5,7)$ and $(3,9)$ is also the mid-point of the line segment joining the points $(8,6)$ and $(0,10)$ .

We have mid point formula,

$P(X,Y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

So, Mid point of line joining the points (5,7) and (3,9) is

$\cfrac {5+3}{2} , \cfrac {7+9}{2}$

$=(4,8 )$ -----(1)

As midpoint

$(x_{1} , y_{1})$ of line joining two points

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

$(x_{3} , y_{3})$

$x_{1} = \cfrac {x_{2} + x_{3}}{2}$ and

$y_{1} = \cfrac {y_{2} + y_{3}}{2}$

So midpoint of line joining two points

$(8,6)$ and

$(0,10)$

$\Rightarrow$ $\cfrac {8+0}{2} , \cfrac{6+10}{2}$ $= (4,8)$ -----(2)

From (1) and (2) we can say that the mid-point of the line segment joining the points (5,7) and (3,9) is also the mid-point of the line segment joining the points (8,6) and (0,10)

(0,10).

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

Given that,

$R(x,y)$ divides

$PQ$ in the ratio

$k:1$

Then we have,

$R(X,Y)=\left(\dfrac{kx_1+x_2}{k+1}, \dfrac{ky_1+y_2}{k+1}\right)$

Here, $x_1=a,y_1=b, \ x_2=b,y_2=a$

Then

$P( {x,y} ) = \left( {\dfrac{{bk + a}}{{k + 1}},\dfrac{{ak + b}}{{k + 1}}} \right)$

$\Rightarrow x = \dfrac{{bk + a}}{{k + 1}}{\text{ }}$ and

${\text{y = }}\left( {\dfrac{{ak + b}}{{k + 1}}} \right)$

$\Rightarrow kx + x = bk + a{\text{ }}$ and

${\text{yk + y = ak + b}}$

$\Rightarrow k\left( {x - b} \right) = a - x{\text{ }}$

$\Rightarrow {\text{k}}\left( {y - a} \right) = b - y$

$\Rightarrow k = \dfrac{{a - x}}{{x - b}}{\text{ }}$ ---(i)

$\Rightarrow {\text{k = }}\left( {\dfrac{{b - y}}{{y - a}}} \right)$ ---(ii)

from (i) and (ii)

$\dfrac{{a - x}}{{x - b}} = \dfrac{{b - y}}{{y - a}}$

$\Rightarrow ay - {a^2} - xy + ax = bx - {b^2} + by - xy$

$\Rightarrow \left( {a - b} \right)y + \left( {a - b} \right)x - \left( {{a^2} - {b^2}} \right) = 0$

$\Rightarrow \left( {a - b} \right)[y + x - \left( {a + b} \right)] = 0$

$\Rightarrow x + y - \left( {a + b} \right) = 0$

$\Rightarrow x + y = a + b$

Hence proved

Determine the equation of the right bisector of the line joining the points $A(-1,2)$ & $B(-3,4)$

Given points

$A(-1,2)$ &

$B(-3,4)$

The slope of a line given by these points is

$m_1=-1$

Hence slope of the line perpendicular to a given line is

$m_2=1$

Also, the midpoint of given points by using formula

$X=\left ( \dfrac{x_1+x_2}{2}\right ) and \ Y=\left ( \dfrac{y_1+y_2}{2}\right )$

$(-2,3)$

So, equations of a line, which is the right bisector of the line given by the points are:

$l:y-3=1(x+2)$

Hence,

$l:x-y+5=0$ .

The sides of a triangle $ABC$ are as under $: AB : 2x+3y=29$ and $AC : x+2y=16$ . If the mid-point of $BC$ is $(5,6)$ then find the equation of $BC$ .

The line

$BC$ is

$(y-6)=m(x-5)$

$\dfrac { x-5 }{ \cos { \theta } } =\dfrac { y-6 }{ \sin { \theta } } =\overset { { r }_{ , } }{ C } \overset { { \quad \quad -r } }{ \quad B } \quad$

$\therefore\quad C\left( r\cos { \theta } +5,r\sin { \theta } +6 \right)$ lies on

$AC$

$B\left( -r\cos { \theta } +5,-r\sin { \theta } +6 \right)$ lies on

$AB$

$\therefore\quad r\left( \cos { \theta } +2\sin { \theta } \right) =-1$
and

$r\left( 2\cos { \theta } +3\sin { \theta } \right) =-1$
Dividing, we get

$\cos { \theta } +2\sin { \theta } =2\cos { \theta } +3\sin { \theta }$

$\therefore\quad \tan { \theta } =-1$ and hence the line

$BC$ is

$x+y=11$

Find the co - ordinates of the circles $x^2 \, + \, y^2 \, - \, 4x \, - \, 2y \, = \, 4$ and $x^2 \, + \, y^2 \, - \, 12x \, - \, 8y \, = \, 12$ touch each other. Also find equations of common tangents touching the circles in distinct points.

From given equation, we get

$C_1(2 \, , \, 1) \, , \, r_1 \, = \, 3 \, , \,C_2(6 \, , \, 4) \, , \, r \, = \, 8 \, , \, C_1C_2 \, = \, 5 \, = \, r_2 \, - \, r_1$

Hence the circles touch internally. The common tangent being

$S_1\, - \, S_2$ = 0 or 4x + 3y + 4 =0.

and common normal

$C_1 \, C_2$ is 3x - 4y - 2 = 0

Solving these, the point of contact is

$\left (-\dfrac{2}{5} \, , \, -\dfrac{4}{5} \right )$

The point of contact can also be found by ratio formula. Since the circle touch internally there will be no direct common tangents.

1035187_1007357_ans_b43b5011ae904612bb2e50575fd77f7c.png

Find the length of the medians of a $\triangle$ ABC whose vertices are A(7, -3), B(5, 3) and C(3, -1).

Let D,E,F be the mid-points of the sides BC, CA and AB respectively. Then, the coordinates of D, E and F are

$D\left( \frac { 5+3 }{ 2 } ,\frac { 3+1 }{ 2 } \right) =D\left( 4,1 \right) ,\quad E\left( \frac { 3+7 }{ 2 } ,\frac { -1-3 }{ 2 } \right) =E\left( 5,-2 \right)$

$and,\quad \quad \quad F\left( \frac { 7+5 }{ 2 } ,\frac { -3+3 }{ 2 } \right) =F\left( 6,0 \right)$

$\therefore \quad \quad AD=\sqrt { { \left( 7-4 \right) }^{ 2 }+{ \left( -3-1 \right) }^{ 2 } } =\sqrt { 9+16 } =5units$

$BE=\sqrt { { \left( 5-5 \right) }^{ 2 }+{ \left( -2-3 \right) }^{ 2 } } =\sqrt { 0+25 } =5units$

$and,\quad \quad \quad CF=\sqrt { { \left( 6-3 \right) }^{ 2 }+{ \left( 0+1 \right) }^{ 2 } } =\sqrt { 9+1 } =\sqrt { 10 } units.$

If $A$ and $B$ are $(-2, -2)$ and $(2, -4)$ , respectively, find the coordination of $P$ such that $AP = \dfrac{3}{7}AB$ and $P$ lies on the line segment $AB$ .

Let $\left( {x,y} \right)$ be the coordinates of P.

As, ${\rm{AP}} = \dfrac{3}{7}{\rm{AB}}$ , then,

${\rm{AP}} = \dfrac{3}{7}\left( {{\rm{AP}} + {\rm{PB}}} \right)$

$7{\rm{AP}} = 3{\rm{AP}} + 3{\rm{PB}}$

${\rm{4AP}} = {\rm{3PB}}$

$\dfrac{{{\rm{AP}}}}{{{\rm{PB}}}} = \dfrac{3}{4}$

So, the point P divides the line segment in the ratio $3:4$ .

By section formula,

$x = \dfrac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}},y = \dfrac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}}$

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

$x = \dfrac{{ - 2}}{7},y = \dfrac{{ - 20}}{7}$

So, the coordinates of P are $\left( { - \dfrac{2}{7}, - \dfrac{{20}}{7}} \right)$ .

976702_1025501_ans_f09999cb2199432c965798b0adace62b.PNG

Find the co-ordinate of the midpoint of the line segment joining points $(4, 7)$ and $(2, -3)$ .

Mid point of line joining points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$\left(\cfrac {x_{1} + x_{2}}{2} , \cfrac {y_{1} + y_{2}}{2}\right).$

Let the midpoint of the given points be

$(h,k),$

$\begin{aligned}{}\left( {h,k} \right) &= \left( {\frac{{4 + 2}}{2},\frac{{7 - 3}}{2}} \right)\\& = \left( {\frac{6}{2},\frac{4}{2}} \right)\\ &= \left( {3,2} \right)\end{aligned}$

Hence, midpoint of line segment joining

$(4,7)$ and

$(2,-3)$ is

$(3,2).$

Show that $A(6,4),B(5,-2)$ and $C(7,-2)$ are the vertices of an isosceles triangle. Also, find the length of the median through $A$ .

We have

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

$AC=\sqrt { { \left( 6-7 \right) }^{ 2 }+{ \left( 4+2 \right) }^{ 2 } } =\sqrt { 37 }$

$\therefore$ AB=ACSo,

$\triangle$ ABC is isosceles.Let D be the mid-point of BC. Then, coordinates of D are

$\left( \dfrac { 5+7 }{ 2 } ,\dfrac { -2-2 }{ 2 } \right) i.e\quad \left( 6,-2 \right)$

$\therefore$

$AD\quad =\sqrt { { \left( 6-6 \right) }^{ 2 }+{ \left( 4+2 \right) }^{ 2 } } =\sqrt { 36 } =6$

Find the point on the x-axis which is equidistant from (5,7) & (2 ,4).

$\quad \quad \quad \quad .\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ .\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ .\\ \quad \quad \quad (5,7)\quad \quad \quad X(X1,Y1)\quad \quad (2,4)$

${ X }_{ }=\dfrac { M_{ 1 }{ X }_{ 1 }+_{ }M_{ 2 }{ X }_{ 2 } }{ M_{ 1 }{ +M }_{ 2 } } \\$

${ X }_{ 1 }=\dfrac { 1\times 2+1\times 5 }{ 1+1 } =\dfrac { 7 }{ 2 } \\ { Y }_{ 1 }=\dfrac { 1\times 4+1\times 7 }{ 1+1 } =\dfrac { 11 }{ 2 }$

Find the image of the point (3,8) with respect to the line $x+3y=7$ assuming the line to be a plane mirror.

Let line AB be

$x+3y=7$ and point P be

$(3,8)$ .

Let

$Q(h,k)$ be the image of point

$P(3,8)$ in the line

$x+3y=7$ .

Since line

$AB$ is a mirror,

1) Point

$P$ and

$Q$ are at equal distance from line

$AB$ , i.e.,

$PR = QR$ , i.e.,

$R$ is the mid-point of

$PQ.$

2) Image is formed perpendicular to mirror i.e., line PQ is perpendicular to line AB.

Since R is the midpoint of PQ.

Mid point of

$PQ$ joining

$(3,8)$ and

$(h,k)$ is

$\left (\dfrac{h+3}{2},\dfrac{k+8}{2}\right )$

Coordinate of point R =

$\left (\dfrac{h+3}{2},\dfrac{k+8}{2}\right )$

Since point R lies on the line AB.

Therefore,

$\left (\dfrac{3+h}{2}\right )+3\left (\dfrac{8+k}{2}\right )=7$

$h+3k=-13$ ....(1)

Also,

$PQ$ is perpendicular to

$AB.$

Therefore,

Slope of

$PQ$

$\times$ Slope of

$AB = -1$

Since, slope of

$AB =$

$-\dfrac{1}{3}$

Therefore, slope of

$PQ =$

$3$

Now, PQ is line joining

$P(3,8)$ and

$Q(h,k).$

Slope of PQ =

$3=\dfrac{k-8}{h-3}$

$3h-k=1$ ........(2)

Solving equation 1 and 2, we get,

$h=-1$ and

$k=-4$

Hence, image is

$Q(-1,-4)$ .

1327170_1052764_ans_4931dea31bff4fd5a1c5b9ee02f7a095.png

If $A(3,y)$ is equidistant from points $P(8,-3)$ and $Q(7,6)$ , find the vaue of $y$ and find the distance $AQ$

Since $A\left( {3,y} \right)$ is equidistant from points $P\left( {8, - 3} \right)$ and $Q\left( {7,6} \right)$ , then,

$AP = AQ$

$\sqrt {{{\left( {3 - 8} \right)}^2} + {{\left( {y - \left( { - 3} \right)} \right)}^2}} = \sqrt {{{\left( {3 - 7} \right)}^2} + {{\left( {y - 6} \right)}^2}}$

$\sqrt {{{\left( 5 \right)}^2} + {{\left( {y + 3} \right)}^2}} = \sqrt {{{\left( 4 \right)}^2} + {{\left( {y - 6} \right)}^2}}$

$\sqrt {25 + {y^2} + 9 + 6y} = \sqrt {16 + {y^2} + 36 - 12y}$

$\sqrt {{y^2} + 6y + 34} = \sqrt {{y^2} - 12y + 52}$

Squaring both sides,

${y^2} + 6y + 34 = {y^2} - 12y + 52$

$18y = 18$

$y = 1$

$AQ = \sqrt {{{\left( {3 - 7} \right)}^2} + {{\left( {1 - 6} \right)}^2}}$

$= \sqrt {{{\left( 4 \right)}^2} + {{\left( 5 \right)}^2}}$

$= \sqrt {16 + 25}$

$= \sqrt {41}$

Therefore, the value of $y$ is 1 and the distance $AQ$ is $\sqrt {41} \;{\rm{sq}}\;{\rm{units}}$ .

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 $BM$ .

$(i)$ Given that

$AD$ is the median

We know that Median is the line drawn from a vertex to the midpoint of opposite side.

So,

$D$ is the midpoint of

$B,C$

The midpoint of

$(x_{1},y_{1})\space and \space (x_{2},y_{2})=(\dfrac{x_{1}+x_{2}}{2},\dfrac{y_{1}+y_{2}}{2})$

Therefore,

$D=(\dfrac{-3+1}{2},\dfrac{-2-8}{2})$

$\Rightarrow D=(\dfrac{-2}{2},\dfrac{-10}{2})$

$\Rightarrow D=(-1,-5)$

So, Now we need the slope of

$AD$

The slope of

$(x_{1},y_{1})\space and \space (x_{2},y_{2})=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Therefore Slope of

$AD=\dfrac{-5-4}{-1-5}=\dfrac{-9}{-6}$

$\Rightarrow$ Slope of median

$AD=\dfrac{3}{2}$

$(ii)$ Slope of altitude

$BM$

$\Rightarrow BM\perp AC$

So,

$(Slope\space of \space BM)\times (Slope\space of \space AC)=-1$

$\Rightarrow (Slope\space of \space BM)=\dfrac{-1}{ (Slope\space of \space AC)}$ .........

$(1)$

So, we first find the slope of

$AC$

The slope of

$(x_{1},y_{1})\space and \space (x_{2},y_{2})=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Therefore Slope of

$AC=\dfrac{-8-4}{1-5}=\dfrac{-12}{-4}$

$\Rightarrow$ Slope of

$AC=3$

Substituting this in

$(1)$ , we get,

$\Rightarrow Slope\space of \space BM=\dfrac{-1}{ 3}$

$\Rightarrow$ Slope of altitude

$BM=\dfrac{-1}{3}$

The point which divides the line joining the points $(1, 3)$ and $(2, 7)$ in the ratio $3:4$ is

Let P be the required point.

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

Since,

$m:n=3:4$

Therefore, we have

$P=\dfrac {6+4}{3+4}, \dfrac {21+12}{3+4}$

$P=\dfrac {10}{7}, \dfrac {33}{7}$

1457449_1075363_ans_3965e44e54c4469ab2132754598c2df1.png

Find the coordinates of the point which divides the line joining the points $(4, -3)$ and $(8, 5)$ in the ratio $3:1$ internally.

We are given,

$(x_{1}, y_{1}) = (4, -3)$ &

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

Let (x,y)coordinates which divides the line joining the point

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

and

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

$m:n = 3:1$ internally.

So,

$(x,y) = \left ( \dfrac{mx_{2}+nx_{1}}{m+n},\dfrac{my_{2}+ny_{1}}{m+n} \right )$

$= \left ( \dfrac{3(8)+1(4)}{3+1}, \dfrac{3(5)+1(-3)}{3+1} \right )$

$= \left ( \dfrac{28}{4},\dfrac{12}{4} \right )$

$(x,y) = (7,3)$

1180635_1068938_ans_9033742c03c04c3e9198d71fd8145647.jpg

The coordinates of the midpoint of the sides of a $\triangle ABC$ are $D(2,1),\ E(5,3),\ F(3,7)$ . Find the length and equation of its sides.

Given, the coordinates of the midpoints of

$\triangle ABC$ are $$D(2,1), E(5,3), F(3,7)$$ respectively.

Let the coordinates of the vertices be

$A=({ x }_{ 1 },{ y }_{ 1 }),$

$B=({ x }_{ 2 },{ y }_{ 2 }),$

$C=({ x }_{ 3 },{ y }_{ 3 })$

$\therefore$ Coordinates of

$D$ is

$\left( \dfrac { { x }_{ 1 }+x_{ 2 } }{ 2 } ,\dfrac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) =(2,1)$

$\Rightarrow \dfrac { { x }_{ 1 }+x_{ 2 } }{ 2 } =2$

$\Rightarrow { x }_{ 1 }+x_{ 2 }=4\quad \quad \longrightarrow (1)$

and,

$\dfrac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } =1$

$\Rightarrow { y }_{ 1 }+{ y }_{ 2 }=2\quad \quad \longrightarrow (2)$

Coordinates of

$E$ is

$\left( \dfrac { { x }_{ 2 }+x_{ 3 } }{ 2 } ,\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } \right) =(5,3)$

$\Rightarrow \dfrac { { x }_{ 2 }+x_{ 3 } }{ 2 } =5$

$\Rightarrow { x }_{ 2 }+x_{ 3 }=10\quad \quad \longrightarrow (3)$

and,

$\dfrac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } =3$

$\Rightarrow { y }_{ 2 }+{ y }_{ 3 }=6\quad \quad \longrightarrow (4)$

Coordinates of

$F$ is

$\left( \dfrac { { x }_{ 1 }+x_{ 3 } }{ 2 } ,\dfrac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 } \right) =(3,7)$

$\Rightarrow \dfrac { { x }_{ 1 }+x_{ 3 } }{ 2 } =3$

$\Rightarrow { x }_{ 1 }+x_{ 3 }=6\quad \quad \longrightarrow (5)$

and,

$\dfrac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 } =7$

$\Rightarrow { y }_{ 1 }+{ y }_{ 3 }=14\quad \quad \longrightarrow (6)$

Solving equations

$(1)$ ,

$(3)$ ,

$(5)$ we get,

${ x }_{ 1 }=0,\ { x }_{ 2 }=4,\ { x }_{ 3 }=6$

Solving equations

$(2)$ ,

$(4)$ ,

$(6)$ we get,

${ y }_{ 1 }=5,\ { y }_{ 2 }=-3,\ { y }_{ 3 }=9$

$\therefore$ coordinates of the vertices will be

$A=(0,5)$

$B=(4,-3)$

$C=(6,9)$

$\therefore$ Lengths of

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

$=2\sqrt { 5 }$

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

$=2\sqrt { 37 }$

$CA=\sqrt { { 6 }^{ 2 }+{ 4 }^{ 2 } }$

$=2\sqrt { 13 }$

The equations of the lines

$AB$

$\Rightarrow (y-5)=\dfrac { -3-5 }{ 4-0 } (x-0)$

$\Rightarrow 2x+y=20$

$BC\Rightarrow (y+3)=\dfrac { 9+3 }{ 6-4 } (x-4)$

$\Rightarrow 6x-y=27$

$CA\Rightarrow (y-9)=\dfrac { 5-9 }{ 0-6 } (x-6)$

$\Rightarrow 2x-3y=-15.$

If $Q$ is mid-point of line segment $AB$ and $P$ is mid-point of $QB$ , then show that $PB=\dfrac14AB$

$Q$ is the midpoint of

$\overline{AB}\Rightarrow\,\overline{AQ}+\overline{QB}=\overline{AB}$

$\Rightarrow\,\overline{AB}=2\overline{AQ}=2\overline{QB}$ .........

$(1)$

$P$ is the midpoint of

$\overline{QB}\Rightarrow\,\overline{QP}+\overline{PB}=\overline{QB}$

$\Rightarrow\,\overline{QB}=2\overline{QP}=2\overline{PB}$ .........

$(2)$

From

$(1)$ and

$(2)$ we have

$\overline{AB}=2\overline{QB}$ from

$(1)$

$\overline{AB}=2\left(2\overline{PB}\right)$ from

$(2)$

$\Rightarrow\,\overline{AB}=4\overline{PB}$

$\therefore\,\overline{PB}=\dfrac{1}{4}\overline{AB}$

$(-2.4),(4.8),(10.7)$ 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.

${\textbf{Step -1: Find the slopes of a parallel sides}}{\textbf{.}}$

${\text{Let }}EFGH{\text{ be the mid points of the quadrilateral }}ABCD.$

${\text{Coordinate of the mid - point formula are found using the formula }}\left( {\dfrac{{{x_2} - {x_1}}}{2},\dfrac{{{y_2} - {y_1}}}{2}} \right).$

${\text{Slope of }}FG = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

$= \dfrac{{\dfrac{{15}}{2} - 6}}{{7 - 1}}$

$= \dfrac{3}{{2\left( 6 \right)}}$

$= \dfrac{1}{4}$

${\text{Slope of }}EH = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

$= \dfrac{{1 + \dfrac{1}{2}}}{{\dfrac{{21}}{2} - \dfrac{9}{2}}}$

$= \dfrac{3}{{\left( {\dfrac{{12}}{2}} \right)}}$

$= \dfrac{1}{4}$

$\therefore FG\parallel EH{\text{ }}\left[ {\because {\text{ slopes are equal}}} \right]$

${\textbf{Step -2: Now, find out the slopes for the other pair of parallel sides}}{\textbf{.}}$

${\text{Slope of }}EF = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

$= \dfrac{{6 + \dfrac{1}{2}}}{{1 - \dfrac{9}{2}}}$

$= \dfrac{{13}}{2} \times - \dfrac{2}{7}$

$= - \dfrac{{13}}{7}$

${\text{Slope of }}FG = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

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

$= \dfrac{{13}}{2} \times - \dfrac{2}{7}$

$= - \dfrac{{13}}{7}$

$\therefore EP\parallel GH{\text{ }}\left[ {\because {\text{ slopes are equal}}} \right]$

${\textbf{Hence, the slope of opposite sides slope are equal, i}}{\textbf{.e}}{\textbf{. }}\mathbf{\therefore EFGH}{\textbf{ is a parallelogram}}{\textbf{.}}$

Find the area of triangle formed by joining the mid point sides of the whose vertices are A (0,-1) B (2,1) C(0,3) Find the ratio of this area to the area of $\bigtriangleup ABC$ .

Let

$P, Q$ and

$R$ be the mid-points on the sides

$AB, BC$ and

$CA$ respectively.

Therefore, by mid-point formula,

Coordinate of

$P = \left( \cfrac{0 + 2}{2}, \cfrac{-1 + 1}{2} \right) = \left( 1, 0 \right)$

Coordinate of

$Q = \left( \cfrac{0 + 2}{2}, \cfrac{3 + 1}{2} \right) = \left( 1, 2 \right)$

Coordinate of

$R = \left( \cfrac{0 + 0}{2}, \cfrac{-1 + 3}{2} \right) = \left( 0, 1 \right)$

Now,

$ar{\left( \triangle{ABC} \right)} = \cfrac{1}{2} \times \begin{vmatrix} 0 & -1 & 1 \\ 2 & 1 & 1 \\ 0 & 3 & 1 \end{vmatrix}$

$\Rightarrow ar{\left( \triangle{ABC} \right)} = \cfrac{1}{2} \times \left[ 0 - 2 \left( -1 - 3 \right) + 0 \right] = 4 \text{ sq. units}$

Again,

$ar{\left( \triangle{PQR} \right)} = \cfrac{1}{2} \times \begin{vmatrix} 1 & 0 & 1 \\ 1 & 2 & 1 \\ 0 & 1 & 1 \end{vmatrix}$

$\Rightarrow ar{\left( \triangle{PQr} \right)} = \cfrac{1}{2} \times \left[ 1 \left( 2 - 1 \right) - 0 + 1 \left( 1 - 0 \right) \right] = 1 \text{ sq. units}$

Therefore,

$\cfrac{ar{\left( \triangle{PQR} \right)}}{ar{\left( \triangle{ABC} \right)}} = \cfrac{1}{4}$

Thus area of triangle formed by joining the mid point sides of the

$\triangle{ABC}$ is

$1$ sq. unit and the ratio between the triangle formed by joining the mid point sides of the

$\triangle{ABC}$ and area of

$\triangle{ABC}$ is

$1 : 4$ .

If the mid-point of the line joining $(3,4)$ and $(k,7)$ is $(x,y)$ and $2x+2y+1=0$ , find the value of $k$ .

Since

$(x,y)$ is the mid-point,

By midpoint formula,

$x=\dfrac{3+k}{2}$

$y=\dfrac{4+7}{2}=\dfrac{11}{2}$

Again,

the above coordinates should satisfy this line,

$2x+2y+1=0$

So,

$2\times\dfrac{3+k}{2}+2\times\dfrac{11}{2}+1=0$

$3+k+11+1=0$

$3+k+12=0$

$k+15=0$

$k=-15$

1086396_1188026_ans_42ffee7dac9b483a9b8d899552b42dd8.png

The coordinates of $A$ and $B$ are $(-3,\alpha)$ and $(1,\alpha+4)$ and the mid point of $\overline {AB}$ is $P(-1,1)$ . Find the value of $\alpha$

$A=(-3,\alpha)$

$B=(1,\alpha +4)$

Mid point of

$AB$ is

$P(-1,1)$

$\cfrac { -3+1 }{ 2 } =\cfrac { -2 }{ 2 } =-1\quad ;\quad \quad \cfrac { \alpha +\alpha +4 }{ 2 } =1\\ 2\alpha +4=2\\ \Rightarrow 2\alpha =-2\\ \Rightarrow \alpha =-1$

Find the ratio in which $(-8,3)$ divides the join of points $(2,-2)$ and $(-4,1).$

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

& for mid points we have,

$m:n = 1:1$ .

Let Point

$(-8, 3)$ divides the line joining

$(2, -2)$ and

$(-4, 1)$ in

$\lambda :1$

$\Rightarrow$ Using section formulas.

$-8=\dfrac{-4\lambda+2}{\lambda +1}$

$\Rightarrow -8\lambda -8=-4\lambda +2$

$-8-2=8\lambda -4\lambda$

$\Rightarrow 4\lambda=-10$

$\Rightarrow \lambda=-\dfrac{5}{2}$

$\Rightarrow$ Point

$(-8, 3)$ divides the given line in

$-5:2$ on

$5:2$ externally.

Point $B$ on line segment joining the points $A(1,2)$ and $B(3,-2)$ divides $AB$ in the ratio $1:1$ . Then the coordinates of $D$ are __________

1354579_1204440_ans_60a31716c58f4fe7bfed0befcba164d4.png

If $C$ is a point lying on the line segment $AB$ joining $A(1, 1)$ and $B(2, - 3)$ such that $3AC = CB$ , then find the coordinates of $C$ .

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

$3AC=CB$

$\dfrac{AC}{CB}=\dfrac{1}{3}$

Since,

$A(1, 1)$ and

$B(2, 3)$ &

$m:n = 1:3$

Therefore, we have

$C\left(\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}\right)$

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

$C\rightarrow \left(\dfrac{5}{4}, \dfrac{3}{2}\right)$ .

The coordinates of the mid-point of the line joining the points $(3p, 4)$ and $(-2, 2q)$ are $(5, p)$ . Find the values of $p$ and $q$ .

Midpt. of

$(3p,4) \& (-2,2q)$ is

$(5,p)$

$\Rightarrow \left(\dfrac{3p-2}{2},\dfrac{4+2q}{2}\right)=(5,p)$

$\Rightarrow \dfrac{3p-2}{2}=5$

$\Rightarrow 3p=12$

$\Rightarrow p=4$

$\dfrac{4+2q}{2}=p$

$2+q=p=4$

$\Rightarrow q=2$

$\therefore$ the values of

$p,q$ are

$4,2$

The mid-points of the sides $BC, CA$ and $AB$ of $\triangle ABC$ are $D(2, 1), E(-1, -3)$ and $P(4, 0)$ respectively. Find the co-ordinates of $A, B$ and $C$ .

Let

$A\equiv \left( { x }_{ 1 },{ y }_{ 1 } \right)$ ;

$B\equiv \left( { x }_{ 2 },{ y }_{ 2 } \right)$ ;

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

So,

$\dfrac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } =4$ and

$\dfrac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } =0$

$\dfrac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } =2$ and

$\dfrac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } =1$

$\dfrac { { x }_{ 3 }+{ x }_{ 1 } }{ 2 } =-1$ and

$\dfrac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 } =-3$

On solving equations in

${ x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 }$ and

${ y }_{ 1 },{ y }_{ 2 },{ y }_{ 3 }$ , we get

${ x }_{ 1 }=1$ ,

${ x }_{ 2 }=7$ ,

${ x }_{ 3 }=-3$ and

${ y }_{ 1 }=-4$ ,

${ y }_{ 2 }=4$ ,

${ y }_{ 3 }=-2$

So,

$A\equiv \left( 1,-4 \right)$ ;

$B\equiv \left( 7,4 \right)$ ;

$C\equiv \left( -3,-2 \right)$

The line joining the points $A(-3,-10)$ and $B(-2,6)$ is divided by the point $P$ such that $\dfrac {PB}{AB}=\dfrac {1}{5}$ , find the co-ordinate of $P$ .

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

Given

$PB:AB=1:5$

$\therefore$

$PB:PA=1:4$

Coordinates of

$P$ are

$(x,y)=(\cfrac{4\times (-2)+1\times (-3)}{5},\cfrac{4\times 6+1\times (-10)}{5})$

$=\left( -\cfrac { 11 }{ 5 } ,\cfrac { 14 }{ 5 } \right)$

Find the coordinates of the points which divides initially the join of the points.
$(a)\,(8,9)$ and $(-7,4)$ in the ratio $2:3;$
$(b)\,(1,-2)$ and $(4,7)$ in the ratio $1:2$

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

(a) Points are

$(8, 9)$ and

$(-7, 4)$

Also,

$m:n = 2 : 3$

Therefore, we have

$(x, y) = \left(\dfrac{8 \times 3 - 7 \times 2}{5} , \dfrac{9 \times 3 + 4 \times 2}{5} \right)$

$= \left(\dfrac{24 - 14}{5} , \dfrac{27 + 8}{5} \right)$

$= (2, 7)$

(b) Points are

$(1, -2)$ and

$(4, 7)$

Also,

$m:n = 1:2$

Therefore, we have

$(x, y) = \left(\dfrac{1 \times 2 + 4 \times 1}{1 + 2} , \dfrac{-2 \times 2 + 7 \times 1}{1 + 2} \right)$

$= (2, 1)$

Find the point on the x-axis which is a equidistant from $(5 ,4)$ and $(-2, 3)$ .

Let the required point be

$(x, 0)$ . Then,

$(x-5)^2+(0-4)^2=(x+2)^2+(0-3)^2$

$(x-5)^2+16=(x+2)^2+9$

$x^2+25-10x+16=x^2+4+4x+9$

$14x=25+16-13$

$14x=28$

$x=2$

Hence, the required point is

$(2, 0)$ .

If a vertex of a triangle is $A\left(1,1\right)$ and the midpoints of two sides through $A$ are $\left(-1,2\right)$ and $\left(3,2\right)$ , then find its centroid.

1338038_1276504_ans_c3ce8d0d3aed46e187a9a748cbd9c458.PNG

If the midpoint of the line segment joining the points p(b,-2) and Q(-2,4)is (2,3). Find the value of b

The given points are

$P(b,-2)$ and

$Q(-2,4)$

The mid point is given by

$x=\dfrac{x_1+x_2}{2}\\y=\dfrac{y_1+y_2}{2}$

Here

$(x,y)=(2,3)$

$\implies 2=\dfrac{b-2}{2}\\4=b-2\\b=6$

If the midpoints of sides of a triangle are $(3,4),(4,6)$ and $(5,7)$ , find the vertices of the triangle.

Let the vertices of triangle be

$\left( { { x_{ 1 } }, { y_{ 1 } } } \right) ,\, \left( { { x_{ 2 } }, { y_{ 2 } } } \right)$ and

$\left( { { x_{ 3 } },{ y_{ 3 } } } \right)$ .

By using the given information in the question,

$\dfrac { { { x_{ 1 } }+{ x_{ 2 } } } }{ 2 } =3 \\ \Rightarrow { x_{ 1 } }+{ x_{ 2 } }=6\quad\quad\quad \dots(1)$

Similarly,

${ x_{ 2 } }+{ x_{ 3 } }=8\quad\quad\quad\dots(2) \\ { x_{ 3 } }+{ x_{ 1 } }=10\quad\quad\quad \dots(3)$

By using equation

$(1),$

${ x_{ 2 } }=6-{ x_{ 1 } }$

Substitute the above value of

$x_2$ in equation

$(2)$

$6-{ x_{ 1 } }+{ x_{ 3 } }=8 \\ \Rightarrow { x_{ 3 } }-{ x_{ 1 } }=2\quad\quad\quad \dots(4)$

Add equations

$(4)$ and

$(3)$ ,

$2{ x_{ 3 } }=12\, \\ \therefore { x_{ 3 } }=6$

And,

${ x_{ 1 } }=4 \left( { \text{from}\;\; 4 } \right) \\ { x_{ 2 } }=6-{ x_{ 1 } }$

$=6-4$

$=2$

By using the same method value of

$y_1=5,y_2=3$ and

$y_3=9$ are calculated.

Vertices of triangle are

$(4,5),(2,3),(6,9).$

1217434_1302588_ans_591eb5e34d5b4432a42ccf8a01265c24.png

Three vertices of a parallelogram $ABCD$ taken in order are $A(3, 6), B(5, 10)$ and $C(3, 2)$
the coordinate of the fourth vertex $D$ .

The coordinate of fourth point be

$\left( x, y \right)$ .

Given three coordinates of parallelogram-

$A \left( 3, 6 \right)$

$B \left( 5, 10 \right)$

$C \left( 3, 2 \right)$

Now,

Mid-point of

$AC$ will be equal to mid-point of

$BD$ .

Therefore,

Mid-point of

$AC = \left( \cfrac{3 + 3}{2}, \cfrac{6+2}{2} \right) = \left( 3, 4 \right)$

Mid-point of

$BD = \left( \cfrac{x + 5}{2}, \cfrac{y + 10}{2} \right)$

Now,

$\cfrac{x+5}{2} = 3 \Rightarrow x = 1\\$

$\cfrac{y + 10}{2} = 4 \Rightarrow y = -2$

Hence the coordinate of fourth point will be

$\left( 1, -2 \right)$ .

Determine whether the points $\mathrm { L } ( 2,5 ) , \mathrm { M } ( 3,3 ) ,$ $N ( 5,1 )$ are collinear.

L(2,5),M(3,3),N(5,1)
Slope of LM =

$\dfrac{3-5}{3-1}=\dfrac{-2}{1}=-2$
Slope of MN =

$\dfrac{1-3}{5-3}=\dfrac{-2}{2}=-1$
Slope of LM is not equal Slope of MN.
Thus, the given points are not collinear.

One end of the diameter of a circle is $(6,9)$ and the centre is $(0,0)$ . Find the co-ordinate of the other end of the diameter.

Coordinate of one end of diameter of circle

$= \left( 6, 9 \right)$

Centre of circle

$= \left( 0, 0 \right)$

Let

$\left( h, k \right)$ be the other end of the diameter of crcle.

As we know that centre of circle is the mid-point of diameter of circle.

Therefore,

$\left( \cfrac{h + 6}{2}, \cfrac{k + 9}{2} \right) = 0$

$\Rightarrow \cfrac{h + 6}{2} = 0 \Rightarrow h = -6$

$\Rightarrow \cfrac{k + 9}{2} = 0 \Rightarrow k = -9$

Hence the other end of diameter of circle is

$\left( -6, -9 \right)$ .

Find the co-ordinates of the point which divides the line segment joining points $(1, -3)$ and $(-3, 9)$ in the ratio $1:3$ internally.

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

Let

$\left( h, k \right)$ be the point which divides the line segment joining points

$\left( 1, -3 \right)$ and

$\left( -3, 9 \right)$ in the ratio

$1 : 3$ internally.

By section formula,

$\left( h, k \right) = \left( \cfrac{1 \times 3 + 1 \times \left( -3 \right)}{3 + 1}, \cfrac{-3 \times 3 + 9 \times 1}{1 + 3} \right) = \left( 0, 0 \right)$

Hence the origin divides the line segment joining points

$\left( 1, -3 \right)$ and

$\left( -3, 9 \right)$ in the ratio

$1 : 3$ internally.

In the figure $\triangle ABC, D, E, F$ are the midpoint of sides $BC, CA$ and $AB$ respectively. Show that
$BDEF$ is parallelogram

Given:-

$\triangle{ABC}$ in which

$D, E$ and

$F$ are the mid-points of sides

$BC, CA$ and

$AB$ respectively.

To prove:-

$BDEF$ is a parallelogram.

Proof:-

We know that line segments joining the mid-point of two sides of a triangle is parallel to the third side.

Now, in

$\triangle{ABC}$ ,

$F$ is mid-point of

$AB$ .

$E$ is mid-point of

$AC$ .

$\therefore EF \parallel BC$

$\Rightarrow EF \parallel BD \quad \left( \because \text{Parts of parallel lines are parallel} \right)$

Similarly,

$DE \parallel BF$

Now, in

$BDEF$ both pairs of opposite sides are parallel.

$\therefore BDEF$ is a parallelogram.

Hence proved.

The midpoints of the sides $BC, CA$ and $AB$ of a $\triangle ABC$ are $D(3, 4), E(8, 9)$ and $F(6, 7)$ respectively. Find the coordinates of the vertices of the triangle.

$\textbf{Step 1: Form equations using the given information.}$

$\text{Let the coordinates of the vertices A,B and C are}$

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

$\text{and}$

$(x_3,y_3)$

$\text{respectively.}$

$\text{Midpoint of BC is D(3,4)}$

$\therefore\left(\dfrac{x_2+x_3}{2},\dfrac{y_2+y_3}{2}\right) =(3,4).......(1)$

$\text{(Using the midpoint formula)}$

$\text{Midpoint of CA is E(8,9)}$

$\therefore\left(\dfrac{x_1+x_3}{2},\dfrac{y_1+y_3}{2}\right) =(8,9).......(2)$

$\text{(Using the midpoint formula)}$

$\text{Midpoint of AB is F(6,7)}$

$\therefore\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right) =(6,7).......(3)$

$\text{(Using the midpoint formula)}$

$\text{From (1)}$

$x_2+x_3=6.....(4)$

$\text{and}$

$y_2+y_3=8.....(5)$

$\text{From (2)}$

$x_1+x_3=16.....(6)$

$\text{and}$

$y_1+y_3=18.....(7)$

$\text{From (3)}$

$x_1+x_2=12.....(8)$

$\text{and}$

$y_1+y_2=14.....(9)$

$\textbf{Step 2: Solve the obtained equation to get the vertices.}$

$\text{Adding eq(4), eq(6) and eq(8)}$

$\Rightarrow2(x_1+x_2+x_3)=34$

$\Rightarrow x_1+x_2+x_3=17......(10)$

$\text{Now, subtracting eq(4) from eq(10)}$

$\Rightarrow (x_1+x_2+x_3)- (x_2+x_3)=17-6$

$\Rightarrow x_1=11$

$\text{Putting the value in eq(6)}$

$\Rightarrow 11+x_3=16$

$\Rightarrow x_3=5$

$\text{Again putting the value in eq(8)}$

$\Rightarrow 11+x_2=12$

$\Rightarrow x_2=1$

$\text{Now, adding eq(5), eq(7) and eq(9)}$

$\Rightarrow2(y_1+y_2+y_3)=40$

$\Rightarrow y_1+y_2+y_3=20......(11)$

$\text{Now, subtracting eq(5) from eq(11)}$

$\Rightarrow (y_1+y_2+y_3)- (y_2+y_3)=20-8$

$\Rightarrow y_1=12$

$\text{Putting the value in eq(7)}$

$\Rightarrow 12+y_3=18$

$\Rightarrow y_3=6$

$\text{Again putting the value in eq(9)}$

$\Rightarrow 12+y_2=14$

$\Rightarrow y_2=2$

$\textbf{Hence, the vertices are }$

$\textbf{A(11,12), B(1,2)}$

$\textbf{and }$

$\textbf{C(5,6)}$

The base of an equilateral triangle with side $2a$ lies along the $y$ -axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Let

$ABC$ be the given equilateral triangle with side

$2a$ .

Accordingly,

$AB= BC = CA = 2a$

Assume that base

$BC$ lies along the y-axis such that the mid-point of

$BC$ is at the origin.

i.e.,

$BO = OC = a$ , where O is the origin.

Now, it is clear that the coordinates of point

$C$ are

$(0, a)$ , while the coordinates of point B are

$(0, -a)$ .

It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.

Hence, vertex A lies on the y-axis.

REF.IMAGE

On applying pythagoras theorem to

$\Delta AOC$ , we obtain

$(AC)^2 = (OA)^2 + (OC)^2$

$\Rightarrow (2a)^2 = (OA)^2 + a^2$

$\Rightarrow 4a^2 - a^2 = (OA)^2$

$\Rightarrow (OA)^2 = 3a^2$

$\Rightarrow OA = \sqrt{3} a$

$\therefore$ Coordinates of point

$A = (\pm \sqrt{3}a, 0)$

Thus, the vertices of the given equilateral are

$(O, a), (0, -a)$ and

$(\sqrt{3}a, 0)$ or

$(0, a), (0, -a)$ and

$(-\sqrt{3}a , 0)$

1825056_1373540_ans_eac2a615655f4208b07a78263d7e082b.png

Show that $\left(3,2\right),\left(2,-3\right),\left(0,0\right)$ are the vertices of a right angled triangle using distance formula.

Let

$A\left({x}_{1},{y}_{1}\right)=\left(3,2\right),\,\,\, B\left({x}_{2},{y}_{2}\right)=\left(2,-3\right)$ and

$C\left({x}_{3},{y}_{3}\right)=\left(0,0\right)$

$\therefore$ Distance between two points

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

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

$BC=\sqrt{{\left(0-2\right)}^{2}+{\left(0+3\right)}^{2}}=\sqrt{4+9}=\sqrt{13}$ units

$CA=\sqrt{{\left(0-3\right)}^{2}+{\left(0-2\right)}^{2}}=\sqrt{9+4}=\sqrt{13}$ units

${\left(\sqrt{13}\right)}^{2}+{\left(\sqrt{13}\right)}^{2}={\left(\sqrt{26}\right)}^{2}$

$\therefore\,\,{BC}^{2}+{CA}^{2}={AB}^{2}$

Hence the given points form a right-angled triangle.

Find the midpoint of the line segment joining $\left(1,2\right)$ and $\left(3,4\right)$

Let

$A\left({x}_{1},{y}_{1}\right)=\left(1,2\right),\,\,\, B\left({x}_{2},{y}_{2}\right)=\left(3,4\right)$ be the points of a line segment

Midpoint of

$AB=\left(\dfrac{{x}_{1}+{x}_{2}}{2},\,\dfrac{{y}_{1}+{y}_{2}}{2}\right)$

Midpoint of

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

Midpoint of

$AB=\left(\dfrac{4}{2},\,\dfrac{6}{2}\right)$

Midpoint of

$AB$

$=\left(2,\,3\right)$

Hence the midpoint of

$AB$ is

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

If the three vertices of a parallelogram are $\left(3,1\right),\left(2,4\right)$ and $\left(3,5\right)$ , find the fourth vertex.

Let

$A\left(3,1\right),B\left(2,4\right), C\left(3,5\right)$ and

$D\left(x,y\right)$ be the vertices of the parallelogram.

Midpoint of

$AC=\left(\dfrac{3+3}{2},\,\dfrac{1+5}{2}\right)$

$=\left(\dfrac{6}{2},\,\dfrac{6}{2}\right)$

$=\left(3,3\right)$

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.

$\Rightarrow \dfrac{x+2}{2}=3$ and

$\dfrac{y+4}{2}=3$

$\Rightarrow x+2=6$ and

$y+4=6$

$\Rightarrow x=4$ and

$y=2$

Hence the fourth vertex is

$\left(4,2\right).$

In what ratio is the line segment joining the points $\left(-4,2,5\right)$ and $\left(1,-7,6\right)$ is divided by the $XZ$ plane ?Also, find the co-ordinates of the point of division.

Let the line segment formed by joining the points

$\left(-4,2,5\right)$ and

$\left(1,-7,6\right)$ is divided by the

$XZ$ plane in the ratio

$k:1$ at the point

$P$ , then the co-ordinates of

$P$ are

$\left(\dfrac{k-4}{k+1},\dfrac{-7k+2}{k+1},\dfrac{6k+5}{k+1}\right)$

But

$P$ lies in the

$XZ$ plane,

$\therefore\,y-$ co-ordinate is zero

$\Rightarrow\,\dfrac{-7k+2}{k+1}=0$

$\Rightarrow\,-7k+2=0$

$\Rightarrow \,7k=2$

$\Rightarrow\,k=\dfrac{2}{7}$

Hence,

$XZ$ plane divides the line segment

$AB$ in the ratio

$\dfrac{2}{7}:1$

i.e.,

$2:7$ internally and the coordinates of the point of division are

$\left(\dfrac{\dfrac{2}{7}-4}{\dfrac{2}{7}+1},\dfrac{-7\times \dfrac{2}{7}+2}{\dfrac{2}{7}+1},\dfrac{6\times \dfrac{2}{7}+5}{\dfrac{2}{7}+1}\right)$

$\left(\dfrac{\dfrac{2-28}{7}}{\dfrac{2+7}{7}},0,\dfrac{\dfrac{12+35}{7}}{\dfrac{2+7}{7}}\right)$

$\left(\dfrac{-26}{9},0,\dfrac{47}{9}\right)$

In a $\triangle{ABC},\,c{\cos}^{2}{\dfrac{A}{2}}+a{\cos}^{2}{\dfrac{C}{2}}=\dfrac{3b}{2},\,$ then show that $a,b,c$ are in A.P

Here,

$\dfrac{c}{2}\left(1+\cos{A}\right)+\dfrac{a}{2}\left(1+\cos{C}\right)=\dfrac{3b}{2}$

$a + c + b = 3b$

$a + c = 2b$

which shows

$a + c = 2b$

Find the centre of circle passing through $A\left(0,1\right),B\left(2,3\right),C\left(-2,5\right)$ is

Let the circle be

${x}^{2}+{y}^{2}+Dx+Ey+F=0$ .....

$\left(1\right)$

$A\left(0,1\right)$ passes through equation

$\left(1\right)$

$\Rightarrow 0+1+0+E+F=0$

$\Rightarrow E+F=-1$ .....

$\left(1\right)$

$B\left(2,3\right)$ passes through equation

$\left(1\right)$

$\Rightarrow 4+9+2D+3E+F=0$

$\Rightarrow 2D+3E+F=-13$ ....

$\left(3\right)$

$C\left(-2,5\right)$ passes through equation

$\left(1\right)$

$\Rightarrow 4+25-2D+5E+F=0$

$\Rightarrow -2D+5E+F=-29$ ....

$\left(4\right)$

$\left(3\right)\Rightarrow 2D+3E+F=-13$

$\Rightarrow 2D+2E+E+F=-13$

$\Rightarrow 2D+2E-1=-13$ from

$\left(2\right)$

$\Rightarrow 2D+2E=-13+1=-12$

$\Rightarrow D+E=-6$ ........

$\left(5\right)$

$\left(4\right)\Rightarrow -2D+4E+E+F=-29$

$\Rightarrow -2D+4E-1=-29$ from

$\left(2\right)$

$\Rightarrow -2D+4E=-29+1=-28$

$\Rightarrow -D+2E=-14$ .....

$\left(6\right)$

Solving equations

$\left(5\right)$ and

$\left(6\right)$ we get

$\left(5\right)+\left(6\right)\Rightarrow D+E-D+2E=-6+\left(-14\right)=-20$

$\Rightarrow 3E=-20$

$\therefore E=\dfrac{-20}{3}$

Substituting for

$E=\dfrac{-20}{3}$ in eqn

$\left(5\right)$ we get

$D=-6-E=-6-\left(\dfrac{-20}{3}\right)=-6+\dfrac{20}{3}=\dfrac{-18+20}{3}=\dfrac{2}{3}$

Substituting the values of

$D=\dfrac{2}{3}$ and

$E=\dfrac{-20}{3}$ in

eqn

$\left(4\right)$ we get

$\left(4\right)\Rightarrow -2D+5E+F=-29$

$\Rightarrow -2\left(\dfrac{2}{3}\right)+5\left(\dfrac{-20}{3}\right)+F=-29$

$\Rightarrow \left(\dfrac{-4}{3}\right)+\left(\dfrac{-100}{3}\right)+F=-29$

$\Rightarrow F=-29-\left(\dfrac{-4}{3}\right)-\left(\dfrac{-100}{3}\right)$

$\Rightarrow F=-29+\dfrac{4}{3}+\dfrac{100}{3}=-29+\dfrac{104}{3}=\dfrac{-87+104}{3}=\dfrac{17}{3}$

$\therefore D=\dfrac{2}{3}, E=\dfrac{-20}{3}, F=\dfrac{17}{3}$

$\therefore$ Equation of the circle is

${x}^{2}+{y}^{2}+Dx+Ey+F=0$ becomes

${x}^{2}+{y}^{2}+\dfrac{2}{3}x+\dfrac{-20}{3}y+\dfrac{17}{3}=0$

is of the form

${x}^{2}+{y}^{2}+2gx+2fy+c=0$

Centre is

$\left(-g,-f\right)$

$\Rightarrow 2g=\dfrac{2}{3}\Rightarrow g=\dfrac{1}{3}$

$\Rightarrow 2f=\dfrac{-20}{3}\Rightarrow f=\dfrac{-10}{3}$

Hence centre is at

$\left(\dfrac{-1}{3},\dfrac{10}{3}\right)$

If the mid-point of the line segments joining the points $(6,a-2)$ and $(-2,4)$ is $(2,-4)$ , then find the value of $a$

Given the mid point of

$(6,a-2)$ and

$(-2,4)$ is

$(2,-4)$

Then we have,

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

or,

$a+2=-8$

or,

$a=-10$ .

If $PQ=6$ cm is a straight line and a perpendicular bisector is drawn to the line $PQ$ and $A$ is the midpoint of $PQ$ and if $PA=3$ cm, then find $AQ$

A perpendicular bisector of a line segment is a line segment perpendicular to and passing through the midpoint of the line.

$PA=AQ=3$ cm

The centroid of a triangle $ABC$ is at the point $\left(1,2,1\right)$ .If the coordinates of $A$ and $B$ are $\left(2,-4,6\right)$ and $\left(-1,6,-5\right)$ respectively.Find the coordinates of the point $C$

Two vertices of a triangle are

$A\left(2,-4,6\right)$ and

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

Let the co-ordinates of the third vertex be

$C\left(a,b,c\right)$ , then the co-ordinates of the centroid of a triangle

$ABC$ are

$\left(\dfrac{2-1+a}{3},\dfrac{-4+6+b}{3},\dfrac{6-5+c}{3}\right)$

$=\left(\dfrac{1+a}{3},\dfrac{2+b}{3},\dfrac{1+c}{3}\right)$

But the centroid of the triangle

$ABC$ is

$\left(1,2,1\right)$

$\therefore\,\dfrac{a+1}{3}=1,\,\,\dfrac{b+2}{3}=2,\,\,\dfrac{c+1}{3}=1$

$\Rightarrow\,a+1=3,\,\,b+2=3,\,\,c+1=3$

$\Rightarrow\,a=3-1=2,\,\,b=3-2=1,\,\,c=3-1=2$

Hence, the co-ordinate of

$C$ are

$\left(2,1,2\right)$

The point $P$ has a co-ordinate $(2, -5)$ and Mid point of $PQ$ is $(4,3)$
Write down the co-ordinates of $Q$

Let the coordinates of

$Q$ be

$(x,y)$

Mid point of

$PQ$ is given by

$\left(\dfrac{2+x}2,\dfrac {y-5}2\right )=(4,3)\\(2+x,y-5)=(8,6)\\(x,y)=(6,11)$

Coordinates of

$Q$ are

$(6,11)$

Find the value of $p$ if the point $A(2,2)$ is mid point of $(3,p)$ and $(p,3)$

Given points are

$(3,p)$ and

$(p,3)$

The mid point is given as

$\dfrac{3+p}2,\dfrac{p+3}2$

Considering ,

$\Rightarrow \dfrac {3+p}2=2$

$\Rightarrow 3+p=4$

$\Rightarrow p=4-3=1$

If the midpoint of the line joining the points P(6, b-2) and Q(-2,4) is (2,-3), find the value of 'b'.

Given that the midpoint of the line joining points

$P(6,b-2)$ and

$Q(-2,4)$ is

$(2,-3)$

We know that, the coordinates of the midpoint of the line segment joining

$(x_1,y_1) \ and\ (x_2,y_2)$ are:

$(x,y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

$\therefore \ (2,-3)=\left( \dfrac{6-2}2,\dfrac {b-2+4}2\right)$

$\therefore \ \dfrac{b+2}{2}=-3$

$\Rightarrow b+2=-6$

$\therefore \ b=-8$

Hence, the value of

$b$ is

$-8$ .

If the points $A(k+1, 2k)$ , $B(3k, 2k+3)$ and $C(5k-1, 5k)$ are collinear, then find the value of k.

1319301_1585559_ans_44c608a224ec4570a69576d87729d573.jpg

If point (-4,6) divides the line seqment AB with A (-6,10) & B (r,s) in the ratio 2 : 1, then find $r$ and $s$ .

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

& for mid points we have,

$m:n = 1:1$ .

Therefore, we have

$(-4,6)=(\frac{2r-6}{2+1},\frac{25+10}{2+1})$

$\Rightarrow \frac{2r-6}{3}=-4$

$\frac{25+10}{3}=6$

$\Rightarrow 2r-6=-12, 25+10=18$

$\Rightarrow 2r=-12+6 , 2s=18-10$

$\Rightarrow 2r=-6, 2s=8$

$\therefore r=-3, s=4$

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

Point

$P(x, y)$ is equidistant from the points

$A(5, 1)$ and

$B(-1, 5)$ , means

$PA=PB$ or

$PA^2=PB^2$

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

$(25+x^2-10x)+(1-y^2-2y)=(1+x^2+2x+25+y^2-10y)$

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

$12x=8y$

$3x=2y$

Hence proved.

If the point $A(x, 2)$ is equidistant from the points $B(8, -2)$ and $C(2, -2)$ , find the value of x. Also, find the length of AB.

Given: Point

$A(x, 2)$ is equidistant from

$B(8, -2)$ and

$C(2, -2)$

Which implies:

$AB=AC$

Squaring both sides

$AB^2=AC^2$

Using distance formula:

Distance formula:

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

We have,

$(8-x)^2+(-2-2)^2=(2-x)^2+(-2-2)^2$

$(8-x)^2+(-4)^2=(2-x)^2+(-4)^2$

$64-16x+x^2=4-4x+x^2$

$64-4=-4x+16x$

$12x=60\Rightarrow x=\dfrac{60}{12}=5$

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

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

$=\sqrt{25}=5$ units

$x=5, AB=5$ units.

If the points $A(4, 3)$ and $B(x, 5)$ lie on a circle with the centre $O(2, 3)$ , find the value of x.

Given: Points

$A(4, 3)$ and

$B(x, 5)$ lie on a circle with centre

$O(2, 3)$

To find: value of x

$OA=OB$

$OA^2=OB^2$

$OA=OB\Rightarrow OA^2=OB^2$

$OA^2=(2-4)^2+(3-3)^2$ and

$OB^2=(2-x)^2+(3-5)^2$

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

$(-2)^2+0^2=4-4x+x^2+(-2)^2$

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

$x^2-4x+4=0\Rightarrow (x-2)^2=0$

$x=2$

The value of x is

$2$ .

Find the point on the y-axis which is equidistant from the points $A(6, 5)$ and $B(-4, 3)$ .

Let point

$P(0, y)$ is on the y-axis, then

$PA=PB$

$PA^2=PB^2$

$(6-0)^2+(5-y)^2=(-4-0)^2+(3-y)^2$

$36+25-10y+y^2=16+9-6y+y^2$

$61-10y+25-6y$

$61-25=-6y+10y$

$36=4y$

$y=9$

The required point is

$(0, 9)$ .

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

Let point

$P(x, 0)$ is on x-axis and equidustant from

$A(2, -5)$ and

$B(-2, 9)$

$PA=PB$

$PA^2=PB^2$

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

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

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

$56=-8x$

$x=-7$

The point on x-axis is

$(-7, 0)$ .

If $P(x, y)$ is a point equidistant from the points $A(6, -1)$ and $B(2, 3)$ , show that $x-y=3$ .

Solve

$P(x, y)$ is equidistant from

$A(6, -1)$ and

$B(2, 3)$ , then

$PA=PB$

$PA^2=PB^2$

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

$(36+x^2-12x)+(1+y^2+2y)=(4+x^2-4x+9+y^2-6y)$

$37-12x+2y=13-4x-6y$

$8x=8y+24$

$x-y=3$

Hence proved.

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

Since,

$P(x, y)$ is equidistant from

$A(5, 1)$ and

$B(-1, 5)$ , then

$PA=PB$

$PA^2=PB^2$

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

$(25+x^2-10x)+(1+y^2-2y)=(1+x^2+2x+25+y^2-10y)$

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

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

$12x=8y$

$3x=2y$

Hence proved.

Using the distance formula, show that the given points are collinear.
$(-1, -1), (2, 3)$ and $(8, 11)$ .

Points are collinear if sum of any two of distances is equal to the distance of the third.
Let

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

$C(8, 11)$
A, B and C are collinear if

$AB+BC=AC$

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

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

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

$BC=\sqrt{(8-2)^2+(11-3)^2}=\sqrt{6^2+8^2}$

$=\sqrt{36+64}=\sqrt{100}=10$ units

$CA=\sqrt{(8+1)^2+(11+1)^2}=\sqrt{9^2+12^2}$

$=\sqrt{81+144}=\sqrt{225}=15$ units
From above, we can see that

$AB+BC=5+10=15=AC$
Therefore, A, B and C are collinear.

If the point $A(0, 2)$ is equidistant from the points $B(3, p)$ and $C(p, 5)$ , find the value of p. Also, find the length of AB.

Given:

$A(0, 2)$ is equidistance from

$B(3, p)$ and

$C(p, 5)$
which implies:

$AB=AC$
or

$AB^2=AC^2$
Using distance formula:
Distance formula:

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

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

$(3)^2+(p-2)^2=p^2+(3)^2$

$p^2=p^2-4p+4$

$4p=4$

$\Rightarrow p=\dfrac{4}{4}=1$
and

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

$=\sqrt{9+1}=\sqrt{10}$ units.

Find the coordinates of the point which divides the join of $A(-1, 7)$ and $B(4, -3)$ in the ratio $2:3$ .

If a point

$P(x, y)$ divides a line segment having end points coordinates

$(x_1, y_1)$ and

$(x_2, y_2)$ , then coordinates of the point P can be find using formula:

$x=\dfrac{m_1x_2+m_2x_1}{m_1+m_2}$

$y=\dfrac{m_1y_2+m_2y_1}{m_1+m_2}$

Let

$P(x, y)$ be the point which divides the line joining the points

$A(-1, 7)$ and

$B(4, -3)$ in the ratio

$2:3$ , then

$x=\dfrac{2\times 4+3\times (-1)}{(2+3)}$

$=\dfrac{(8-3)}{5}$

$=1$

$\therefore x=1$ .

$y=\dfrac{(2\times -3+3\times 7)}{5}$

$=\dfrac{(-6+21)}{5}$

$=3$

$\therefore y=3$

Therefore, required point is

$(1, 3)$ .

The line segment joining $A(-2, 9)$ and $B(6, 3)$ is a diameter of a circle with centre C. Find the coordinates of C.

The line segment joining the points

$A(-2, 9)$ and

$B(6, 3)$ is a diameter of a circle with centre C.

Which means C is the midpoint of AB.

Let

$(x, y)$ be the coordinates of C, then

$x=(-2+6)/2=2$ and

$y=(9+3)/2=6$

So, coordinates of C are

$(2, 6)$ .

The midpoint of the line segment joining $A(2a, 4)$ and $B(-2, 3b)$ is $C(1, 2a+1)$ . Find the values of a and b.

Mid point of the line segment joining the points

$A(2a, 4)$ and

$B(-2, 3b)$ is

$C(1, 2a+1)$

Mid point of

$AB=((2a-2)/2, (4+3b)/2)$ ........

$(1)$

Mid point of

$AB=(1, 2a+1)$ .....

$(2)$ (given)

Now, from

$(1)$ and

$(2)$

$1=(2a-2)/2$

$\Rightarrow a=2$

and

$2a+1=(4+3b)/2$

$10-4=3b$

$b=2$

Answer:

$a=2$ and

$b=2$ .

Using the distance formula, show that the given points are collinear.
$(-2, 5), (0, 1)$ and $(2, -3)$ .

Points are collinear if sum of any two of distances is equal to the distance of the third.

Let

$A(-2, 5), B(0, 1)$ and

$C(2, -3)$

A, B and C are collinear if

$AB+BC=AC$

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

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

$=\sqrt{4+16}=\sqrt{20}$

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

$=\sqrt{4+16}=\sqrt{20}$

$=sqrt{4\times 5}=2\sqrt{5}$ units

$CA=\sqrt{(2+2)^2+(-3-5)^2}=\sqrt{4^2+(-8)^2}$

$=\sqrt{16+64}=\sqrt{80}=\sqrt{16\times 5}=4\sqrt{5}$

From above, we can see that

$AB+BC=\sqrt{20}+2\sqrt{5}=2\sqrt{5}+2\sqrt{5}=4\sqrt{5}=AC$

Therefore, A, B and C are collinear.

Find the coordinated of the midpoint of the line segemnt joining $A(3, 0)$ and $B(-5, 4)$ .

Midpoint of the line segment joining

$A(3, 0)$ and

$B(-5, 4)$ :

Midpoint

$=((x_1+x_2)/2, (y_1+y_2)/2)$

$=((3-5)/2, (0+4)/2)$

$=(-1, 2)$ .

Find the coordinates of the midpoint of the line segment joining $P(-11, -8)$ and $Q(8, -2)$ .

Mid pint of the points

$\left( {{x_1},y{ _1}} \right)$ ,

$\left( {{x_2},y{ _2}} \right)$ is given by,

$\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$ .

By using the above formula midpoint of the line segment joining

$P(-11, -8)$ and

$Q(8, -2)$ will be,

$\begin{aligned}{}\left( {\frac{{ - 11 + 8}}{2},\frac{{ - 8 - 2}}{2}} \right)& = \left( {\frac{{ - 2}}{2},\frac{{ - 10}}{2}} \right)\\ &= \left( { - 1, - 5} \right)\end{aligned}$

Point A lies on the line segment PQ joining $P(6, -6)$ and $Q(-4, -1)$ in such a way that $PA/PQ=2/5$ . If the point A also lies on the line $3x+k(y+1)=0$ , find the value of k.

Let the point

$A(x, y)$ which lies on line joining

$P(6, -6)$ and

$Q(-4, -1)$ such that

$PA/PQ=2/5$

Line segment PQ is divided by the point A in the ratio

$2:3$ .

Step

$1$ : Find coordinates of

$A(x, y)$

$x=\dfrac{m_1x_2+m_2x_1}{m_1+m_2}$

$x=(2(-4)+3(6))/(2+3)$

$=(-8+18)/5$

$=10/5=2$

$y=\dfrac{m_1y_2+m_2y_1}{m_1+m_2}$

$y=(2(-1)+3(-6))/5$

$=(-2-18)/5$

$=-20/5$

$=-4$

Step

$2$ : Point A also lies on the line

$3x+k(y+1)=0$

$3(2)+k(-4+1)=0$

$6-3k=0$

$k=2$ .

Find the coordinates of the point which divides the join of $A(-5, 11)$ and $B(4, -7)$ in the ratio $7:2$ .

If a point

$P(x, y)$ divides a line segment having end points coordinates

$(x_1, y_1)$ and

$(x_2, y_2)$ , then coordinates of the point P can be find using formula:

$x=\dfrac{m_1x_2+m_2x_1}{m_1+m_2}$

$y=\dfrac{m_1y_2+m_2y_1}{m_1+m_2}$

Let

$P(x, y)$ be the point which divides the line joining the points

$A(-5, 11)$ and

$B(4, -7)$ in the ratio

$7:2$ .

$x=(7\times 4+2\times (-5))/(7+2)$

$=(28-10)/9$

$=18/9$

$=2$

$y=(7\times (-7)+2\times 11)/9$

$=(-49+22)/9$

$=-27/9$

$=-3$

Therefore, required point is

$(2, -3)$ .

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

Given:

$(2, p)$ is the mid point of the line segment joining the points

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

To find: the value of p

$p=(-5+11)/2=6/2=3$ .

Find the coordinates of a point A, where AB is a diameter of a circle with centre $C(2, -3)$ and the other end of teh diameter is $B(1, 4)$ .

AB is diameter of a circle with centre C.

Coordinates of

$C(2, -3)$ and other point is

$B(1, 4)$

Point C is the midpoint of AB.

Let

$(x, y)$ be the coordinates of A, then

$2=(x+1)/2$

$4=x+1$

$x=3$

and

$-3=(y+4)/2$

$-6=y+4$

$y=-10$

So, coordinates of A are

$(3, -10)$ .

For what value of y are points $P(1, 4), Q(3, y)$ and $R(-3, 16)$ are collinear?

Given: Points

$P(1, 4), Q(3, y)$ and

$R(-3, 16)$ are collinear.

Which means area of triangle PQR

$=0$

Area of triangle

$=\dfrac12[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$=\dfrac{1}{2}[1(y-16)+3(16-4)+(-3)(4-y)]$

$=\dfrac{1}{2}[y-16+3\times 12-12+3y]$

$=\dfrac{1}{2}[4y-16+36-12]=\dfrac{1}{2}[4y+8]$

Since points are collinear:

$\dfrac12(4y+8)=0$

$4y = -8$

$y=-2$ .

For what values of k are the points $A(8, 1), B(3 -2k)$ and $C(k, -5)$ collinear.

Points are

$A(8, 1), B(3, -2k)$ and

$C(k, -5)$ are collinear.

Which means area of triangle ABC

$=0$

Area of a triangle

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

$=\dfrac{1}{2}[8(-2k+5)+3(-5-1)+k(1+2k)]$

$=\dfrac{1}{2}[-16k+40-3(-6)+k+2k^2]$

$=\dfrac{1}{2}[-16k+40-18+k+2k^2]$

$=\dfrac{1}{2}[22-15k+2k^2]$

Since points are collinear:

$\dfrac12(2k^2-15k+22)=0$

$2k^2-15k+22=0$

$2k^2-11k-4k+22=0$

$k(2k-11)-2(2k-11)=0$

$(k-2)(2k-11)=0$

$k=2$ or

$k=\dfrac{11}{2}$

For what value of x are the points $A(-3, 12)$ , $B(7, 6)$ and $C(x, 9)$ collinear?

Points are

$A(-3, 12), B(7, 6)$ and

$C(x, 9)$ are collinear.

Which means area of triangle ABC

$=0$

Area of a triangle

$=\dfrac12[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$=\dfrac{1}{2}[-3(6-9)+7(9-12)+x(12-6)]$

$=\dfrac{1}{2}[-3\times (-3)+7\times (-3)+x\times 6]$

$=\dfrac{1}{2}[9-21+6x]$

$=\dfrac{1}{2}[6x-12]$

Since points are collinear:

$\dfrac12(6x-12)=0$

$x=2$ .

Points $A(-1, y)$ and $B(5, 7)$ lie on a circle with centre $O(2, -3y)$ . Find the values of y.

Points

$A(-1, y)$ and

$B(5, 7)$ lie on a circle with centre

$O(2, -3y)$ .

Which means:

$OA=OB$ or

$OA^2=OB^2$

using distance formula, we get

$\Rightarrow (-1-2)^2+(y-(-3y))^2=(5-2)^2+(7-(-3y))^2$

$\Rightarrow 9+16y^2+9+(7+3y)^2$

$\Rightarrow 16y^2=49+42y+9y^2$

$\Rightarrow 7y^2-42y-49=0$

$\Rightarrow 7(y^2-6y-7)=0$

$\Rightarrow y^2-7y+y-7=0$

$\Rightarrow y(y-7)+1(y-7)=0$

$\Rightarrow (y+1)(y-7)=0$

Therefore

$y=7$ or

$y=-1$

Possible values of y are

$7$ or

$-1$ .

Find a relation between x and y, if the points $A(2, 1), B(x, y)$ and $C(7, 5)$ are collinear.

Points are

$A(2, 1), B(x, y)$ and

$C(7, 5)$ are collinear.

Which means area of triangle ABC

$=0$

Area of a triangle

$=\dfrac12[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$=\dfrac{1}{2}[2(y-5)+x(5-1)+7(1-y)]$

$=\dfrac{1}{2}[2y-10+4x+7-7y]$

$=\dfrac{1}{2}[4x-5y-3]$

Since points are collinear:
Area of triangle

$=0$

$\dfrac12(4x-5y-3)=0$

$4x-5y-3=0$

Relationship between x and y is given as

$4x-5y-3 =0$

If $a\neq b\neq c$ , prove that $(a, a^2), (b, b^2), (0, 0)$ will not be collinear.

Let

$A(a, a^2), B(b, b^2)$ and

$C(0, 0)$ are the vertices of a triangle.

Let us assume that the points are collinear, then area of

$\Delta ABC$ must be zero.

Now, area of

$\Delta ABC$

$=\dfrac{1}{2}[a(b^2-0)+b(0-a^2)+0(a^2-b^2)]$

$=\dfrac{1}{2}(ab^2-bc^2)$

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

$\neq 0$

Which is contraction to our assumption.

This implies points are not be collinear. Hence proved.

Find the value of y for which the points $A(-3, 9), B(2, y)$ and $C(4, -5)$ are collinear.

Points are

$A(-3, 9), B(2, y)$ and

$C(4, -5)$ are collinear.

Which means area of triangle ABC

$=0$

Area of a triangle

$=\dfrac12[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$=\dfrac{1}{2}[-3(y+5)+2(-5-9)+4(9-y)]$

$=\dfrac{1}{2}[-3y-15+2\times (-14)+36-4y]$

$=\dfrac{1}{2}[-7y-15-28+36]$

$=\dfrac{1}{2}[-7y-7]$

Since points are collinear:

$\dfrac12(-7y-7)=0$

$y=-1$ .

Prove that the points $A(a, 0), B(0, b)$ and $C(1, 1)$ are collinear, if $\dfrac1a+\dfrac1b=1$ .

Points are

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

$C(1, 1)$ are collinear.

Which means area of triange ABC

$=0$

Area of

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

$=\dfrac{1}{2}[a(b-1)+0(1-0)+1(0-b)]$

$=\dfrac{1}{2}[ab-a+0-b]$

Since points are collinear:

$\dfrac12(ab-a-b)=0$

$ab-a-b=0$

Divide each term by "ab", we get

$1-\dfrac1b-\dfrac1a=0$

$\dfrac1a+\dfrac1b=1$ .

Hence proved.

If the points $P(-3, 9), Q(a, b)$ and $R(4, -5)$ are collinear and $a+b=1$ , find the values of a and b.

Points are

$P(-3, 9)$ ,

$Q(a, b)$ and

$R(4, -5)$ are collinear.

Which means area of triangle

$=0$

Area of

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

$=\dfrac{1}{2}[-3(b+5)+a(-5-9)+4(9-b)]$

$=\dfrac{1}{2}[-3b-15-5a-9a+36-4b]$

$=\dfrac{1}{2}[-14a-7b+21]$

Since points are collinear, we have

$\dfrac12(-14a-7b+21)=0$

$-14a-7b+21=0$

$2a+b=3$ ..........

$(1)$

$a+b=1$ .......

$(2)$ (given)

From

$(1)$ and

$(2)$

$a=2$ and

$b=-1$ .

If the point $P(k-1, 2)$ is equidistant from the points $A(3, k)$ and $B(k, 5)$ , find the values of k.

Point P

$(k-1, 2)$ is equidistant from the points A

$(3, k)$ and B

$(k, 5)$ .

$PA=PB$ or

$PA^2=PB^2$

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

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

$\Rightarrow 16-8k+k^2+k^2-4k+4=1+9$

$\Rightarrow 2k^2-12k+20=10$

$\Rightarrow 2k^2-12k+20-10=0$

$\Rightarrow 2k^2-12k+10=0$

$\Rightarrow k^2-6k+5=0$

$\Rightarrow k^2-k-5k+5=0$

$\Rightarrow k(k-1)-5(k-1)=0$

$\Rightarrow (k-5)(k-1)=0$

$k=1$ or

$k=5$ .

If the points $C(k, 4)$ divides the join of $A(2, 6)$ and $B(5, 1)$ in the ratio $2:3$ , then find the value of k.

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

$C(k, 4)$ divides the join of

$A(2, 6)$ and

$B(5, 1)$ in the ratio

$2:3$ .

Using section formula:

$k=\dfrac{\dfrac{2}{3}\times (5)+1\times (2)}{\dfrac{2}{3}+1}$

After simplifying, we get

$k=\dfrac{16}{5}$

The value of k is

$\dfrac{16}{5}$ .

If $P(x, y)$ is equidistant from the points $A(7, 1)$ and $B(3, 5)$ , find the relation between $x$ and $y.$

Given,

$P(x, y)$ is equidistant from the point

$A(7, 1)$ and

$B(3, 5)$

$\Rightarrow PA=PB$

$\Rightarrow PA^2=PB^2$

$\Rightarrow (x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2$

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

$\Rightarrow -8x+8y=-16$

$\Rightarrow x-y=2$

$\Rightarrow$ Relation between

$x$ and

$y$ is

$x-y=2$

Find the lengths of the medians AD and BE of $\Delta ABC$ whose vertices are $A(7, -3), B(5, 3)$ and $C(3, -1)$ .

Vertices of

$\Delta ABC$ are

$A(7, -3), B(5, 3)$ and

$C(3, -1)$

From figure: BE and AD are the medians of triangle.

Find Coordinates of E and D:

Coordinates of E

$=\left(\dfrac{3+7}{2}, \dfrac{-1-3}{2}\right)$

$=(5, -2)$

Coordinates of

$D=\left(\dfrac{3+5}{2}, \dfrac{-1+3}{2}\right)$

$=(4, 1)$

Find AD and BE using distance formula:

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

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

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

$=\sqrt{9+16}$

$=\sqrt{25}$

$=5$ units.

and

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

$=\sqrt{25}=5$ units

Therefore,

$BE=5$ units and

$AD=5$ units.(both are equal)

1602640_1714503_ans_1b74d31297ed449e9ef9e7b041b51c4d.png

If the points $A(4, 3)$ and $B(x, 5)$ lie on the circle with centre $O(2, 3)$ , find the value of x.

Points

$A(4, 3)$ and

$B(x, 5)$ lie on the circle with centre

$O(2, 3)$

Which means:

$OA=OB$

$\Rightarrow OA^2=OB^2$

$\Rightarrow (2-4)^2+(3-3)^2=(2-x)^2+(3-5)^2$

$\Rightarrow (-2)^2+0^2=(2-x)^2+(-2)^2$

$\Rightarrow (2-x)^2=0$

$\Rightarrow 2-x=0$

$x=2$

The value of x is

$2$ .

Three vertices of a rectangle $\displaystyle ABCD$ are $\displaystyle A (3, 1), B (-3, -1)$ and $\displaystyle C (-3, 3)$ . Plot these points on the graph paper and find the coordinates of the fourth vertex $\displaystyle D$ . Also, find the area of the rectangle $\displaystyle ABCD$ .

From the graph,

The coordinates of point

$\displaystyle D$ are

$\displaystyle (3, 3)$ .

Area of rectangle

$\displaystyle ABCD = l \times b =AB \times BC$

On further calculation we get

Area of rectangle

$\displaystyle ABCD = 6 \times 2$

Area of rectangle

$\displaystyle ABCD = 12$ sq. units

1565077_1715365_ans_fb44c206de8b43bea174e1ee40f5006d.png

Prove that the diagonals of a rectangle ABCD with vertices $A(2, -1), B(5, -1), C(5, 6)$ and $D(2, 6)$ are equal and bisect each other.

Vertices of a rectangle ABCD are

$A(2, -1), B(5, -1), C(5, 6)$ and

$D(2, 6)$

To prove: Diagonals of the rectangle are equal and bisect each other.

Diagonal AC

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

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

$=\sqrt{9+49}$

$=\sqrt{58}$

Diagonal BD

$\sqrt{(5-2)^2+(-1-6)^2}$

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

$=\sqrt{58}$

AC and BD are equal in length. Thus, Diagonals are equal.

Now,

Consider that O is the midpoint of AC then its cooridnate are

Midpoint formula:

$(x, y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

$=((2+5)/2, (-1+6)/2)$

$=(7/2, 5/2)$

If point O divides AC in the ratio

$m:n$ , then

$\dfrac{7}{2}=\dfrac{mx_2+nx_1}{m+n}=\dfrac{m\times 2+n\times 5}{m+n}$

$=\dfrac{2m+5n}{m+n}$

$7m+7n=4m+10n$

$7m-4m=10n-7n$

$\Rightarrow 3m=3n$

$m=n$

Which shows, O is the midpoint of diagonals.

1602610_1714502_ans_40278f441d4f4da3a2d606e9faab340c.png

Find the point on x-axis which is equidistant from points $A(-1, 0)$ and $B(5, 0)$ .

Since point lies on x-axis, y-coordinate of the point will be zero.

Let

$P(x, 0)$ be on x-axis which is equidistant from

$A(-1, 0)$ and

$B(5, 0)$

Using section formula:

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

$x=2$

Thus, the required point is

$(2, 0)$ .

If the points $A(2, 3), B(4, k)$ and $C(6, -3)$ are collinear, find the value of k.

Points

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

$C(6, -3)$ are collinear.

Area of triangle having vertices A, B and C

$=0$

Area of a triangle

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

Area of given

$\Delta ABC=0$

$\Rightarrow \dfrac{1}{2}[2(k-(-3))+4(-3-3)+6(3-k))]=0$

$\Rightarrow 2k+6-24+18-6k=0$

$\Rightarrow -4k=0$

$k=0$

The value of k is zero.

Prove that the coordinates, $x$ and $y$ , of the middle point of the line joining the point $(2, 3)$ to the point $(3, 4)$ satisfy the equation $x - y + 1 = 0.$

1563962_1740528_ans_cf24c899cb8049a2870aa59709bfdd4e.png

Find the areas of the quadrilaterals the coordinate of whose angular points, taken in order, are
$(1 , 1) , (3 , 4) , (5 , -2) ,$ and $(4 , -7).$

1563878_1740539_ans_d124e162e35843079eebd5800659e8f7.png

Find the areas of the quadrilaterals the coordinate of whose angular points, taken in order, are
$(-1 , 6) , (-3 , -9) , (5 , -8) ,$ and $(3 , 9).$

1563887_1740540_ans_b830eb1776a4404f8240bd8ad6e7baad.png

Find the locus of the foot of the perpendicular let fall from the origin upon any chord of the circle $x^{2} + y^{2} + 2gx + 2fy + c = 0$ which subtends a right angle at the origin.
Find also the locus of the middle points of these chords.

1979428_1740754_ans_70b5972598744c78b7db6ec42f5e7df4.jpeg

If the points $A(1,2), B(0,0)$ and $C(a, b)$ are collinear, then what is the relation between $a$ and $b?$

Given,

Points

$A,B$ and

$C$ are collinear

$\therefore$ Area of

$\triangle ABC=0$

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

$\triangle ABC=\dfrac{1}{2}[1(0-b)+0(b-2)+a(2-0)]$

$\Rightarrow 0=\dfrac{1}{2}[1(0-b)+0(b-2)+a(2-0)]$

$\Rightarrow 0= -b+0+2a$

$\Rightarrow 2a-b=0$

Hence, the relation between

$a$ and

$b$ is

$2a-b=0$ .

Write the coordinates of a point on $x-$ axis which is equidistant from the points $(-3,4)$ and $(2,5)$ ,

Let the required point be

$(x, 0)$

Since,

$(x,0)$ is equidistant from the points

$(-3,4)$ and

$(2,5)$

$\therefore \sqrt{(-3-x)^{2}+(4-0)^{2}}=\sqrt{(2-x)^{2}+(5-0)^{2}}\qquad\text{ [ Using distance formula ]}\\ \Rightarrow sqrt{9+x^{2}+6 x+16}=\sqrt{4+x^{2}-4 x+25} \\\Rightarrow x^{2}+6 x+25=x^{2}-4 x+29 \\ \Rightarrow 10 x=4\\ \Rightarrow x=\dfrac{4}{10}=\dfrac{2}{5}$

$\therefore$ Required point is

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

If $(1,2),(4, y),(x, 6)$ and (3,5) are the vertices of a parallelogram taken in order, find $x$ and $y$

Let

$A(1,2), B(4, y), C(x, 6)$ and

$D(3,5)$ be the vertices of a parallelogram

$A B C D$

We know that,

The diagonals of a parallelogram bisect each other.

$\therefore\quad\left(\dfrac{x+1}{2}, \dfrac{6+2}{2}\right)=\left(\dfrac{3+4}{2}, \dfrac{5+y}{2}\right)\\ \Rightarrow \dfrac{x+1}{2}=\dfrac{7}{2} \quad\text{and}\quad 4=\dfrac{5+y}{2}\\\Rightarrow x+1=7 \quad\text{and}\quad 5+y=8\\ \Rightarrow x=6\quad\text{and}\quad y= 3$

Hence,

$x=6$ and

$y=3$

If $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x, 6),$ find the value of $x$ . Also, find the distances $QR$ and $PR$ .

Since, point

$Q(0,1)$ is equidistant from

$P(5,-3)$ and

$R(x, 6)$

Therefore,

$QP=QR$

Squaring both sides,

$\Rightarrow QP^{2}=QR^{2} \\\Rightarrow(5-0)^{2}+(-3-1)^{2}=(x-0)^{2}+(6-1)^{2}\\\Rightarrow 25+16=x^{2}+25\\ \Rightarrow x^{2}=16\\ \therefore x=\pm 4$

Thus,

$R$ is (4,6) or (-4,6)

Now,

$QR=\sqrt{(4-0)^{2}+(6-1)^{2}}=\sqrt{16+25}=\sqrt{41}$

$Q R=\sqrt{(-4-0)^{2}+(6-1)^{2}}=\sqrt{16+25}=\sqrt{41}$

And,

$P R=\sqrt{(4-5)^{2}+(6+3)^{2}}=\sqrt{1+81}=\sqrt{82}$

$P R=\sqrt{(-4-5)^{2}+(6+3)^{2}}=\sqrt{81+81}=9 \sqrt{2}$

of the points $A$ and $B$ . Also, find the coordinates of another point $D$ such that $BACD$ is a rhombus.

Since,

$O$ is the mid-point of the base

$BC$ .

$\therefore$ Coordinates of point

$B$ are

$(0,3)$

So,

${BC}=6\text { units }$

Let the coordinates of point

$A$ be

$(x, 0)$

Using distance formula,

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

$BC=\sqrt{(0-0)^{2}+(-3-3)^{2}}=\sqrt{36}$

Also,

$A B=B C \quad[\therefore \Delta{ABC}$ is an equilateral triangle

$]$

$\Rightarrow \sqrt{x^{2}+9}=\sqrt{36}$

$\Rightarrow x^{2}+9=36$

$\Rightarrow x^{2}=27$

$\Rightarrow x=-3 \sqrt{3} \text { or } x=3 \sqrt{3}$

$\Rightarrow x=\mp 3 \sqrt{3}$

$\therefore$ Coordinates of point

$A=(3 \sqrt{3}, 0)$

And, Coordinates of point

$D=(-3 \sqrt{3}, 0)$

1709556_1785769_ans_faaad3296d744910aceb791f7d7ec38c.png

Find the coordinates of a points $A,$ where $A B$ is the diameter of a circle whose centre is $(2,-3)$ and $B$ is $(1,4)$ .

Let,

The coordinates of

$A$ be

$(x, y)$

Now,

$C$ is the centre of the circle,

Therefore, the coordinates of

$C=\left(\dfrac{x+1}{2},\dfrac{y+4}{2}\right)$

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

$\Rightarrow\dfrac{x+1}{2}=2 \quad\text{and}\quad \dfrac{y+4}{2}=-3\\\Rightarrow x+1=4\quad\text{and}\quad y+4=-6\\\Rightarrow x=3\quad\text{and}\quad y=-10$

Hence, coordinates of

$A$ are

$(3,-10)$

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

Given,

$x_{1}=-2, y_{1}=1, x_{2}=a, y_{2}=b, x_{3}=4, y_{3}=-1$

$a-b=1\quad\dots(i)$

Since the given points

$A,B$ and

$C$ are collinear,

Therefore, the area of

$\triangle ABC=0$

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

Substituting the values in equation

$(i)$ ,

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

$\Rightarrow -2 b-2-2 a+4-4 a b=0 \quad \Rightarrow 2 a+6 b=2$

$\Rightarrow a+3 b=1\quad\dots(iii)$

Subtracting equation

$(i)$ from

$(iii)$ , we get

$4 b=0 \Rightarrow b=0$

Substituting the value of

$b$ in equation

$(i)$ , we get

$a=1$

Hence,

$a=1$ and

$b=0$

Find the coordinates of the point which divides the line segment joining $(-1,7)$ and $(4,-3)$ in the ratio $2: 3$ .

Let

$P(x,y)$ be the required point.

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

Therefore,

$x=\dfrac{2\times 4+3 \times(-1)}{2+3}=\dfrac{8-3}{5}=\dfrac{5}{5}=1 \quad \text { and }$

$y=\dfrac{2\times(-3)+3 \times 7}{2+3}=\dfrac{-6+21}{5}=\dfrac{15}{5}=3$

So, the coordinates of

$P$ are

$(1,3)$ .

Aadya and Nitya planted some trees in a square garden as shown in the Figure, both arguing that they have planted them in a straight line. Find out who is correct? Justify your decision. (N stands for Nitya and A for Aadya)

Given:

Aadya planted the tress at

$(2,1),(4,3)$ and

$(6,5)$

Area of the triangle (if any) formed by these points is,

$A_1=\dfrac{1}{2}[2(3-5)+4(5-1)+6(1-3)]$

$=\dfrac{1}{2}(-4+16-12)$

$=0$

$\therefore$ Given points are collinear

Now, Nitya planted the trees at

$(2,3),(3,4),(4,6)$

Area of the triangle (if any) formed by these points is,

$A_2=\dfrac{1}{2}[2(4-6)+3(6-3)+4(3-4)]$

$=\dfrac{1}{2}(-4+9-4)$

$=\dfrac{1}{2} \text { sq. units }$

$\therefore$ Given points are not collinear.

Hence, Aadya is correct. As the area of the triangle formed by the trees planted by Nitya is non-zero which implies that they are non-collinear.

If $A$ and $B$ are $(-2,-2)$ and $(2,-4),$ respectively, find the coordinates of $P$ such that $A P=\dfrac{3}{7} A B$ and $P$ lies on the line segment $A B$ .

In the given figure, we have,

$A P=\dfrac{3}{7} A B\Rightarrow\dfrac{A B}{A B}=\dfrac{3}{7} \\\Rightarrow \dfrac{A B}{A P}=\dfrac{7}{3}\\\Rightarrow \dfrac{A P+P B}{A P}=\dfrac{7}{3} \\ \Rightarrow \dfrac{A P}{A P}+\dfrac{PB}{A P} =\dfrac{7}{3}\\ \Rightarrow 1+\dfrac{P B}{A P}=\dfrac{7}{3} \\ \Rightarrow \dfrac{P B}{AP}=\dfrac{7}{3}-1=\dfrac{4}{3}\\ \Rightarrow\dfrac{A P}{P B}=\dfrac{3}{4} \Rightarrow A P: P B=3: 4$

Let

$P(x,y)$ be the point which divides the join of

$A(-2,-2)$ and

$B(2,-4)$ in the ratio

$3: 4$

$\therefore$ By section formula,

$x=\dfrac{3 \times 2+4 \times-2}{3+4}=\dfrac{6-8}{7}=\dfrac{-2}{7}$

And,

$y=\dfrac{3 \times-4+4 \times-2}{3+4}=\dfrac{-12-8}{7}=\dfrac{-20}{7}$

Hence, the coordinates of the point

$P$ are

$\left(\dfrac{-2}{7}, \dfrac{-20}{7}\right)$

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

Let

$P(x,0)$ be any point on

$x$ -axis.

Given,

$P(x,0)$ is equidistant from point

$A(2,-5)$ and

$B(-2,9)$

$\therefore AP=BP$

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

Squaring both sides, we have

$(x-2)^{2}+25=(x+2)^{2}+81$

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

$\Rightarrow-8 x=56$

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

$\therefore$ The point on the

$x$ -axis equidistant from given points is

$(-7,0)$

The point $P(4,1)$ divides the line segment joining the points $A(2,2)$ and $B$ in the ratio of $3:5$ . Find the point $B$ .

Let the co-ordinates of

$B$ be

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

Given co-ordinates of

$A=(2,-2)$

Co-ordinates of

$P=(-4,1)$

Ratio

$m:n = 3:5$

$x_{1}=2,y_{1}=-2,x=-4,y=1$

$P$ divides

$AB$ in the ratio

$3:5$

$\therefore$ By section formula,

$x=(mx_{2}+nx_{1})/(m+n)$

$\therefore -4 = (3\times x_{2}+5\times 2)/(3+5)$

$\therefore -4 = (3x_{2}+10)/8$

$\therefore -32 = 3x_{2}+10$

$\therefore 3x_{2} = -32-10 = -42$

$\therefore x_{2} = -42/3 = -14$

$\therefore$ By Section formula

$y=(my_{2}+ny_{1})/(m+n)$

$\therefore 1= (3y_{2}+5 \times (-2))/(3+5)$

$\therefore 1 = (3y_{2}-10)/8$

$\therefore 8 = 3y_{2}-10$

$\therefore 3y_{2} = 8+10 = 18$

$\therefore y = 18/3 = 6$

Hence the co-ordinates of

$B$ are

$(-14,6)$ .

The mid-point of the line segment joining the points $(3m,6)$ and $(-4,3n)$ is $(1,2m-1)$ . Find the values of $m$ and $n$ .

Let the midpoint of line joining the points

$A(3m,6)$ and

$B(-4,3n)$ be

$C(1,2m-1)$ .

Here

$x_{1}=3m,y_{1}=6,x_{2}=-4,y_{2}=3n$

$x=1, y=2m-1$

$\therefore$ By Midpoint formula,

$x = (x_{1}+x_{2})/2$

$\therefore 1 = (3m+-4)/2$

$\therefore 3m-4 = 2$

$\therefore 3m = 2+4$

$\therefore 3m = 6$

$\therefore m = 6/3 = 2$

$\therefore$ By Midpoint formula,

$y = (y_{1}+y_{2})/2$

$\therefore 2m-1 = (6+3n)/2$

$\therefore 4m-2 = 6+3n$

Put

$m = 2$ in above equation,

$\therefore 4\times2-2 = 6+3n$

$\therefore 8-2-6 = 3n$

$\therefore 3n = 0$

$\therefore n = 0$

Hence the value of

$m$ and

$n$ are

$2$ and

$0$ respectively.

$AB$ is a diameter of a circle with centre $C(2,5)$ . If point $A$ is $(3,7)$ . Find:
(i) the length of radius $AC$ .
(ii) the coordinates of $B$ .

(i) Length of radius

$AC = d(A,C)$

Co-ordinates of

$A = (3,-7)$

Co-ordinates of

$C = (-2,5)$

Here

$x_{1} = 3, y_{1} = -7, x_{2} = -2, y_{2} = 5$

By distance formula,

$d(A,C) = \sqrt{[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]}$

$\sqrt{[(-2-3)^{2}+(5-(-7))^{2}]}$

$\sqrt{[(-5)^{2}+(12)^{2}]}$

$\sqrt{[25+144]}$

$\sqrt{169}$

$13$

Hence the radius is

$13$ units.

(ii) Given

$AB$ is the diameter and

$C$ is the centre of the circle.

$\therefore$ By midpoint formula,

$-2 = (x+3)/2$

$\Rightarrow -4 = x+3$

$\Rightarrow x = -4-3 = -7$

$\therefore$ By midpoint formula,

$5 = (-7+y)/2$

$\Rightarrow 10 = -7+y$

$\Rightarrow y = 10+7 = 17$

Hence the co-ordinates of

$B$ are

$(-7,17)$ .

1765443_1787795_ans_b9681a558ce2417f8cf4e930efec42a9.png

The co-ordinates of two points $A$ and $B$ are $(3,3)$ and $(12,7)$ respectively. $P$ is a point on the line segment $AB$ such that $AP : PB = 2 : 3$ . Find the co-ordinates of $P$ .

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

Let the co-ordinates of

$P(x,y)$ divides

$AB$ in the ratio

$m:n$ .

$A(-3,3)$ and

$B(12,-7)$ are the given points.

Given

$m:n = 2:3$

$\therefore x_{1}=-3, y_{1}=3, x_{2}=12, y_{2}=-7, m=2$ and

$n=3$

$\therefore$ By Section formula, we have

$x = (2\times 12+3\times -3)/(2+3)$

$\therefore x = (24-9)/5$

$\therefore x = 15/5$

$\therefore x = 3$

And

$y = (2\times -7+3\times 3)/5$

$\therefore y = (-14+9)/5$

$\therefore y = -5/5$

$\therefore y = -1$

Hence the co-ordinate of point

$P$ are

$(3,-1)$ .

The point $P$ divides the line segment joining the points $A(– 2,1)$ and $B(1,4)$ in the ratio of $2 : 1$ . Calculate the co-ordinates of the point $P$ .

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

Let the point

$P(x, y)$ divides

$AB$ in the ratio

$m:n$ .

$A(-2,1)$ and

$B(1,4)$ are the given points.

Given

$m:n = 2:1$

$\therefore x_{1}=-2, y_{1}=1, x_{2}=1, y_{2}=4, m=2$ and

$n=1$

$\therefore$ By Section formula

$x=\dfrac{2\times1+1\times-2}{2+1}$

$\therefore x=\dfrac{2-2}{3}$

$\therefore x = \dfrac{0}{3}$

$\therefore x = 0$

And

$y=\dfrac{2\times4+1\times1}{2+1}$

$\therefore y=\dfrac{8+1}{3}$

$\therefore y = \dfrac{9}{3}$

$\therefore y = 3$

Hence the co-ordinates of point

$P$ are

$(0,3)$ .

Find the co-ordinates of the mid-point of the line segments joining the following pairs of points:
$(a + 3, 5b),(2a 1, 3b + 4)$ .

Co-ordinates of midpoint of line joining the points

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

$(x_{2},y_{2}) = {(x_{1}+x_{2})/2 ,(y_{1}+y_{2})/2}$

Co-ordinates of midpoint of line joining the points

$(a+3,5b)$ and

$(2a-1,3b+4) = {(a+3+2a-1)/2, (5b+3b+4)/2}$

${(3a+2)/2, (8b+4)/2}$

${(3a+2)/2, (4b+2)}$

Hence the co-ordinates of midpoint of line joining the points

$(a+3,5b)$ and

$(2a-1,3b+4)$ are

${(3a+2)/2,(4b+2)}$ .

Find the co-ordinates of the mid-point of the line segments joining the following pair of points:
$(2, 3)$ , $( 6,7)$ .

Co-ordinates of midpoint of line joining the points

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

$(x_{2},y_{2}) = {(x_{1}+x_{2})/2 ,(y_{1}+y_{2})/2}$

Co-ordinates of midpoint of line joining the points

$(2,-3)$ and

$(-6,7)={(2+(-6))/2, (-3+7)/2}$

$(-4/2, 4/2)$

$(-2, 2)$

Hence the co-ordinates of midpoint of line joining the points

$(2,-3)$ and

$(-6,7)$ is

$(-2, 2)$ .

Find the co-ordinates of the mid-point of the line segments joining the following pairs of points:
$(5,11),(4,3)$ .

Co-ordinates of midpoint of line joining the points

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

$(x_{2},y_{2}) = {(x_{1}+x_{2})/2 ,(y_{1}+y_{2})/2}$

Co-ordinates of midpoint of line joining the points

$(5,-11)$ and

$(4,3)= {(5+4)/2,(-11+3)/2}$

$(9/2, -8/2)$

$(9/2, -4)$

Hence the co-ordinates of midpoint of line joining the points

$(5,-11)$ and

$(4,3)$ is

$(9/2, -4)$ .

The coordinates of the midpoint of the line segment $PQ$ are $(1,–2)$ . The coordinates of $P$ are $(–3,2)$ . Find the coordinates of $Q$ .

Given that,

The coordinates of

$P$ are

$(-3,2)$ .

The coordinates of midpoint of line segment

$PQ$ are

$(1,-2)$ .

To find out,

The coordinates of

$Q$ .

Let the coordinates of

$Q$ be

$(h,k)$ .

We know that the coordinates of the midpoint of the line segment joining

$(x_1,y_1) \ and\ (x_2,y_2)$ are:

$P(x,y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

Here,

$x_{1}=-3,\ y_{1}=2,\ x_2=h,\ y_2=k,\ x=1,\ y=-2$

Hence,

$1=\dfrac{-3+h}{2}$ and

$-2=\dfrac{2+k}{2}$

$\Rightarrow -3+h=2$ and

$2+k=-4$

$\Rightarrow h=2+3$ and

$k=-4-2$

$\Rightarrow h=5$ and

$k=-6$

$\therefore \ (h,k)=(5,-6)$

Hence, the coordinates of

$Q$ are

$(5, -6)$ .

The line segment joining $A(3,1)$ and $B(5,4)$ is a diameter of a circle whose center is $C$ . find the co-ordinates of the point $C$ .

Given Co-ordinates of

$A=(-3,1)$ ,

Co-ordinates of

$B=(5,-4)$ ,

Here

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

Let

$C(x,y)$ be the midpoint of

$AB$ ,

$\therefore$ By midpoint formula,

$x = (x_{1}+x_{2})/2$

$y = (y_{1}+y_{2})/2$

$\therefore x = (-3+5)/2 = 2/2 = 1$

$\therefore y = (1+(-4))/2 = -3/2$

Hence the co-ordinate of midpoint of

$AB$ is

$C(1,-3/2)$ .

In the above figure, $l \parallel m$ and $M$ is the mid-point of a line segment $AB$ . Show that $M$ is also the mid-point of any line segment $CD$ , having its end points on $l$ and $m$ , respectively.

$\triangle \ AMC$ and

$\triangle \ BMD$ , we have

$\angle 1= \angle 3$ [Alt.

$\angle s$ because

$l \parallel m$ ]

$\angle 2 = \angle 4$ [Vert. opp.

$\angle s$ ]

$AM=BM$ [Given]

$\therefore$

$\triangle \ AMC \cong \triangle \ BMD$ [By ASS congruence rule]

$\therefore$

$CM=DM$ [CPCT]
Hence,

$M$ is also the mid-point of

$CD$ .

1809253_1794805_ans_5ae5b98d7d674013b41f200875130604.png

Find a point which is equidistant from the points $A(-5 , 4)$ and $B(-1 , 6)$ . How many such points are there ?

Let

$P(x , y)$ is equidistant from

$A (-5 , 4)$ and

$B(-1 , 6)$ .

Then,

$PA = PB$

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

$\Rightarrow \,(x + 5)^2 + (y - 4)^2 = (x + 1)^2 + (y - 6)^2$

$\Rightarrow \, x^2 + 25 + 10x + y^2 + 16 - 8y = x^2 + 1 + 2x + y^2 + 36 - 12y$

$\Rightarrow \,41 + 10x - 8y = 37 + 2x - 12y$

$\Rightarrow \, 8x + 4y + 4 = 0$

$\Rightarrow \,2x + y + 1 = 0$

The above equation shows that infinite points are equidistant from

$AB$ , because all the points on perpendicular bisector of

$AB$ will be equidistant from

$AB$ .

$\Rightarrow \,$ One such point which is equidistant from

$A$ and

$B$ is the midpoint

$M$ of

$AB$ i.e.,

$M \left ( \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \right )$

$M \left ( \dfrac{-5 - 1}{2} , \dfrac{4 + 6}{2} \right )$

$M \left ( \dfrac{-6}{2} , \dfrac{10}{2} \right )$

$M (-3 , 5)$

So ,

$(-3 , 5)$ is equidistant from point

$A$ and

$B$ .

If the point $A(2 , -4)$ is equidistant from $P(3 , 8)$ and $Q(-10 , y)$ then find the value of $y$ . Also find distance PQ.

According to the questions ,

$PA = QA$

$\Rightarrow \,PA^2 = QA^2$

$\Rightarrow \,(3 - 2)^2 + (8 + 4)^2 = (-10 - 2)^2 + (y + 4)^2$

$\Rightarrow \,1^2 + 12^2 = (-12)^2 + y^2 + 16 + 8y$

$\Rightarrow \,y^2 + 8y + 16 - 1 = 0$

$\Rightarrow \,y^2 + 8y + 15 = 0$

$\Rightarrow \,y^2 + 5y + 3y + 15 = 0$

$\Rightarrow \,y(y + 5) + 3(y + 5) = 0$

$\Rightarrow \,(y + 5) (y + 3) = 0$

$\Rightarrow \,y + 5 = 0$ or

$y = -3$

So , the coordinates are

$P(3 , 8) , Q_1(-10 , -3) , Q_2(-10 , -5)$ .

Now ,

$PQ_{1}^{2} = (3 + 10)^2 + (8 + 3)^2 = 13^2 + 11^2$

$\Rightarrow \,PQ_{1}^{2} = 169 + 121$

$\Rightarrow \,PQ_{1} = \sqrt{290}$ units

and ,

$PQ_{2}^{2} = (3 + 10)^2 + (8 + 5)^2 = 13^2 + 13^2$

$= 13^2 [1 + 1]$

$\Rightarrow \, PQ_{2}^{2} = 13^2 \times 2$

$\Rightarrow \,PQ_{2} = 13\sqrt{2}$ units

Hence ,

$y = -3 , -5$ and

$PQ = \sqrt{290}$ units and

$13 \sqrt{2}$ units.

Plot the points $(0,2), (3,0), (0, -2)$ and $(-3,0)$ on a graph paper. Join these points (in order). Name the figure so obtained and find the area of the figure obtained.

Given points are

$(0,2), (3,0), (0,-2)$ and

$(-3,0).$
The points are plotted in the graph below.
The quadrilateral obtained is a rhombus.

$BD$ and

$AC$ are the diagonals of the rhombus.
Area of a rhombus

$= 1/2 ×d_1×d_2$
Where

$d_1$ and

$d_2$ are the length of diagonals.

$AC = 4$ units [from graph]

$BD = 6$ units [from graph]
Area of rhombus

$ABCD = 1/2 ×BD×AC$

$= 1/2 ×6×4$

$= 12 sq. units.$
Hence the area is

$12 sq. units.$
1721733_1811879_ans_3775c80af1754fe6903aa3b71f91af05.PNG

Plot the points $A (1,2), B (-4,2), C (-4, -1)$ and $D (1, -1).$ What kind of quadrilateral is $ABCD$ ? Also find the area of the quadrilateral $ABCD.$

Given points are

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

$D (1, -1).$
The points are plotted in the graph below.

$ABCD$ is a rectangle.
Area of rectangle

$ABCD =$ length ×breadth

$= AB×AD$

$= (1-(-4))×(2-(-1))$

$= 5×3$

$= 15 sq. units.$
1721706_1811875_ans_f9f19e9095f74b2eb9f5e0c1511a67b6.PNG

If $Q(0, 1)$ is equidistant from $P (5, -3)$ and $R (x, 6)$ find the values of $x$ .

$Q(0,1)$ is equidistant from

$P(5,-3)$ and

$R(x,6)$ .
So

$PQ = QR$
By distance formula,

$\sqrt{[(5-0)^2+(-3-1))^2]} = \sqrt{[(x-0)^2+(6-1)^2}$

$\sqrt{[(5)^2+(-4^2]} = \sqrt{[(x^2+(5)^2]}$

$\sqrt{[(25+16)]} = \sqrt{[x^2+25]}$

$\sqrt{(41)} = \sqrt{[x^2+25]}$
Squaring both sides,

$41 = x^2+25$

$x^2+25-41 = 0$

$x^2-16= 0$

$(x-4)(x+4) = 0$

$(x-4) = 0$ or

$(x+4) = 0$

$x = 4$ or

$x = -4$
Hence the value of

$x$ is

$4$ or

$-4$ .

The $x-coordinate$ of a point $P$ is twice its y-coordinate. If $P$ is equidistant from the points $Q (2, -5)$ and $U (-3, 6)$ , then find the coordinates of $P.$

Let the y co-ordinate be

$x$ .
Then x coordinate is

$2x.$
So coordinates of

$P$ are

$(2x,x)$ .

$P$ is equidistant from the points

$Q (2, -5)$ and

$U (-3, 6)$ .

$PQ = PU$
By distance formula,

$\sqrt{(2-2x)^2+(-5-x))^2]} = \sqrt{[(-3-2x)^2+(6-x)^2]}$

$\sqrt{[(4-8x+4x^2+25+10x+x^2]} = \sqrt{[9+12x+4x^2+36-12x+x^2]}$

$\sqrt{[29+2x+5x^2]} = \sqrt{[45+5x^2]}$
Squaring both sides,

$29+2x+5x^2 = 45+5x^2$

$2x+29-45 = 0$

$2x-16 = 0$

$2x = 16$

$x = 16/2$

$x = 8$
So

$2x = 2×8 = 16$

$P(2x,x) = P(16,8)$
Hence the coordinates of

$P$ are

$(16,8)$ .

Write the co-ordinates of the vertices of a rectangle which is $6$ units long and $4$ units wide if the rectangle is in the first quadrant, its longer side lies on the $x-axis$ and one vertex is at the origin.

The rectangle which is

$6$ units long and

$4$ units wide is shown in the graph.
Rectangle is in the first quadrant.
Longer side lies on

$x -axis$ and one vertex is at origin.
Coordinates of the rectangle are

$A(0,0), B(6,0), C(6,4)$ and

$D(0,4).$

1721741_1811895_ans_f77c2f453a7542e99223b22121e02b29.PNG

Find the co-ordinates of point which divides the line segment joining the points $A (4,-3)$ and $B(8,5)$ in the ratio 3: 1 internally.

If point

$P$ divides the line joining the points

$(x_1,y_1)$ and

$(x_2,y_2)$ in ratio of

$m:n$ , then coordinates of the point

$P$ are given using Section Formula as follows:

$P=\left(\dfrac{x_{2} m+x_{1} n}{m+n}, \dfrac{y_{2} m+y_{1} n}{m+n}\right)$

Here

$x_{1}=4, y_{1}=-3, x_{2}=8, y_{2}=5, m=3, n=1$

$P=\left(\dfrac{8(3)+4(1)}{3+1}, \dfrac{5(3)+(-3) 1}{3+1}\right)$

$P=\left(\dfrac{24+4}{4}, \dfrac{15-3}{4}\right)$

$P=\left(\dfrac{28}{4}, \dfrac{12}{4}\right)$

$P=(7,3)$

Find the coordinates of the point $R$ on the line segment joining the points $P(-1 , 3)$ and $Q(2 , 5)$ such that $PR = \dfrac{3}{5} PQ$ .

$PR = \dfrac{3}{5} PQ$

$\Rightarrow \, \dfrac{5}{3} = \dfrac{PQ}{PR}$

$\Rightarrow \, \dfrac{5}{3} = \dfrac{PR + RQ}{PR}$

$\Rightarrow \, \dfrac{3}{5} = \dfrac{PR}{PR} + \dfrac{RQ}{PR}$

$\Rightarrow \, \dfrac{QR}{PR} = \dfrac{5}{3} - 1 = \dfrac{5 - 3}{3}$

$\Rightarrow \, \dfrac{QR}{PR} = \dfrac{2}{3}$

$\dfrac{PR}{QR} = \dfrac{3}{2}$ or

$PR : QR = 3 : 2$

$\therefore \, m_1 = 3$ and

$m_2 = 2$

Now , the coordinates of point

$R$ are given by

$x = \dfrac{m_1 (x_2) + m_2(x_1)}{m_1 + m_2}$ and

$y = \dfrac{m_1(y_2) + m_2(y_1)}{m_1 + m_2}$

$\Rightarrow \, x = \dfrac{3(2) + 2(-1)}{3 + 2} = \dfrac{6 - 2}{5}$

$\Rightarrow \,y = \dfrac{3(5) + 2(3)}{3 + 2} = \dfrac{15 + 6}{5}$

$\Rightarrow \, x = \dfrac{4}{5}$

$\Rightarrow \, y = \dfrac{21}{5}$

Hence , the required coordinates of

$R$ are

$\left ( \dfrac{4}{5} , \dfrac{21}{5} \right )$

1765143_1796135_ans_854c44e2d8f2436988abd84e3c4423e7.png

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.

Given

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

$S (-3, -2)$ be four points in a plane.
By distance formula,

$PQ$ =

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

$PQ$ =

$\sqrt{[(3-2)^2+(4-(-1))^2]}$

$PQ$ =

$\sqrt{(1)^2+(5)^2}$

$PQ$ =

$\sqrt{(1+25)}$

$PQ$ =

$\sqrt{26}$
By distance formula,

$QR$ =

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

$QR$ =

$\sqrt{[(-2-3)^2+(3-4)^2]}$

$QR$ =

$\sqrt{[(-5)^2+(-1)^2]}$

$QR$ =

$\sqrt{[25+1]}$

$QR$ =

$\sqrt{26}$
By distance formula,

$RS$ =

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

$RS$ =

$\sqrt{[(-3-(-2))^2+(-2-3)^2]}$

$RS$ =

$\sqrt{[(-1)^2+(-5)^2]}$

$RS$ =

$\sqrt{[1+25]}$

$RS$ =

$\sqrt{26}$
By distance formula,

$PS$ =

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

$PS$ =

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

$PS$ =

$\sqrt{[(-5)^2+(-1)^2]}$

$PS$ =

$\sqrt{[25+1]}$

$PS$ =

$\sqrt{26}$
Here

$PQ = QR = RS = PS$ .
So it can be a rhombus or a square.
Diagonal,

$PR$ =

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

$PR$ =

$\sqrt{[(-4))^2+(4)^2]}$

$PR$ =

$\sqrt{(16+16)} = \sqrt{32} = \sqrt{(16×2)} = 4\sqrt{2}$
Diagonal,

$QS$ =

$\sqrt{[(-3-3)^2+(-2-4)^2]}$ [Distance formula]

$QS$ =

$\sqrt{[(-6))^2+(-6)^2]}$

$QS$ =

$\sqrt{[36+36]} = \sqrt{(2×36)} = 6\sqrt{2}$
Here diagonals are not equal. So

$PQRS$ is not a square. It is a rhombus.
Area of rhombus

$PQRS$ =

$^{1}/_{2} \times PR \times QS$
=

$^1/_2 \times 4\sqrt{}2 \times 6\sqrt{}2$
=

$24$ sq units.
Hence the ares of the rhombus

$PQRS$ is

$24$ sq. units.

Show that the points $(0, 1), (- 2, 3), (6, 7)$ and $(8, 3)$ , taken in order, are the vertices of a rectangle.
Also find its area.

Let the points

$A(0, – 1), B(- 2, 3), C(6, 7)$ and

$D(8, 3)$ be the vertices of a rectangle.
By distance formula,

$AB$ =

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

$AB$ =

$\sqrt{[(-2-0)^2+(3-(-1))^2]}$

$AB$ =

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

$AB$ =

$\sqrt{(4+16)}$

$AB$ =

$\sqrt{20}$

$AB$ =

$\sqrt{(4×5)}$

$AB$ =

$2\sqrt{5}$
By distance formula,

$BC$ =

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

$BC$ =

$\sqrt{[(6-(-2))^2+(7-3)^2]}$

$BC$ =

$\sqrt{[(8)^2+4^2]}$

$BC$ =

$\sqrt{(64+16)}$

$BC$ =

$\sqrt{(80)}$

$BC$ =

$\sqrt{(5×16)}$

$BC$ =

$4\sqrt{5}$
By distance formula,

$CD$ =

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

$CD$ =

$\sqrt{[(8-6)^2+(3-7)^2]}$

$CD$ =

$\sqrt{[(2)^2+(-4)^2]}$

$CD$ =

$\sqrt{(4+16)}$

$CD$ =

$\sqrt{(20)}$

$CD$ =

$\sqrt{(4×5)}$

$CD$ =

$2\sqrt{5}$
By distance formula,

$AD$ =

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

$AD$ =

$\sqrt{[(8-0)^2+(3-(-1))^2]}$

$AD$ =

$\sqrt{[(8)^2+(4)^2]}$

$AD$ =

$\sqrt{(64+16)}$

$AD$ =

$\sqrt{(80)}$

$AD$ =

$\sqrt{(5×16)}$

$AD$ =

$4\sqrt{5}$
Here

$AB = CD$ and

$BC = AD$ .
Hence these are the vertices of a rectangle.
Area of

$□ABCD = AB×BC$
=

$2√5×4√5$
=

$40$ sq. units.
Hence the area of

$□ABCD$ is

$40$ sq. units.

Examine whether the points $(1, -1), (-5, 7)$ and $(2, 6)$ are equidistant from the point $(-2, 3)$ ?

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

1. Three vertices of a rectangle are $A (2, -1), B (2, 7)$ and $C(4, 7)$ . Plot these points on a graph and hence use it to find the co-ordinates of the fourth vertex $D$
Also find the co-ordinates of
(i) the mid-point of $BC$
What is the area of the rectangle?

Given three vertices of a rectangle are

$A (2, -1), B (2, 7) and C(4, 7).$
These points are marked on the graph shown below.
Join the points to form rectangle

$ABCD.$
Also join the diagonals

$AC$ and

$BD$
The coordinates of fourth vertex

$D$ is

$(4,-1)$
(i) The midpoint of

$BC$ is

$(3,7).$
Area of rectangle

$=AB\times BC$

$=8\times 2$

$16sq.units$
1727425_1812056_ans_616ee5c543c34f3b8aba24e3ad5041d2.PNG

Find the point on x-axis which is equidistant from the following pair of points:
$(3, 2)$ and $(-5, -2)$

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

1. Three vertices of a rectangle are $A (2, -1), B (2, 7)$ and $C(4, 7)$ . Plot these points on a graph and hence use it to find the co-ordinates of the fourth vertex $D$
Also find the co-ordinates of
(iii) the point of intersection of the diagonals.
What is the area of the rectangle?

Given three vertices of a rectangle are

$A (2, -1), B (2, 7) and C(4, 7).$
These points are marked on the graph shown below.
Join the points to form rectangle

$ABCD.$
Also join the diagonals

$AC$ and

$BD$
The coordinates of fourth vertex

$D$ is

$(4,-1)$
(iii)The point of intersection of diagonals is

$(3,3).$
Area of the rectangle

$ABCD = AB×BC$

$= 8×2$

$= 16 sq.$ units.

1724796_1812064_ans_f5552341a0cc45a0aa64ac5d72e4ae0b.PNG

2. Three vertices of a parallelogram are $A (3, 5), B (3, -1)$ and $C (-1, -3)$ . Plot these points on a graph paper and hence use it to find the coordinates of the fourth vertex $D$ . Also find the coordinates of the mid-point of the side $CD$ . What is the area of the parallelogram?

Given

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

$C (-1, -3)$ are the three vertices of a parallelogram.
These points are marked on the graph shown below.
Join the points to form parallelogram

$ABCD.$
The coordinates of fourth vertex

$D$ is

$(-1,3).$
The coordinates of midpoint of

$CD$ is

$(-1,0)$ .
Area of parallelogram

$ABCD =$ Base ×height

$= AB×EF$

$= 6×4$

$= 24 sq.$ units.
Hence the area of the parallelogram is

$24 sq.$ units.
1723675_1812072_ans_f42997816e5b484e804b1c5d8dc6f31f.PNG

If the point $(x, y)$ on the tangent is equidistant from the points $(2, 3)$ and $(6, -1),$ find the relation between x and y.

If $A(4, -8), B(3, 6)$ and $C(5, -4)$ are the vertices of a $\Delta ABC, D$ is the mid-point of BC and P is a point on AD joined such that $\dfrac{AP}{PD} = 2,$ find the coordinates of P.

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

If the point $(x, y)$ be equidistant from the points $(a + b, b - a)$ and $(a - b, a + b),$ prove that $\dfrac{a - b}{a + b} = \dfrac{x - y}{x + y}$

Find the coordinates of the point which divides the join of $(-1, 7)$ and $(4, -3)$ internally in the ratio $2 : 3.$

Let P(x, y) be the point which divides the line segment in ternally.
Using the section formula for the internal division, i.e.

$(x, y) = \left ( \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \right )$ .....(i)
Here,

$m_1 = 2, m_2 = 3$

$(x_1, y_1) = (-1, 7)$ and

$(x_2, y_2) = (4, -3)$
Putting the above values in the above formula, we get

$\Rightarrow x = \dfrac{2(4) + 3(-1)}{2 + 3}, y = \dfrac{2(-3) + 3(7)}{2 + 3}$

$\Rightarrow x = \dfrac{8 - 3}{5}, y = \dfrac{-6 + 21}{5}$

$\Rightarrow x = \dfrac{5}{5}, y = \dfrac{15}{5}$

$\Rightarrow x = 1, y = 3$
Hence,

$(1, 3)$ is the point which divides the line segment internally.

If p divides the join of $A(-2, -2)$ and $B(2, -4)$ such that $\dfrac{AP}{AB} = \dfrac{3}{7},$ find the coordinates of P.

Given:

$\dfrac{AP}{AB} = \dfrac{3}{7}$

$\Rightarrow AP = \dfrac{3}{7} AB$

$\Rightarrow AP = \dfrac{3}{7} (AP + PB)$

$\Rightarrow 7AP = 3AP + 3PB$

$\Rightarrow 7AP - 3AP = 3PB$

$\Rightarrow 4AP = 3PB$

$\Rightarrow \dfrac{AP}{PB} = \dfrac{3}{4}$
Hence, the point P divides AB in the ratio

$3 : 4$

$\Rightarrow m_1 = 3$ and

$m_2 = 4$ and

$(x_1, y_1) = (-2, -2); (x_2, y_2) = (2, -4)$
Using the section formula for the internal division, i.e.

$(x, y) = \left ( \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2} \right )$ ...(i)

$\Rightarrow (x, y) = \dfrac{3(2) + 4(-2)}{3 + 4}, y = \dfrac{3(-4) + 4(-2)}{3 + 4}$

$\Rightarrow x = \dfrac{6 - 8}{7}, y = \dfrac{-12 - 8}{7}$

$\Rightarrow x = \dfrac{-2}{7}, y = \dfrac{-20}{7}$
Hence, the coordinates of P are

$P(x, y) = P \left ( \dfrac{-2}{7}, \dfrac{-20}{7} \right )$

The coordinates of the middle points of the sides of a triangle are $(1, 1), (2, 3)$ and $(4, 1),$ find the coordinates of its vertices.

Consider a

$\Delta ABC$ with

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

$C(x_3, y_3).$ If

$P(1, 1), Q(2, 3)$ and

$R(4, 1)$ are the midpoints of AB, BC, and CA. Then,

$1 = \dfrac{x_1 + x_2}{2} \Rightarrow x_1 + x_2 = 2$ .......(i)

$1 = \dfrac{y_1 + y_2}{2} \Rightarrow y_1 + y_2 = 2$ ......(ii)

$2 = \dfrac{x_2 + x_3}{2} \Rightarrow x_2 + x_3 = 4$ ....(iii)

$3 = \dfrac{y_2 + y_3}{2} \Rightarrow y_2 + y_3 = 6$ ......(iv)

$4 = \dfrac{x_1 + x_3}{2} \Rightarrow x_1 + x_3 = 8$ .......(v)

$1 = \dfrac{y_1 + y_3}{2} \Rightarrow y_1 + y_3 = 2$ ...(vi)
Adding (i), (iii) and (v) , we get

$x_1 + x_2 + x_2 + x_3 + x_1 + x_3 = 2 + 4 + 8$

$\Rightarrow 2(x_1 + x_2 + x_3) = 14$

$\Rightarrow x_1 + x_2 + x_3 = 7$ ..(vii)
From (i) and (vii), we get

$x_3 = 7 - 2 = 5$
From (iii) and (vii), we get

$x_1 = 7 - 4 = 3$
From (v) and (vii), we get

$x_2 = 7 - 8 = -1$
Now adding (ii), (iv) and (vi), we get

$y_1 + y_2 + y_2 + y_3 + y_1 + y_3 = 2 + 6 + 2$

$\Rightarrow 2(y_1 + y_2 + y_3) = 10$

$\Rightarrow y_1 + y_2 + y_3 = 5$ .......(viii)
From (ii) and (viii), we get

$y_2 = 5 - 2 = 3$
From (iv) and (vii), we get

$y_1 = 5 = 6 = -1$
From (vi) and (vii), we get

$y_2 = 5 - 2 = 3$
Hence, the vertices of

$\Delta ABC$ are

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

$C(5, 3)$

If the points $(10, 5), (8, 4)$ and $(6, 6)$ are the mid-points of the sides of a triangle, find its vertices.

Consider a

$\Delta ABC$ with

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

$C(x_3, y_3).$ If

$P(10, 5), Q(8, 4)$ and

$R(6, 6)$ are the midpoints of AB, BC, and CA. Then,

$10 = \dfrac{x_1 + x_2}{2} \Rightarrow x_1 + x_2 = 20$ ....(i)

$5 = \dfrac{y_1 + y_2}{2} \Rightarrow y_1 + y_2 = 10$ ......(ii)

$8 = \dfrac{x_2 + x_3}{2} \Rightarrow x_2 + x_3 = 16$ ...(iii)

$4 = \dfrac{y_2 + y_3}{2} \Rightarrow y_2 + y_3 = 8$ .........(iv)

$6 = \dfrac{x_1 + x_3}{2} \Rightarrow x_1 + x_3 = 12$ ......(v)

$6 = \dfrac{y_1 + y_3}{2} \Rightarrow y_1 + y_3 = 12$ ......(vi)
Adding (i), (iii) and (v), we get

$x_1 + x_2 + x_2 + x_3 + x_1 + x_3 = 20 + 16 + 12$

$\Rightarrow 2(x_1 + x_2 + x_3) = 48$

$\Rightarrow x_1 + x_2 + x_3 = 24$ .......(vii)
From (i) and (vii), we get

$x_3 = 24 - 20 = 4$
From (iii) and (vii), we get

$x_1 = 24 - 16 = 8$
From (v) and (vii), we get

$x_2 = 24 - 12 = 12$
Now adding (ii), (iv) and (vi), we get

$y_1 + y_2 + y_2 + y_3 + y_1 + y_3 = 10 + 8 + 12$

$\Rightarrow 2(y_1 + y_2 + y_3) = 30$

$\Rightarrow y_1 + y_2 + y_3 = 15$ .......(viii)
From (ii) and (viii), we get

$y_1 = 15 - 8 = 7$
From (vi) and (vii), we get

$y_2 = 15 - 12 = 3$
Hence, the vertices of

$\Delta ABC$ are

$A(8, 7), B(12, 3)$ and

$C(4, 5).$

Find the coordinates of a point A, where AB is the diameter of a circle whose center is $(2, -3)$ , and B is $(1, 4)$

A quadrilateral has the vertices at the point $(-4, 2), (2, 6), (8, 5)$ and $(9, -7).$ Show that the mid-point of the sides of this quadrilateral are the vertices of a parallelogram.

1961879_1816213_ans_a6900ad3a1264440b9484f0463a00025.jpg

One end of a diameter of a circle is at $(2, 3)$ and the center is $(-2, 5),$ find the coordinates of the other end of the diameter.

Find the area of the quadrilateral whose vertices are
Given $(0, 0), (6, 0), (4, 3)$ and $(0, 3)$

To Find: Find the area of quadrilateral.

Let us Assume A(x1,y1)=(0,0)

Let us Assume B(x2,y2)=(6,0)

Let us Assume C(x3,y3)=(4,3)

Let us Assume D(x4,y4)=(0,3)

Let us join Ac to from two triangles ΔABC and ΔACD

Now Area of triangle

=12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]

Then, Area of given triangle ABC

=12|0(0−3)+6(3−0)+4(0−0)|

=12|0+18+0|

=12|18|

=9 Square units

Area of given triangle ABC

=12|0(3−3)+4(3−0)+0(0−3)|

=12|0+12+0|

=12|12|

=6 square units

Area of quadrilateral ABCD = Area of ABC + Area of ACD

=9+6

Hence, Area =15 sq units

Find the area of the quadrilateral whose vertices are
Given $(1, 0), (5, 3), (2, 7)$ and $(-2, 4)$

Find the area of the quadrilateral whose vertices are $(-4, 5), (0, 7), (5, -5)$ and $(-4, -2)$ .

Given:-

$\;(-4, 5), (0, 7), (5, -5)$ and

$(-4, 2)$ are the vertices of quadrilateral.

To Find:-

Find the area of the quadrilateral.

Let us Assume,

$A(x_1, y_1) = (-4, 5)$

$B(x_2, y_2) = (0, 7)$

$C(x_3, y_3) = (5, -5)$

$D(x_4, y_4) = (4, -2)$

Let us join

$AC$ to form two triangles

$\triangle ABC$ and

$\triangle ACD$ .

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

Then, area of the triangle

$ABC$ ,

$ar(\triangle ABC) = \dfrac{1}{2} | -4 (7 + 5) + 0(-5 - 5) + 5(5 - 7) |$

$= \dfrac{1}{2} [-48 + 0 - 10]$

$= \dfrac{1}{2} [-58]$

$= \dfrac{58}{2}\text{ sq. units}$ [Negative sign will be ignored as area cannot be negative]

Area of the triangle

$ACD$ ,

$ar(\triangle ACD)= \dfrac{1}{2} [ -4 (-5 + 2) + 5(-2 - 5) -4(5 + 5)]$

$= \dfrac{1}{2} [ -4(-3) + 5(-7) - 4(10)]$

$= \dfrac{1}{2} [ 12 - 35 - 40 ]$

$= \dfrac{1}{2} [ -63]$ [Negative sign will be ignored as area cannot be negative]

$= \dfrac{63}{2}\text{ sq. units}$

$ar(ABCD) \;= ar(ABC)+ ar(ACD)$

$= \dfrac{58}{2} + \dfrac{63}{2}$

Hence, area of quadrilateral

$ABCD$ is equal to

$\dfrac{121}{2}\text{ sq. units}.$

1795245_1816238_ans_baf9383a25d14ae4af25ad4b343334bf.png

Find the area of the quadrilateral whose vertices are
$(1, 1), (7, -3), (12, 2)$ and $(7, 21)$

Find the area of the quadrilateral whose vertices are
Given $(-5, 7), (-4, -5), (-1, -6)$ and $(4, 5)$

The line joining the points $A (-3 , -10)$ and $B(-2 , 6)$ is divided by the point P such that $\dfrac{PB}{AB} = \dfrac{1}{5}$ . Find the co-ordinates of $P$ .

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

Let the coordinates of point P be taken as

$(x , y)$ .
Given

$PB : AB = 1 : 5$
So,

$PB : PA = 1 : 4$
Hence, the coordinates of

$P$ are

$(x , y) = \left ( \dfrac{4 \times (-2) + 1 \times (-3)}{5} , \dfrac{4 \times 6 + 1 \times (-10)}{5} \right ) = \left ( \dfrac{-11}{5} , \dfrac{14}{5} \right )$ .

Find the mid-point of the line segment joining the points :
$(-6 , 7)$ and $(3 , 5)$

$\text{The mid point of the line joining the points}$

$(\text{x}_1,\text{y}_1) \text{ and}$

$(\text{x}_2, \text{y}_2)$

$\text{is given by}$

$\left(\dfrac{\text{x}_1+\text{x}_2}{2}, \dfrac{\text{y}{_1}+\text{y}_2}{2}\right)$

$\text{So, for the given points:}$

$(5 , -3)$

$\text{and B} (-1 , 7)$

$\text{Hence, The coordinates of the mid-point are:}$

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

Points A and B have co-ordinates $(3, 5)$ and $(x, y)$ respectively. The mid-point of AB is $(2, 3)$ . Find the values of $x$ and $y$ .

Given mid-point of AB

$= (2 , 3)$
Thus,

$\dfrac{3 + x}{2} , \dfrac{5 + y}{2} = (2 , 3)$

$3 + \dfrac{x}{2} = 2$ and

$5 + \dfrac{y}{2} = 3$

$3 + x = 4$ and

$5 + y = 6$

$x = 1$ and

$y = 1$

If the area of the quadrilateral whose angular points taken in order are $(1, 2), (-5, 6), (7, -4)$ and $(h, -2)$ be zero, show that $h = 3.$

Given: vertices of the quadrilateral be

$A(1, 2), B(-5, 6), C(7, -4)$ and

$D(h, -2).$
Let join AC to form two triangles,
Now, We know that
Area of triangle

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

$= \dfrac{1}{2}[ 1(6 + 4) - 5(-4 - 2) + 7(2 - 6)]$

$= \dfrac{1}{2}[10 + 30 - 28]$

$= \dfrac{12}{2}$

$= 6$ sq units
Now, Area of triangle ADC

$= \dfrac{1}{2}[1 (-2 + 4) + h(-4 - 2) + 7(2 + 2)]$

$= \dfrac{1}{2} [2 - 6h + 28]$

$= \dfrac{12}{2} [-6h + 30]$

$= 3h - 15$
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC

$= 3h - 15 + 6$

$= 3h = 9$

$= h = 3$
Hence, h is

$3.$

Calculate the co-ordinates of the point P which divides the line segment joining the points $A(-4 , 6)$ and $B(3 , -5)$ is the ratio $3 : 2$ .

Let's assume the co-ordinates of the point P be

$(x , y)$
Then by section formula , we have
P

$(x , y) = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2} , \dfrac{m_1 y_2 + m_2y_1}{m_1 + m_2}$

$x = \dfrac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \dfrac{3 \times 3 + 2 \times (-4)}{3 + 2} = \dfrac{1}{5}$

$y = \dfrac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \dfrac{3 \times (-5) + 2 \times 6}{3 + 2} = \dfrac{-3}{5}$ .

Hence, the co-ordinates of point P are

$\dfrac{1}{5} , \dfrac{-3}{5}$ .

P is a point on the line joining $A (4 , 3)$ and $B ( -2 , 6)$ such that $5AP = 2BP$ . Find the co-ordinates of $P$ .

Using the section formula, if a point

$(x,y)$ divides the line joining the points

$({ x }_{ 1 },{ y }_{ 1 })$ and

$({ x }_{ 2 },{ y }_{ 2 })$ in the ratio

$m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

$5 AP = 2BP$
So,

$\dfrac{AP}{BP} = \dfrac{2}{5}$
Hence, the co-ordinates of the point

$P$ are

$\dfrac{2 \times (-2) + 5 \times 4}{2 + 5} , \dfrac{2 \times 6 + 5 \times 3}{2 + 5}$

$\dfrac{16}{7} , \dfrac{27}{7}$

Calculate the co-ordinates of the point P which divides the line segment joining thr points $A(1 , 3)$ and $B(5 , 9)$ is the ratio $1 : 2$ .

Let's assume the co-ordinates of the point P be

$(x , y)$

Given

$A(1,3)$ and

$B(5,9)$
Then by section formula , we have
P

$(x , y) = \dfrac{m_1 x_2 + m_2x_1}{m_1 + m_2} , \dfrac{m_1 y_2 + m_2 y_1}{m_1 + m_2}$

$x = \dfrac{m_1 x_2 + m_2x_1}{m_1 + m_2} = \dfrac{1 \times 5 + 2 \times 1}{1 + 2} = \dfrac{7}{3}$

$y = \dfrac{m_1y_2 + m_2 y_1}{m_1 + m_2} = \dfrac{1 \times 9 + 2 \times 3}{1 + 2} = \dfrac{15}{3} = 5$

Hence, the co-ordinates of point P are

$(\dfrac{7}{3} , 5)$ .

Find the mid-point of the line segment joining the points :
$(5 , -3)$ and $(-1 , 7)$

Let

$A (5 , -3)$ and

$B (-1 , 7)$
So the -point of

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

Find the coordinates of midpoint of the segment joining the points (22, 20) and (0, 16).

Let the given points be A (22, 20) and B(0, 16).
Let the coordinate of the mid point be (x, y).
Using the midpoint formula

$x=\dfrac{22+0}{2}$

$\Rightarrow 2x=22$

$\Rightarrow x=11$

$y=\dfrac{20+16}{2}$

$\Rightarrow 2y=36$

$\Rightarrow y=18$

$(x,y)=(11,18)$

Determine whether the point is collinear.
$A(1,-3),B(2,-5),C(-4,7)$

$A(1,-3),B(2,-5),C(-4,7)$

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

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

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

$\sqrt{36+144}$

$\sqrt{180}$

$=6\sqrt{5}$

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

$=\sqrt{25+100}$

$=\sqrt{125}$

$=5\sqrt{5}$

$AB+AC=BC$

$\sqrt{5}+5\sqrt{5}=6\sqrt{5}$
Thus,the given points lie on the same line.Hence,they are collinear.

Determine whether the point is collinear.

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

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

$=\sqrt{9+1}=\sqrt{10}$

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

$=\sqrt{9+1}$

$=\sqrt{10}$

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

$=\sqrt{36+4}$

$=\sqrt{40}$

$=2\sqrt{10}$

$PQ+QR=PR$
So, the given points lie on the same line. Hence, the given points are collinear.

Determine whether the point is collinear.
$R(0, 3), D(2, 1), S(3, -1)$

$R(0, 3), D(2, 1), S(3, -1)$

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

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

$=2\sqrt{2}$

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

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

$=\sqrt{5}$

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

$=\sqrt{9+16}$

$=\sqrt{25}$

$=5$
Sum of two sides is not equal to the third side.
Hence, the given points are not collinear.

Point P is the centre of the circle and AB is a diameter . Find the coordinates of point B if coordinates of point A and P are (2,-3) and (-2, 0) respectively.

Centre

$= P(-2, 0)$

Diameter has

$A(2, -3)$ on one end and

$B(x, y)$ on the other.

The centre point

$P$ divides

$AB$ in two equal parts.

So, using the midpoint formula,

$-2=\dfrac{2+x}{2}$

$\Rightarrow -4=2+x$

$\Rightarrow x=-6$

And also,

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

$\Rightarrow 0=-3+y$

$\Rightarrow y=3$

$(x,y)=(-6,3)$

1805710_1852899_ans_5e8caace9eb142a0b7cca10b0bc495e6.png

Find the co-ordinate of point $A$ which divides segment $PQ$ in the ratio $a : b,$ where $P(2, 6), Q(-4, 1), a : b = 1 : 2$ .

Using the section formula, if a point $(x,y)$ divides the line joining the points $({ x }_{ 1 },{ y }_{ 1 })$ and $({ x }_{ 2 },{ y }_{ 2 })$ in the ratio $m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

Let the coordinates of point

$A$ be

$(x, y).$
Since,

$P(2, 6), Q(-4, 1), a : b = 1 : 2$
Using section formula -

$x=\dfrac{1\times (-4)+2\times 2}{1+2}=\dfrac{-4+4}{3}=0$

$y=\dfrac{1 \times 1+2\times 6}{1+2}=\dfrac{1+12}{3}=\dfrac{13}{3}$

$(x,y)=(0,\dfrac{13}{3})$

Find the point on the X-axis which is equidistant from A(-3, 4) and B(1, -4).

Let the point on the x-axis be P(a, 0).

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

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

$PB=\sqrt{(a-1)^2+(0-(-4))^2}$

$=\sqrt{(a-1)^2+16}$

$PA^2=PB^2$

$\Rightarrow (a+3)^2+16=(a-1)^2+16$

$\Rightarrow a^2+6a+9=a^2-2a+1$

$\Rightarrow 8a=-8$

$\Rightarrow a=-1$

$(a,0)=(-1,0)$

Determine whether the following point is collinear. A(-1, -1), B(0, 1), C(1, 3)

A(-1, -1), B(0, 1), C(1, 3)
Slope of AB =

$\dfrac{1-(-1)}{0-(-1)}=\dfrac{2}{1}=2$
Slope of BC =

$\dfrac{3-1}{1-0}=\dfrac{2}{1}=2$
Slope of AB =Slope of BC=2
Thus, the given points are collinear.

Point P is the midpoint of seg CD. If $CP = 2.5$ , find $l(CD)$ .

We have $l(CP) = 2.5.$

Since, P is the midpoint of seg CD, then

∴ $l(CD) = 2 \times l(CP) = 2 \times 2.5 = 5$

So, the length of CD is $5$ .

1817525_1855464_ans_564d875efce6454ba5e28942de29c5c5.png

Determine whether the given point is collinear.
$P(1,2),Q\left(2,\dfrac{8}{5}\right),R\left(3,\dfrac{6}{5}\right)$

Given points are

$P(1,2),Q\left(2,\dfrac{8}{5}\right),R\left(3,\dfrac{6}{5}\right)$

Slope of PQ =

$\dfrac{\dfrac{8}{5}-2}{2-1}=\dfrac{\dfrac{-2}{5}}{1}=\dfrac{-2}{5}$

Slope of QR =

$\dfrac{\dfrac{6}{5}-\dfrac{8}{5}}{3-2}=\dfrac{\dfrac{-2}{5}}{1}=\dfrac{-2}{5}$

So, the slope of PQ = slope of QR.

Point Q lies on both the lines.

Hence, the given points are collinear.

Determine whether the following point is collinear. $A(-4,4),K(-2,\dfrac{5}{2}),N(4,-2)$

$A(-4,4),K(-2,\dfrac{5}{2}),N(4,-2)$
Slope of AK =

$\dfrac{\dfrac{5}{2}-4}{-2-(-4)}=\dfrac{\dfrac{-3}{2}}{2}=\dfrac{-3}{4}$
Slope of KN =

$\dfrac{-2-\dfrac{5}{2}}{4-(-2)}=\dfrac{-3}{4}$
Slope of AK = Slope of KN.
Thus, the given points are collinear.

Determine whether the following point is collinear. P(2,-5),Q(1,-3),R(-2,3)

P(2,-5),Q(1,-3),R(-2,3)
Slope of PQ =

$\dfrac{-3-(-5)}{1-2}=\dfrac{2}{-1}=-2$
Slope of QR =

$\dfrac{3-(-3)}{-2-1}=\dfrac{6}{-3}=-2$
Slope of PQ = Slope of QR.
Thus, the given points are collinear.

Find the coordinates of the midpoint of the line segment joining $P(0,6)$ and $Q(12,20).$

The given points are

$P(0,6)$ and

$Q(12,20).$

Midpoint of

$PQ = \left(\dfrac{0+12}{2},\dfrac{6+20}{2}\right) = (6, 13)$ .

Determine whether the given point is collinear.
$L(1,2),M(5,3), N(8,6)$

Given points are

$L(1,2),M(5,3), N(8,6)$

Slope of

$LM=$

$\dfrac{3-2}{5-1}=\dfrac{1}{4}$

Slope of

$MN=$

$\dfrac{6-3}{8-5}=\dfrac{3}{3}=1$

Thus, the slope of

$LM$ not equal to slope

$MN$

So, the given points are not collinear.

Determine whether the following point is collinear. R(1,-4),S(-2,2),T(-3,4)

R(1,-4),S(-2,2),T(-3,4)
Slope of RS =

$\dfrac{2-(-4)}{-2-1}=\dfrac{6}{-3}=-2$
Slope of ST =

$\dfrac{4-2}{-3-(-2)}=\dfrac{2}{-1}=-2$
Slope of RS = Slope of ST.
Thus, the given points are collinear.

Find the value of x for which the points ( x , -1 ) , (2,1) and (4,5) are collinear

1975254_1865483_ans_01d25a86b69747f1b4be2d83213d2fae.jpg

If A and B are $(-2 , -2)$ and $(2 , -4)$ respectively, find the coordinates of P such that $AP = \dfrac{3}{7}$ AB and P lies on the line segment AB.

Let coordinate of

$P$ are

$(x , y)$ .

$AP = \dfrac{3}{7} AB$

$AP : AB = 3 : 7$
As per section formula ,

$AP : AB = 3 : 7$

$\therefore AP + PB = AB$

$3 + PB = 7$

$\therefore PB = 7 - 3 = 4$

$\therefore$ Let

$AP : PB = 3 : 4 = m_{1} : m_{2}$

Coordinates of

$P$ are

$P(x , y) = \left ( \dfrac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}} , \dfrac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} \right )$

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

$= \left (\dfrac{6 - 8}{7} , \dfrac{-12 - 8}{7} \right )$

$P(x , y) = \left ( - \dfrac{2}{7} , - \dfrac{20}{7} \right )$

$\therefore \, x = - \dfrac{2}{7} , y = - \dfrac{20}{7}$

$\therefore$ Coordinates of P are

$\left (- \dfrac{2}{7} , - \dfrac{20}{7} \right )$

In each of the following find the value of 'k', for which the points are collinear.
$(8, 1) , (k , -4) , (2 , 5)$

$Let$

$A (8 , 1) = (x_1 , y_1)$

$B (k , -4) = (x_{2} , y_{2})$

$C(2, -5) = (x_{3} , y_{3})$

$Area\space of\space Triangle$

$ABC = 0$

$\therefore ABC$

$is\space a\space straight\space line.$

$= \dfrac{1}{2} [x_1 (y_2 - y_3) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$= \dfrac{1}{2}[8 (-4 + 5) + k(-5 - 1) + 2 (1 + 4)] = 0$

$= \dfrac{1}{2} [8(1) + k(-6) + 2(5)] = 0$

$= \dfrac{1}{2}[8 - 6k + 10] = 0$

$\dfrac{1}{2}(18 - 6k) = 0$

$9 - 3k = 0$

$-3k = -9$

$3k = 9$

$\therefore k = \dfrac{9}{3} = 3$

Find a relation between 'x' and 'y' such that the point (x , y) is equidistant from the point $(3 , 6)$ and $(-3 , 4)$ .

$A(3 , 6) = A (x_{1} , y_{1})$

$Given\space AP= PB$ .

$Let \space P(x , y)$

$Let$

$B (-3 , 4) = B(x_{2} , y_{2})$ .

$(i)$

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

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

$= \sqrt{x^{2} - 6x + 9 + y^{2} - 12y + 36}$

$AP = \sqrt{x^{2} - 6x + y^{2} - 12y + 45}$

$(ii)$

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

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

$= \sqrt{9 + 6x + x^{2} + 16 - 8y + y^{2}}$

$PB = \sqrt{x^{2} + y^{2} + 6x - 8y + 25}$

$Here,$

$AP = PB$ .

$\therefore \, AP^{2} = PB^{2}$ .

$x^{2} - 6x + y^{2} - 12y + 45 = x^{2} + y^{2} + 6x - 8y + 25$

$-6x - 6x - 12y + 8y = 25 - 45$

$-12x - 4y = -20$

$12x + 4y = 20$

$3x + y = 5$

$The \space relation \space is \space 3x+y=5$

Find the coordinates of the point which divides the join of $(-1 , 7)$ and $(4 , -3)$ in the ratio $2 : 3$ .

$Let$

$P(x , y)$

$is\space the\space required\space point.$

$P(x , y)$

$divides\space the \space line \space segment \space joining \space the\space points\space$

$A(-1 , 7)$

$B(4 , -3)$

$in\space the\space ratio$

$m_{1} : m_{2} = 2 : 3$

$As \space per\space Section\space Formula\space$ ,

$coordinates\space of$

$P(x , y)$

$are$

$\left ( \dfrac{m_{1} x_{2} + m_{2} x_{1}}{m_{1} + m_{2}} , \dfrac{m_{1} y_{2} + m_{2} y_{1}}{m_{1} + m_{2}} \right )$

$= \left ( \dfrac{2 \times 4 + 3 \times -1}{2 + 3} , \dfrac{2 \times -3 + 3 \times 7}{2 + 3} \right )$

$= \left ( \dfrac{8 - 3}{5} , \dfrac{-6 + 21}{5} \right )$

$= \left ( 1 , \dfrac{15}{5} \right )$

$= (1 , 3)$

$\therefore \, P(x , y) = (1 , 3)$ .

If $Q(0 , 1)$ is equidistant from $P(5 , -3)$ and $R(x , 6)$ , find the value of $x$ . Also find the distances QR and PR.

Point

$Q$ is equidistant from

$P$ and

$R$ , then

$PQ = QR , x =$ ?
(i)

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

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

$= \sqrt{-5^{2} + 4^{2}}$

$= \sqrt{25 + 16}$

$PQ = \sqrt{41}$

(ii)

$QR = \sqrt{(x_{2} - x_{2})^{2} + (y_{3} - y_{2})^{2}}$

$= \sqrt{(x - 0)^{2} + (6 - 1)^{2}}$

$= \sqrt{x^{2} + 5^{2}}$

$QR = \sqrt{x^2 + 25}$

But

$PQ = QR$

$\therefore \, PQ^{2} = QR^{2}$

$41 = x^{2} + 25$

$x^{2} = 41 - 25$

$x^{2} = 16$

$\therefore \, x = 4$ units.

(iii) If

$PR = \sqrt{(x_{3} - x_{1})^{2} + (y_{3} - y_{1})^{2}} , x = 4$ then,

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

$= \sqrt{1 + 81}$

$PR = \sqrt{82}$

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

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

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

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

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

$= \sqrt{4 + 4x + x^{2} + 81 } = \sqrt{x^{2} + 4x + 85}$

We have

$PA = PB$

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

$x^{2} - 4x + 29 = x^{2} + 4x + 85$

$-4x - 4x = 85 - 29$

$-8x = 56$

$\therefore \, x = \dfrac{-56}{8}$

$\therefore \, x = -7$

$\therefore$ Coordinates of P are

$(x , 0)$ , it means

$(-7 , 0)$ .

$\therefore$ Required point

$(x , 0) = (-7 , 0)$

$\therefore \, x = -7$

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

Let

$A (7 , -2) = (x_1 , y_1)$

$B (5 , 1 ) = (x_{2} , y_{2})$

$C(3 , k ) = (x_{3} , y_{3})$

$Points\space are\space collinear\space , therefore\space area\space of\space triangle\space formed\space by\space these\space is\space zero\space (0)$ .

$= \dfrac{1}{2} [x_1 (y_2 - y_3) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

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

$= \dfrac{1}{2} [ 7 - 7k + 5k + 10 - 9] = 0$

$= \dfrac{1}{2} [ -2k + 8] = 0$

$-k + 4 = 0$

$-k = -4$

$\therefore k = 4$

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.

Given :

$A \rightarrow (-1 , 1) , B \rightarrow (-1 , 4) , C \rightarrow (5 , 4)$ &

$D \rightarrow (5 , -1)$ Co-ordinates of midpoint

$= \left ( \dfrac{x_{1} + x_{2}}{2} , \dfrac{y_{1} + y_{2}}{2} \right )$

Co-ordinates of P are

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

Co-ordinates of Q

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

Co-ordinates of R

$= \left ( \dfrac{5 + 5}{2} , \dfrac{-1 + 4}{2} \right ) = \left ( 5 , \dfrac{3}{2} \right )$

Co-ordinates of S

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

Distance formula

$= \sqrt{(x_{2} - x_{1})^{2} + (y_2 - y_1)^{2}}$

$PQ = \sqrt{(2 + 1)^{2} + \left (4 - \dfrac{3}{2} \right )^2} = \sqrt{9 + \dfrac{25}{4}}$

$= \sqrt{\dfrac{36 + 25}{4}} = \dfrac{\sqrt{61}}{2}$

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

$= \sqrt{9 + \dfrac{25}{4}} = \sqrt{\dfrac{36 + 25}{4}} = \dfrac{\sqrt{61}}{2}$

$RS = \sqrt{(2 - 5)^{2} + \left ( -1 - \dfrac{3}{2} \right )^{2}} = \sqrt{9 + \dfrac{25}{4}}$

$= \sqrt{\dfrac{36 + 25}{4}} = \sqrt{\dfrac{61}{4}} = \dfrac{\sqrt{61}}{2}$

$SP = \sqrt{(2 + 1)^{2} + \left (-1 - \dfrac{3}{2} \right )^{2}} = \sqrt{\dfrac{9}{1} + \dfrac{25}{4}}$

$= \sqrt{\dfrac{36 + 25}{4}} = \sqrt{\dfrac{61}{4}} = \dfrac{\sqrt{61}}{2}$

$\therefore \, PQ = QR = RS = SP$

$\therefore PQRS$ is either square (or) Rhombus
Diagonal

$PR$

$PR = \sqrt{(5 + 1)^{2} + \left ( \dfrac{3}{2} - \dfrac{3}{2} \right )^{2}} = \sqrt{(6)^2} = 6$

and Diagonal

$QS$

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

$PR \neq QS$

$\therefore$ Diagonal are not equal
Therefore ,

$PQRS$ is a rhombus
1844477_1868550_ans_7af67c250c4149b99fc68e466edd1f3b.png

You have studied Class IX, (Chapter 9 , Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for $\Delta ABC$ whose vertices are $A(4 , -6) , B(3 , -2)$ and $C((5 , 2)$ .

AD median is drawn to side BC which is the vertex of

$\triangle ABC$ .
Median

$AD$ divides

$\triangle ABC$ into two triangles

$\triangle ABD$ and

$\triangle ADC$ which are equal in area.
Now Median

$AD$ bisects

$BC$ .

$\therefore BD = DC$ .

(i) As per Mid-Points formula,
Coordinates of D are

$= \sqrt{(4)^{2} , (0)^2}$

$= \sqrt{16 , 0}$

$= (4 , 0)$

$\therefore$ Coordinate of D are

$(4 , 0)$

(ii) Now, Area of

$\triangle ABD$ :

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

$= \dfrac{1}{2} [4 (-2 - 0) + 3(0 + 6) + 4(-6 + 2)]$

$= \dfrac{1}{2} [4 (-2) + 3 (6) + 4(-4)]$

$= \dfrac{1}{2} [-8 + 18 - 16]$

Area of

$\triangle ABD = \dfrac{1}{2} (-6)$

$= -3$ sq. units

(iii) Now, Area of

$ADC$ :

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

$= \dfrac{1}{2} [4 (0 - 2) + 4(2 + 6) + 5(-6 + 0)]$

$= \dfrac{1}{2}[ 4(-2) + 4(8) + 5(-6)]$

$= \dfrac{1}{2}[-8 + 32 - 30]$

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

$= -3$ sq. units

$\therefore$ Area of

$\triangle ADC = -3$ sq. units.

$\therefore$ Area of

$\triangle ABC$ :

Area of

$\triangle ABD + Area of \triangle ADC$

$= (-3) + (-3)$

$= -6$ sq. units.

Since area can not be negetive

$=6$ sq. unit

(iv) Now, Area of

$\triangle ABC$ : (Direct Method)

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

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

$= \dfrac{1}{2}[4 (-4) + 3(8) + 5(-4)]$

$= \dfrac{1}{2}[-16 + 24 - 20]$

$= \dfrac{1}{2} [-36 + 24]$

$= \dfrac{1}{2} [-12]$

$= -6$ sq. units.

Since area can not be negetive

$=6$ sq. unit

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

Let

$A \rightarrow (6 , -6) , B \rightarrow (3 , -7)$ and

$C \rightarrow (3 , 3)$

$IA = IB = IC$ [Radii of circle]

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

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

Squaring

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

$x^{2} - 12x + 36 + y^{2} + 12y + 36$

$= x^{2} - 6x + 9 + y^{2} + 14y + 49$

$x^{2} - x^{2} + y^{2} - y^{2} - 12x + 6x + 12y - 14y + 72 - 58 = 0$

$-6x - 2y = -14$

divide by

$-2$

$3x + y = 7 \rightarrow (1)$

$(x - 3)^{2} + (y + 7)^{2} = (x - 3)^{2} + (y - 3)^{2}$

$y^{2} + 49 + 14y = y^{2} + 9 - 6y$

$y^{2} - y^{2} + 14y + 6y = 9 - 49$

$20 y = -40$

$y = -2$
Putting

$y = -2$ in eqn (1)

$3x + y = 7$

$3x - 2 = 7$

$3x = 7 + 2$

$3x = 9$

$x = \dfrac{9}{3} = 3$

$x = 3$
Thus

$I(x , y) \rightarrow (3 , -2)$
Hence, the center of a circle is

$(3 , -2)$

1844457_1868549_ans_420f8df68b0e4c99bf566fc0fb942a5b.png

Find the area of the quadrilateral whose vertices , taken in order are $(-4 , -2) , (-3 , -5) , (3, -2)$ and $(2 , 3)$

Area of quadrilateral ABCD = ?
In the quadrilateral ABCD , diagonal AC is drawn which divides

$\triangle ABC , \triangle ACD$ , Sum of these triangles is equal to the Area of quadrilateral.
(i) Now , Area of

$\triangle ABC$ :

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

$= \dfrac{1}{2} [-4 (-5 + 2) - 3(-2 + 2) + 3 (-2 + 5)]$

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

$= \dfrac{1}{2} [+12 + 0 + 9]$

$= \dfrac{1}{2} \times 21$

$= 10.5$ sq. units

(ii) Area of

$\triangle ACD$

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

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

$= \dfrac{1}{2} [-4 (-5) + 3(5) + 2(0)]$

$= \dfrac{1}{2} [20 + 15 + 0]$

$= \dfrac{1}{2} \times 35$

$= 17.5$ sq. units

$\therefore$ Area of quadrilateral:
= Area of

$\triangle ABC$ + Area of

$\triangle ACD = 10.5 + 17.5 = 28$ sq. units.

$\therefore$ Area of quadrilateral

$ABCD = 28$ sq. units.

Write the coordinates of intersection points of $x=\pm 4$ and $y \pm 3$ and find the area of rectangle so formed .

Given,

$x=\pm 4$
and

$y=\pm 3$
Equation of line

$AB$ is

$y=+3$
Equation of line

$AD$ is

$y=+4$
Then coordinates of point

$A$ of intersection of lines

$AB$ and

$AD=(4,3)$
For the coordinates of point

$B$
Equation of line

$AB \Rightarrow =+3$
Equation of line

$BC \Rightarrow =-4$
Coordinates of point

$B$ of their intersection

$=(-4,3)$
For the coordinates of point

$C$ ,
Equation of line

$BC \Rightarrow x=-4$
Equation of line

$CD \Rightarrow y=-3$
Coordinates of point

$C$ of their intersection

$=(-4,-3)$
For the coordinates of point

$D$ , equation of line

$CD \Rightarrow y=-3$
Equation of line

$AD \Rightarrow x=+4$
Coordinates of point

$C$ of their intersection

$=(4,-3)$
Thus coordinates of point

$B,C$ and

$D$ are respectively

$(4,3),(-4,3),(-4,-3)$ and

$(4,-3)$
Then,
length of

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

$=\sqrt{(-8)^2+0^2}=8$
and length of

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

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

$=6$
Thus area of rectangle

$ABCD$

$=length \times breadth$

$=8 \times 6$

$=48\ sq. units$
1954291_1874311_ans_bccc81a9312f4f6ba590c633454cc2ba.png

Prove that $(x_1,y_1)$ will be coordinates of mid point of section cut by line.
$\dfrac{x}{2x_1}+\dfrac{y}{2y_1}=1$

Let straight line cuts

$x$ and

$y$ axis of point

$B$ and

$A$ coordinates of points

$A$ and

$B$ are

$(0,2y_1)$
and

$(2x_1,0)$ then their mid point bill be

$\left(\dfrac{0+2x_1}{2},\dfrac{0+2y_1}{2}\right)$ or

$(x_1,y_1)$ .
1845124_1874600_ans_8dfcd139184b4c0dbaaeb05c18d6d7b6.png

If points (3,K) and (K,5) are equidistant from point (0,2), then find the value of K.

Let,

$P(0,2)$ and

$A(3,K)$ and

$B(K,5)$
According to the question point P(0,2) is equal distance from A(3,K) and B(K,5)
Now,

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

$=\sqrt {9+K^2+4-4K}$

$=\sqrt {K^2-4k+13}$

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

$=\sqrt{k^2+9}$

$\because PA=PB$

$\sqrt{k^2+9}=\sqrt {K^2-4k+13}$
squaring on both the sides,

$k^2+9=K^2-4k+13$

$4K=13-9$

$4K=4$

$K=1$

Prove that the points (1,1), (-2,7) and (3,-3) are collinear.

Let the given points A(1,1), B(-2,7) and C(3,-3) are lie in a line.

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

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

$=\sqrt{9+36}$

$=\sqrt{45} = 3\sqrt 5$

$BC=\sqrt{(3+2)^2+(-3-7)^2}$

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

$=\sqrt{25+100}$

$=\sqrt{125} = 5\sqrt 5$

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

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

$=\sqrt{4+16}$

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

$\therefore AB+CA = 3\sqrt 5+ 2 \sqrt 5 = 5 \sqrt 5 = BC$

$\therefore BC = AB+CA$
Hence, A, B and C are collinear.

Find the length of that part of line $x \sin a+y \cos a= \sin 2 \alpha$ which cuts axis at mid point. Also, find the coordinate of mid point of this part.

$x \sin a+y \cos a= \sin 2 \alpha$

$\Rightarrow x \sin a+y \cos a= 2\sin \alpha \cos \alpha$

$\Rightarrow \dfrac{x \sin \alpha}{2 \sin \alpha \cos \alpha}+\dfrac{x \cos \alpha}{2 \sin \alpha \cos \alpha}=1$

$\Rightarrow \dfrac{x}{2 \cos \alpha}+\dfrac{y}{2 \sin \alpha}=1$
Thus intercept of

$x-$ axis

$=2$

$\cos \alpha$ and intercept at

$y-axis=2 \sin \alpha$ coordinated of point

$A$ and

$B$ are

$(2 \cos \alpha, 0)$ and

$(0,2 \sin \alpha)$

$\therefore AB=\sqrt{(0-2 \cos \alpha)^2+(2 \sin \alpha-0)^2}$

$=\sqrt{4\cos^2 \alpha+4\sin^2 \alpha}$

$=\sqrt{4(\cos^2 \alpha+\sin^2 \alpha)}$

$(\because cos^2 \alpha+\sin^2 \alpha=1)$

$=\sqrt 4=2$
and coordinates of mid point of

$AB$ are

$= \left( \dfrac{2 \cos \alpha +0}{2}, \dfrac{0+2 \sin \alpha}{2}\right)$ or

$(\cos \alpha, \sin \alpha)$
Thus, length of middle part cut of axis by given line is

$2$ unit and coordinates of mid point are

$(\cos \alpha, sin \alpha)$
1847364_1874332_ans_6be04e34a5724bdea387e19771c14c9c.png

Prove that the points (1,-2),(3,9),(1,2) and (-1,0) are the vertices of a square.

Let, A(1,-2), B(3,9), C(1,2) and D(-1,0) are the vertices of a square
Now,

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

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

$=\sqrt {4+4} = \sqrt {8}$

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

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

$=\sqrt {4+4}= \sqrt {8}$

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

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

$=\sqrt {4+4} = \sqrt {8}$

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

$=\sqrt {4+4} = \sqrt {8}$

Again diagonal,

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

$=\sqrt {0+4^2} = \sqrt{16} = 4$

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

$=\sqrt {0+4^2} = \sqrt{16} = 4$

$\therefore AB=BC=CD=DA = \sqrt{8}$ and diagonal AC = diagonal BD.
Hence ABCD is a square and given points are vertices of a square.
Hence,
1857508_1876429_ans_14dc0d1aaaa548b7a502f195aaaa8e3b.png

1857508_1876429_ans_14dc0d1aaaa548b7a502f195aaaa8e3b.png

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

We know that on x axis the y coordinate of any point is zero. Hence on x axis any point P (x,0) is equal distance from the points

$A(-2,-5)$ and

$B(2,-3).$

$\therefore PA=PB$

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

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

$=\sqrt{x^2+4x+29}$

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

$=\sqrt{x^2+4-4x+9}$

$=\sqrt{x^2-4x+13}$

according to question

$PA = PB$

$\sqrt{x^2-xy+13}=\sqrt{x^2+4x+29}$
squaring on both the sides,

$x^2-4x+13=x^2+4x+29$

$4x+4x=13-29$

$8x=-16$

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

Hence, the required point is

$x(-2,0)$

If points $(2,-3), (\lambda, -2)$ and $(0,5)$ are collinear, then find $\lambda$

Given points

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

$(0,5)$ are collinear, then

$\begin{vmatrix} 2 & -3 & 1 \\ \lambda & -2 & 1 \\ 0 & 5 & 1 \end{vmatrix}=0$

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

$\Rightarrow -14+3 \lambda+5 \lambda=0$

$\Rightarrow 8 \lambda=14$

$\Rightarrow \lambda=\dfrac{14}{8}=\dfrac{7}{4}$

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

We know that on y axis the x coordinate of any point is zero. Hence on y axis any point

$P (0,y)$ is equal distance from the points

$A(-5,-2)$ and

$B(3,2)$ .

$\therefore PA=PB$

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

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

$=\sqrt{25+4+y^2+4y}$

$=\sqrt{y^2+4y+29}$

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

$=\sqrt{9+4+y^2-4y}$

$=\sqrt{y^2-4y+13}$

according to question

$PA = PB$

$\sqrt{y^2-4y+13}=\sqrt{y^2+4y+29}$
squaring on both the sides,

$y^2-4y+13=y^2+4y+29$

$4y+4y=13-29$

$8y=-16$

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

Hence, the required point is

$y(0,-2)$

Find the mid point of joining the points (22,20) and (0,16).

Let coordinates of mid point be P(x,y) of joining the points A(22,20) and B(0,16).

$x_1=22,\space y_1=20,\space x_2=0,\space y_2=16$
By formula of mid point,

$x=\dfrac{x_1+x_2}{2}$

$=\dfrac{22+0}{2} = \dfrac{22}{2} = 11$
and

$y=\dfrac{y_1+y_2}{2}$

$=\dfrac{20+16}{2} = \dfrac{36}{2} = 18$
Hence the coordinate of mid point

$P(x,y) = P(11,18)$

Find the coordinate points of which divides the line segment joining the point (3,5) and (7,9) in the ratio 2:3 internally.

$x_1=3\,,\,y_1=5\,,\,m_1=2$

$x_2=7\,,\,y_2=9\,,\,m_2=3$

Let point P(x,y) divides the line segment joining the points

$A(3,5)$ and

$B(7,9)$ in the ratio

$2:3$ internally.
By the formula of internal division.

$x=\dfrac{m_1x_2+m_2x_1}{m_1+m_2}$

$=\dfrac{(2)(7)+(3)(3)}{2+3}$

$=\dfrac{14+9}{5} = \dfrac {23}{5} = 4\dfrac{3}{5}$
and

$y=\dfrac{m_1y_2+m_2y_1}{m_1+m_2}$

$=\dfrac{(2)(9)+(3)(5)}{2+3}$

$=\dfrac{18+15}{5} = \dfrac{33}{5}=6\dfrac{3}{5}$

Hence, the co-ordinates of P is

$(4\dfrac{3}{5},6\dfrac{3}{5})$
1857576_1876641_ans_60dd914bc095426b83410d945c14a95c.png

If the point $P(3,5)$ divides the line segment which joins $A(-2,3)$ and $B (x,y)$ in the ratio $4:7$ internally, then find the co-ordinates of $B.$

Given: Point

$P(3,5)$ divides the line segment which joins

$A(-2,3)$ and

$B (x,y)$ in the ration

$4:7$ internally

By section formula, we know

$(x,y)=\bigg(\dfrac{mx_2+nx_1}{m+n},$

$\dfrac{my_2+ny_1}{m+n}\bigg)$

Equating the coordinates of point

$P$

$\Rightarrow 3=\dfrac{(4)(x)+(7)(-2)}{4+7}$

$\Rightarrow 3=\dfrac{4x-14}{11}$

$\Rightarrow 4x-14=33$

$\Rightarrow 4x=33+14$

$\Rightarrow x=\dfrac{47}{4}$

And

$\Rightarrow 5=\dfrac{(4)(y)+(7)(3)}{4+7}$

$\Rightarrow 5=\dfrac{4y+21}{11}$

$\Rightarrow 4y = 55-21$

$\Rightarrow y=\dfrac{34}{4} = \dfrac{17}{2}$

Hence the co-ordinates of point

$B=\left (\dfrac{47}{4},\dfrac{17}{2}\right )$

1857755_1876688_ans_b7e163018115453b8007f912debcf53a.png

How we will find the point in side BC of $\Delta ABC$ which is equidistant from sides AB and AC?

$\Delta ABC$ , draw AX, bisector of

$\angle A$ which cuts BC at D.
Take any point P on AX. Draw

$\perp$ PN and PM from P to AB and P to AC respectively.
PN

$\perp$ AB and PM

$\perp$ AC
From

$\Delta APN$ and

$\Delta APM$

$\angle PNA = \angle PMA = 90^o$ (By construction)

$\angle PAN = \angle PAM$ (AX is bisector of

$\angle A$ )
AP = AP (Common)
by A.A.S. rule

$\Delta APN \cong \Delta APM$

$\Rightarrow$ PN = PM (By CPCT)
Thus P is equidistant from AB and AC.

$\therefore$ Any point on AX will be equidistant from AB and AC.
Therefore point D on line BC, will be equidistant from AB and AC.
1860995_1876839_ans_132d72be0d0946c4a8e5ad9a9895ae1e.png

1860995_1876839_ans_132d72be0d0946c4a8e5ad9a9895ae1e.png

Prove that the midpoint of a line segment which joins the points (5,7) and (3,9) is the same as the mid point of the line segment which joins the points (8,6) and (3,10).

Coordinate of mid point joining the live segment of points (5,7) and (3,9) is

$=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right)$

$=\left(\dfrac{5+3}{2}, \dfrac{7+9}{2}\right)$

$=\left(\dfrac{8}{2}, \dfrac{16}{2}\right)$

$=(4,8)$
Again, Coordinate of mid point of line segment joining the points (8,6) and (0,10)

$=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right)$

$=\left(\dfrac{8+0}{2}, \dfrac{6+10}{2}\right)$

$=\left(\dfrac{8}{2}, \dfrac{16}{2}\right)$

$=(4,8)$
Clearly mid point of line segment which joins the points (5,7) and (3,9) is the same as mid point of line segment which joins the points (5,7) and (3,9) .

Prove that, the points (1,5),(2,3), and (-2,-11) are not collinear.

Let the points are

$\mathrm{A}(1,5), \mathrm{B}(2,3)$ and

$\mathrm{C}(-2,-11)$ respectively i.e. co-ordinate of

$\mathrm{A}$ is

$(1,$ 5), co-ordinate of B is ( 2, 3) and co-ordinate of C is (-2, -11).

$\begin{aligned} A B &=\sqrt{(2-1)^{2}+(3-5)^{2}} \\ &=\sqrt{1^{2}+(-2)^{2}}=\sqrt{1+4}=\sqrt{5} \\ B C &=\sqrt{(-2-2)^{2}+(-11-3)^{2}} \\ &=\sqrt{16+196} \\ &=\sqrt{212} \\ C A &=\sqrt{(1-(-2))^{2}+(5-(-11))^{2}} \\ &=\sqrt{9+256} \\ &=\sqrt{265} \end{aligned}$
It is clear that sum of distances of any two sides is not equal to third side. So these points are not collinear.

Find the midpoint of the line segment joining the point (-3,4) and point (7,5).

The mid point of line segment joining the point (-3,4) and (7,5)

$=\left(\dfrac{-3+7}{2}, \dfrac{4+5}{2}\right)$

$=\left(\dfrac{4}{2}, \dfrac{9}{2}\right)=2, \dfrac{9}{2}$

Find the type of quadrilateral, If its vertices are (1,4),(-5,4),(-5,-3) and (1,-3)

Let the vertices of quadrilateral are

$\mathrm{A}(1,4), \mathrm{B}(-5,4) \mathrm{C}(-5,-3)$ and

$\mathrm{D}(1,-3)$ respectively.
Then Hence,

$A B=C D$ , and

$B C$ = DA and Diagonal BD = Diagonal AC

$\begin{aligned} A B &=\sqrt{(-5-1)^{2}+(4-4)^{2}} \\ &=\sqrt{(-6)^{2}+(0)^{2}} \\ &=\sqrt{36}=6 \\ B C &=\sqrt{(-5+5)^{2}+(-3-4)^{2}} \\ &=\sqrt{0+(-7)^{2}}=\sqrt{49}=7 \\ C D &=\sqrt{[1-(-5)]^{2}+[(-3)+(-3)]^{2}} \\ &=\sqrt{(1+5)^{2}+(-3+3)^{2}} \\ &=\sqrt{(6)^{2}} \\ &=6 \\ D A &=\sqrt{(1-1)^{2}+(4+3)^{2}} \\ &=\sqrt{0+(7)^{2}} \\ &=\sqrt{49}=7 \end{aligned}$

$\begin{aligned} \text { Diagonal } \mathrm{BD} &=\sqrt{(1+5)^{2}+(-3-4)^{2}} \\ &=\sqrt{36+49} \\ &=\sqrt{85} \end{aligned}$

$\begin{array}{l}\begin{aligned}\text { Diagonal } A C &=\sqrt{(-5-1)^{2}+(-3-4)^{2}} \\&=\sqrt{(-6)^{2}+(-7)^{2}} \\&=\sqrt{36+49} \\&=\sqrt{85}\end{aligned}\end{array}$
Hence, these points are the vertices of rectangle.
1858099_1877159_ans_cec1e199607f4bb4a54c3a698bbe78a3.png

1858099_1877159_ans_cec1e199607f4bb4a54c3a698bbe78a3.png

If in any plane there are four points $\mathrm{P}(2,-1), \mathrm{Q}(3,4), \mathrm{R}(-2,3)$ and $\mathrm{S}(-3,-2),$ then provethat PQRS is not a square but a rhombus.

Let four points

$\mathrm{P}(2,-1), \mathrm{Q}(3,4), \mathrm{R}(-2,3)$ and

$\mathrm{S}(-3,-2)$ are in a plane.

$\begin{aligned} \mathrm{PQ} &=\sqrt{(3-2)^{2}+(4-(-1))^{2}} \\ &=\sqrt{(1)^{2}+(4+1)^{2}} \\ &=\sqrt{25+1}=\sqrt{26} \\ \mathrm{QR} &=\sqrt{(-2-3)^{2}+(3-4)^{2}} \\ &=\sqrt{(-5)^{2}+(-1)^{2}} \\ &=\sqrt{25+1}=\sqrt{26} \\ \mathrm{RS} &=\sqrt{(-3-(-2))^{2}+(-2-3)^{2}} \\ &=\sqrt{(-3+2)^{2}+(-5)^{2}} \\ &=\sqrt{1+25}=\sqrt{26} \\ \mathrm{SP} &=\sqrt{(2-(-3))^{2}+(-1-(-2))^{2}} \\ &=\sqrt{(2+3)^{2}+(-1+2)^{2}} \\ &=\sqrt{(5)^{2}+(1)^{2}} \\ &=\sqrt{25+1}=\sqrt{26} \end{aligned}$

$\begin{array}{l}\begin{aligned} \text { Diagonal } \mathrm{PR} &=\sqrt{(2-(-2))^{2}+(-1-3)^{2}} \\ &=\sqrt{(2+2)^{2}+(-4)^{2}} \\ &=\sqrt{16+16} \\ &=\sqrt{32} \\ &=4 \sqrt{2} \\ \text { Diogonal } \mathrm{SQ} &=\sqrt{(-3-3)^{2}+(-2-4)^{2}} \\ &=\sqrt{(-6)^{2}+(-6)^{2}} \\ &=\sqrt{36+36} \\ &=\sqrt{72} \\ &=6 \sqrt{2} \end{aligned} \\ \because P Q=\mathrm{QR}=\mathrm{RS}= S P=\sqrt{26}\end{array}$
And diagonal PR

$\neq$ diagonal SQ
since diagonals are also equal in a square but here diagonals are not equal. Hence PQRS is not square but a rhombus.
1858153_1877166_ans_6535706223c0456ebf94737c2a716a96.png

1858153_1877166_ans_6535706223c0456ebf94737c2a716a96.png

If co-ordinate of one end and the midpoint of a line segment are (4,0) and (4,1) respectively, then find the co-ordinate of other ends of the line segment.

Let co-ordinate of side A is (4,0) and co-ordinate of side

$B$ is

$(x, y)$ of line segment

$A B .$ The co-ordinate ofmid point P is (4,1)
By mid point formula

$x=\dfrac{x_{1}+x_{2}}{2}$

$\begin{aligned}4 &=\dfrac{4+x}{2} \\\Rightarrow \quad 4+x &=8 \\x &=8-4=4 \\\text { and } \quad y &=\dfrac{y_{1}+y_{2}}{2}\end{aligned}$

$\begin{aligned}1 &=\dfrac{0+y}{2} \\\Rightarrow \quad y &=2\end{aligned}$
Hence, co-ordinate of point

$B$ is (4,2)
1858134_1877163_ans_6ef6b66139954200aae06240d8dc6472.png

If mid points of sides of a triangle is (1,2),(0,-1) and (2,-1) , then find its vertices.

Let

$P(1,2), Q(0,-1)$ and

$R(2,-1)$ are mid point of sides of given triangle and coordinate of vertices of triangleare

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

$\mathrm{C}\left(\mathrm{x}_{3}, \mathrm{y}_{3}\right)$ respectively.
since, co-ordinate of mid point

$P$ is (1,2) of

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

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

$1=\dfrac{x_{1}+x_{2}}{2}$ and

$2=\dfrac{y_{1}+y_{2}}{2}$

$\Rightarrow x_{1}+x_{2}=2 \ldots$ (i)
and

$y_{1}+y_{2}=4 \ldots$ (ii)
since

$\mathrm{Q}(0,-1)$ is the midpoint of

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

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

$\mathrm{So}^{\dfrac{x_{2}+x_{3}}{2}}=0, \dfrac{y_{2}+y_{3}}{2}=-1$

$\Rightarrow x_{2}+x_{3}=0 \ldots$ (iii)
and

$y_{2}+y_{3}=-2 \ldots$ (iv)
since

$\mathrm{R}(2,-1)$ is the mid point of

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

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

$\dfrac{x_{1}+x_{3}}{2}=2, \dfrac{y_{1}+y_{3}}{2}=-1$

$\Rightarrow x_{1}+x_{3}=4 \ldots(v)$
and

$y_{1}+y_{3}=-2 \ldots$ (vi)
adding equation (i), (iii) and (v)

$2 x_{1}+2 x_{2}+2 x_{3}=6$

$=x_{1}+x_{2}+x_{3}=3 \ldots$ (vii)
Adding equation (ii), (iv) and (vi)

$2 y_{1}+2 y_{2}+2 y_{3}=0$

$y_{1}+y_{2}+y_{3}=0 \ldots$ (viii)
Put the value of equation (i) in equation (vii)

$x_{1}+x_{2}+x_{3}=3$

$2+x_{3}=3$

$x_{3}=3-2=1$
Put the value of equation (ii) in equation (viii)

$y_{1}+y_{2}+y_{3}=0$

$4+y_{3}=0$

$y_{3}=-4$
Hence, co-ordinate of point

$C\left(x_{3}, y_{3}\right)=(1,-4)$
Subtract equation (iii) from equation (vii)

$\begin{array}{lllllll}x_1 &+&x_2&+&x_3&=&3 \\&&x_2&+&x_3&=&0 \\&-&&-&&&- \\\hline x_1&&&&&=&3\end{array}$
Subtract eq (iv) from eq (viii)

$\begin{array}{lllllll}y_1 &+&y_2&+&y_3&=&0 \\&&y_2&+&y_3&=&-2 \\&-&&-&&&+ \\\hline \hline y_1&&&&&=&2\end{array}$
Hence, co-ordinate of point

$A\left(x_{1}, y_{1}\right)=(3,2)$
Again subtract equation (

$\mathrm{v}$ ) from equation (vii)

$\begin{array}{llllllll}&x_1 &+&x_2&+&x_3&=&3 \\&&&x_2&+&x_3&=&4 \\-&&-&&-&&&- \\\hline &&&x_2&&&=&-1\end{array}$
Subtract equation (vi) from equation (viii)

$\begin{array}{lllllll}y_1 &+&y_2&+&y_3&=&0 \\y_1&&&+&y_3&=&-2 \\&-&&-&&&+ \\\hline \hline &&y_2&&&=&2\end{array}$
Hence, co-ordinate (-1,2)
Hence, the vertices of triangle are

$(1,-4),(3,2)$ and

$(-1,2)$ Respectively.
1860893_1877185_ans_bd49a43cec7f40a8a895df33e68193b6.png

$\mathrm{A}(4,2), \mathrm{B}(6,5)$ and $\mathrm{C}(1,4)$ are the vertices of triangle $\mathrm{ABC}, \mathrm{AD}$ is a median through point A meet BC at point D, find the co-ordinate of D.

Co-ordinate of mid point D of BC.

$=\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}\right)$

$=\left(\dfrac{6+1}{2}, \dfrac{5+4}{2}\right)$

$=\left(\dfrac{7}{2}, \dfrac{9}{2}\right)$
Hence, co-ordinate of D is

$\left(\dfrac{7}{2}, \dfrac{9}{2}\right)$

Find the value of $x$ which is equidistant from the points $A(6,5)$ and $B(-4,5)$ .

Let co-ordinate of any point

$P$ on

$x$ -axis

$=(x, 0)$

$\therefore P A=P B$

$\Rightarrow P A^{2}=P B^{2}$

$\Rightarrow(x-6)^{2}+(0-5)^{2}=(x+4)^{2}+(0-5)^{2}$

$\Rightarrow x^{2}+36-12 x+25=x^{2}+16+8 x+25$

$\Rightarrow-12 x+36=8 x+16$

$\Rightarrow-12 x-8 x=16-36$

$\Rightarrow-20 x=-20$

$\Rightarrow x=\dfrac{-20}{-20}=1$
Hence, the co-ordinate of required point

$=(1,0)$

Find the point on $\mathrm{x}$ -axis which is equidistant from (-2,5) and (2,-3)

Let co-ordinate of any point on

$x$ axis

$(x, 0)$ because for

$x$ -axis co-ordinate of y is zero Distance between

$(x, 0)$ and (2,-5)

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

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

$=\sqrt{x^{2}-4 x+29}$
And distance between

$(x, 0)$ and

$(-2,9)$

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

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

$=\sqrt{x^{2}+4 x+85}$
According to question

$\sqrt{x^{2}-4 x+29}=\sqrt{x^{2}+4 x+85}$
Squaring both sides.

$x^{2}-4 x+29=x^{2}+4 x+85$

$-4 x-4 x=85-29$

$-8 x=56$

$x=\dfrac{-56}{8}=-7$
Hence, co-ordinate of required point

$=(-7,0)$

If point $\mathrm{P}(2,4)$ is equidistant from the points $\mathrm{A}(5, \mathrm{K})$ and $\mathrm{B}(\mathrm{K}, 7)$ , then find the value of K.

Point

$P(2,4)$ is equidistant from the points

$A(5, K)$ and

$B(K, 7)$ . So

$P A=P B$

$\Rightarrow P A^{2}=P B^{2}$

$\Rightarrow(5-2)^{2}+(k-4)^{2}=(k-2)^{2}+(7-4)^{2}$

$\Rightarrow 9+k^{2}+16-8 K=k^{2}+4-4 K+9$

$\Rightarrow \mathrm{K}^{2}-8 \mathrm{K}+16-\mathrm{K}^{2}-4+4 \mathrm{K}=0$

$\Rightarrow-4 K+12=0$

$\Rightarrow-4 K=-12$

$\Rightarrow \mathrm{K}=\dfrac{-12}{-4}=3$
Hence,

$K=3$

If the points $\mathrm{A}(6,1), \mathrm{B}(8,2), \mathrm{C}(9,4)$ and $\mathrm{D}(\mathrm{x}, \mathrm{y})$ are vertices of parallelogram respectively, then find the point $D(x, y)$ .

Let point

$\mathrm{A}(6,1), \mathrm{B}(8,2), \mathrm{C}(9,4)$ and

$\mathrm{D}(\mathrm{x}, \mathrm{y})$ are the vertices of parallelogram.

$\therefore$ Diagonals of a parallelogram bisects each other.
So,

$=$ co-ordinate of mid point diagonal AC

$\Rightarrow\left(\dfrac{6+9}{2}, \dfrac{1+4}{2}\right)=\left(\dfrac{8+x}{2}, \dfrac{2+y}{2}\right)$

$\Rightarrow \quad\left(\dfrac{15}{2}, \dfrac{5}{2}\right)=\left(\dfrac{8+x}{2}, \dfrac{2+y}{2}\right)$
comparing

$\dfrac{8+x}{2}=\dfrac{15}{2}$ and

$\dfrac{2+y}{2}=\dfrac{5}{2}$

$\Rightarrow 8+x=15$ and

$2+y=5$

$\Rightarrow x=15-8$ and

$y=5-2$

$\Rightarrow x=7$ and

$y=3$
Hence, co-ordinate of point

$D(7, 3)$

1860481_1877733_ans_ea26331c250c4186adaeccbe475a5840.png

Find the co-ordinate of point P on the line segment joining the points $A(-2,-2)$ and $B(2,-4)$ such that $\mathrm{AP}=\dfrac{3}{7} \mathrm{AB}$ and point $\mathrm{P}$ lies on line segment $\mathrm{AB}$ .

Using the section formula, if a point $(x,y)$ divides the line joining the points $({ x }_{ 1 },{ y }_{ 1 })$ and $({ x }_{ 2 },{ y }_{ 2 })$ in the ratio $m:n$ , then

$(x,y) = \left( \dfrac { m{ x }_{ 2 } + n{ x }_{ 1 } }{ m + n },\dfrac { m{ y }_{ 2 } + n{ y }_{ 1 } }{ m + n } \right)$

Let required point is

$\mathrm{P}(\mathrm{x}, \mathrm{y})$
Given

$A P=\dfrac{3}{7} A B \ldots$ (i)
But

$P B=A B-A P$

$=A B-\dfrac{3}{7} A B$

$=\dfrac{7 A B-3 A B}{7}$

$\mathrm{PB}=\dfrac{4}{7} \mathrm{AB} \ldots$ (ii)
From equation (i) and (ii),

$\mathrm{AP}: \mathrm{PB}=\dfrac{3}{7} \mathrm{AB}: \dfrac{4}{7} \mathrm{AB}$

$A P: P B=3: 4$

$\therefore$ Hence point

$P$ divides point

$A$ and

$B$ in the ratio 3: 4 respectively.

$\therefore$ Hence coordinates of point

$P$ are

$\begin{array}{c}x=\dfrac{3 \times 2+4 \times(-2)}{3+4} \\=\dfrac{6-8}{7}=\dfrac{-2}{7}\end{array}$
and

$y=\dfrac{3 \times(-4)+4 \times(-2)}{3+4}$

$=\dfrac{-12-8}{7}=\dfrac{-20}{7}$
Hence coordinate of point

$P=\left(\dfrac{-2}{7}, \dfrac{-20}{7}\right)$

1860205_1877766_ans_0dfa736c62c941db8db85c63e56449da.png

If the co-ordinates of the midpoints of the sides of the triangle are (3,4),(8,9), and $(6, 7)$ respectively. Find the co-ordinate of vertices of the triangle. (NCERT Exemplar Problem)

Let mid points of sides of triangle are

$P(3,4), Q(8,9)$ and

$R(6,7)$ respectively.
Let

$A\left(x_{1}, y_{1}\right) B\left(x_{2}, y_{2}\right)$ and

$C\left(x_{3}, y_{3}\right)$ be the vertices of the triangle ABC.
since

$P(3,4)$ is the mid point of

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

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

$3=\dfrac{x_{1}+x_{2}}{2}$ and

$4=\dfrac{y_{1}+y_{2}}{2}$

$x_{1}+x_{2}=6$ and

$y_{1}+y_{2}=8 \ldots$ (i)
since

$Q(8,9)$ is the mid point of

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

$C\left(x_{3}, y_{3}\right)$
So

$8=\dfrac{x_{2}+x_{3}}{2}$ and

$9=\dfrac{y_{2}+y_{3}}{2}$

$x_{2}+x_{3}=16 ; y_{2}+y_{3}=18 \ldots . .($ ii

$)$
since

$\mathrm{R}(6,7)$ is the mid point of

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

$\mathrm{C}\left(\mathrm{x}_{3}, \mathrm{y}_{3}\right) .$
So

$6=\dfrac{x_{1}+x_{3}}{2}$ and

$7=\dfrac{y_{1}+y_{3}}{2}$

$\Rightarrow x_{1}+x_{3}=12$ and

$y_{1}+y_{3}=14 \ldots$ (iii)
Adding equation (i), (ii) and (iii)

$2 x_{1}+2 x_{2}+2 x_{3}=34$
and

$2 y_{1}+2 y_{2}+2 y_{3}=40$

$\Rightarrow x_{1}+x_{2}+x_{3}=17 \ldots$ (iv)
and

$y_{1}+y_{2}+y_{3}=20$
Subtract equation (i) from equation (iv).

$x_{1}+x_{2}+x_{3}-x_{1}-x_{2}=17-6$
and

$y_{1}+y_{2}+y_{3}-y_{1}-y_{2}=20-8$

$\Rightarrow \mathrm{x}_{3}=11$ and

$\mathrm{y}_{3}=12$
subtract equation (ii) from equation (iv)

$x_{1}+x_{2}+x_{3}-x_{2}-x_{3}=17-6$
and

$y_{1}+y_{2}+y_{3}-y_{2}-y_{3}=20-8$

$\Rightarrow x_{1}=1$ and

$y_{1}=2$
subtract equation (iii) from equation (iv)

$x_{1}+x_{2}+x_{3}-x_{1}-x_{3}=17-12$
and

$y_{1}+y_{2}+y_{3}-y_{1}-y_{3}=20-14$

$\Rightarrow x_{2}=5$ and

$y_{2}=6$
Hence, the co-ordinate of vertices of triangle

$\mathrm{A}(1,2) \mathrm{B}(5,6)$ and

$\mathrm{C}(11,12)$
1860461_1877739_ans_6e20e814d3754b5d99897e1ec561805f.png

If $\mathrm{M}(4,5)$ is a mid point of line segment $\mathrm{AB}$ and co-ordinate of $\mathrm{A}$ is (3,4) , then find theco-ordinate of point B.

Let co-ordinate of

$\mathrm{B}$ is

$(\mathrm{x}, \mathrm{y})$
Given: co-ordinate of mid point of (3,4) and

$(x, y)$

$4=\dfrac{3+x}{2} \Rightarrow 3+x=8$

$\Rightarrow x=8-3=5$
and

$5=\dfrac{y+4}{2}$

$\Rightarrow y+4=10$

$\Rightarrow y=10-4=6$
Hence, co-ordinate of

$\mathrm{B}=(5,6)$

Prove that points 4(1,1) $\mathrm{B}(-2,7)$ and $\mathrm{C}(3,-3)$ are collinear

Let

$A(1,1), B(-2,7)$ and

$C(3,-3)$ then

$\begin{aligned} \mathrm{AB} &=\sqrt{(-2-1)^{2}+(7-1)^{2}} \\ &=\sqrt{(-3)^{2}+(6)^{2}} \\ &=\sqrt{9+36} \\ &=\sqrt{45}=3 \sqrt{5} \\ \mathrm{BC} &=\sqrt{(3+2)^{2}+(-3+7)^{2}} \\ &=\sqrt{(5)^{2}+(-10)^{2}} \\ &=\sqrt{25+100} \\ &=\sqrt{125} \\ &=5 \sqrt{5} \\ \mathrm{AC} &=\sqrt{(3-1)^{2}+(-3-1)^{2}} \\ &=\sqrt{(2)^{2}+(-4)^{2}} \\ &=\sqrt{20} \\ &=2 \sqrt{5} \\ \therefore A B &+\mathrm{AC}=3 \sqrt{5}+2 \sqrt{5}=5 \sqrt{5}=\mathrm{BC} \end{aligned}$

$\mathrm{So} \ \mathrm{AB}+\mathrm{AC}=\mathrm{BC}$
So these point

$\mathrm{A}, \mathrm{B}$ , and

$\mathrm{C}$ are collinear.

Find the center of a circle passing through the points (6,-6),(3,-7) and (3,3) (NCERT)

Let

$O(x, y)$ is a center of circle.
The circle of center 0 is passes through the points

$P(6,-6), Q(3,-7)$ and

$R(3,3)$ respectively.

$\because$ since radius of circle is same

$\therefore \mathrm{OP}=\mathrm{OQ}=\mathrm{OR}$
or

$\mathrm{OP}^{2}=\mathrm{OQ}^{2}=\mathrm{OR}^{2}$
Now

$\mathrm{OP}^{2}=\mathrm{OQ}^{2}$

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

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

$\Rightarrow-12 x+6 x+12 y-14 y+72-58=0$

$\Rightarrow-6 x-2 y+14=0$

$\Rightarrow 3 x+y-7=0 \ldots$ (i)
Now

$\mathrm{OQ}^{2}=\mathrm{OR}^{2}$

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

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

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

$\Rightarrow 20 y=-40$

$\Rightarrow y=\dfrac{-40}{20}=-2$
Put the value of y in equation (i)

$3 x-2-7=0$

$\Rightarrow 3 x-9=0$

$\Rightarrow 3 x=9$

$\Rightarrow x=\dfrac{9}{3}=3$
Hence, required center is

$(3,-2)$ .
1860221_1877750_ans_c4215429a3fd4106b2565ab560e0cd44.png

$A(a+1, a-1)$ , $B(a^{2}+1, a^{2}-1)$ and $C(a^{3}+1, a^{3}-1)$ are given points. $D(11, 9)$ is the mid point of $AB$ and $E(41, 39)$ is the mid point of $BC$ . If $F$ is the mid point of $AC,$ then $\dfrac{(BF)^{2}}{81}$ is equal to

Using mid-point formula,

$\dfrac{a+1+a^{2}+1}{2}=11, \dfrac{a-1+a^{2}-1}{2}=9$

$\Rightarrow a+a^{2}-20=0\Rightarrow a=-5$ or

$4$
And

$\dfrac{a^{2}+1+a^{3}+1}{2}=41, \dfrac{a^{2}-1+a^{3}-1}{2}=39$

$\Rightarrow a^{ 2 }+a^{ 3 }=80$ which holds for

$a=4$
So, the given points are

$A(5,3), B(17,15), C(65,33)$ and coordinates of

$F$ are

$\left( \dfrac { 65+5 }{ 2 } ,\dfrac { 63+3 }{ 2 } \right) =(35,33)$ and

$(BF)^{2}=(35-17)^{2}+(33-15)^{2}$

$=(18)^{2}+(18)^{2}=648$

Thus the final answer is

$\dfrac{(BF^2)}{81} = \dfrac{648}{81} = 8$

Without using distance formula, show that point $(-2, -1), (4, 0), (3, 3)$ and $(-3, 2)$ are the vertices of a parallelogram.

$\left( { -2,-1 } \right) ,\left( { 4,0 } \right) ,\left( { 3,3 } \right) \, \, and\, \, \left( { -3,2 } \right) \\ are\, \, vertices\, \, of\, \, { { parallelogram } }\, \, \\ { { Inparallelogramdiagonalsareintersect } } \\ { { atsamepoints, } } \\ \therefore { { midpointsdiagonalAC } } \\ \Rightarrow { { AC= } }\frac { { 3-2 } }{ 2 } ,\frac { { 3-1 } }{ 2 } =\left( { \frac { 1 }{ 2 } ,1 } \right) \\ and\, \, mid\, \, { { point } }\, \, { { ofdiagonal3Dis } } \\ \Rightarrow { { BD= } }\left( { \frac { { 4-3 } }{ 2 } } \right) ,\frac { { 2+0 } }{ 2 } =\left( { \frac { 1 }{ 2 } ,1 } \right) \\$

If $M$ is the mid-point of $AB$ , find the co-ordinates of:
a) $A$ if the coordinates of $M$ and $B$ are $M(2,8)$ and $B(-4,19)$ and
b) $B$ if the coordinates of $A$ and $M$ are $A(-1,2)$ , $M(-2,4)$ .

(a) Let the point of

$A$ be

$(x,y)$

Then,

$(\dfrac{x-4}{2},\dfrac{y+19}{2})=(2,8)$

$\dfrac{x-4}{2}=2,\;\dfrac{y+19}{2}=8$

$x=8,\;y=-3$

So, the point

$A$ is

$(8,-3)$

(b) Let the point

$B$ be

$(x,y)$

Then,

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

$\dfrac{x-1}{2}=-2,\;\dfrac{y+2}{2}=4$

$x=-3,\;y=6$

So, point

$B$ is

$(-3,6)$

If $( 10,5 ), ( 8,4 )$ and $( 6,6 )$ be the mid-points of the sides of a triangle, find the coordinates of the vertices.

Let's say

$D,\ E$ and

$F$ are the midpoints of

$AB,\ BC$ and

$AC$ and coordinate of vertices

$A,B,C$ be

$(x,y),\ (a,b)$ and

$(p,q)$ respectively.

For points $(x_1,y_1)$ and $(x_2,y_2)$ we define the midpoint as,

Point $P(X,Y)=\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$

For mid point

$D$ of

$AB,$

$\dfrac{x+a}{2}=10$

$\Rightarrow x+a=20$

$\dfrac{y+b}{2}=5$

$\Rightarrow y+b=10$

For midpoint

$D$ of

$BC,$

$\dfrac{a+p}{2}=6$

$\Rightarrow a+p=12$

$\dfrac{b+q}{2}=6$

$\Rightarrow b+q=12$

For midpoint

$F$ of

$AC,$

$\dfrac{x+p}{2}=8$

$\Rightarrow x+p=16$

$\dfrac{y+q}{2}=4$

$\Rightarrow y+q=8$

Consider,

$x+a=20$ and

$a+p=12$

$x+12-p=20$

$\Rightarrow x-p=8$

Now, we have,

$x-p=8,x+p=16$

On solving;

$x=12$ and

$p=4$

Now, we have

$x=12$ and

$x+a=20$ . So,

$12+a=20$

$\Rightarrow a=8$

Consider,

$y+q=8$ and

$q+b=12$

$y+12-b=8$

$\Rightarrow y-b=-4$

Now, we have,

$y-b=-4,y+b=10$

On solving;

$y=3$ and

$b=7$

Now, we have

$y=3$ and

$y+q=8$ . So,

$3+q=8$

$\Rightarrow q=5$

Hence, the coordinates of the vertices will be

$A(12,3),\ B(8,7),\ C(4,5).$

1086056_1175767_ans_a0a32383dfac49abba2c0744ca8c929b.png

Find k if (0, k) is equidistant from (5, -3) and (4, 6)

2039517_1295848_ans_7043561aaf6c41e1a2c1ebfde206bcbb.JPG

Three consecutive vertices of a parallelogram ABCD are A $\left(3,-1,2\right)$ , B $\left(1,2,-4\right)$ and C $\left(-1,1,2\right)$ . Find the coordinates of the fourth vertex D.

Given that, a parallelogram

$ABCD$ has three consecutive vertices as

$A\left(3,-1,2\right),\,B\left(1,2,-4\right)$ and

$C\left(-1,1,2\right)$ .

Let the coordinates of the fourth vertex

$D$ be

$\left(x,y,z\right)$

We know that, the diagonals of a parallelogram bisect each other.

So, if

$X$ be the point of intersection of both diagonals,

$AC$ and

$BD$ , it will be the midpoint of both

$AC$ and

$BD$ .

We also know that the coordinates of the midpoint of a line joining two points with coordinates

$(x_1, y_1, z_1)$ and

$(x_2, y_2, z_2)$ are

$\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}, \dfrac{z_1+z_2}{2}\right)$

$\therefore \$ Midpoint of

$AC=\left(\dfrac{3-1}{2},\dfrac{-1+1}{2},\dfrac{2+2}{2}\right)=\left(1,0,2\right)$

and midpoint of

$BD=\left(\dfrac{1+x}{2},\dfrac{2+y}{2},\dfrac{-4+z}{2}\right)$

Since, midpoint of

$AC$ and

$BD$ is

$X$ ,

$\therefore\,\left(\dfrac{1+x}{2},\dfrac{2+y}{2},\dfrac{-4+z}{2}\right)=\left(1,0,2\right)$

$\Rightarrow\,\dfrac{1+x}{2}=1,\dfrac{2+y}{2}=0,\dfrac{-4+z}{2}=2$

$\Rightarrow\,x+1=2,\,y+2=0,\,z-4=4$

$\Rightarrow\,x=2-1=1,\,y=0-2=-2,\,z=4+4=8$

Hence, the coordinates of the fourth vertex

$D$ are

$\left(1,-2,8\right)$

Three consecutive vertices of a parallelogram $ABCD$ are $A\left(-2,3,5\right),\,B\left(-3,2,1\right)$ and $C\left(3,5,1\right)$ .Find the fourth vertex $D$

In a parallelogram

$ABCD$ are

$A\left(-2,3,5\right),\,B\left(-3,2,1\right)$ and

$C\left(3,5,1\right)$ .Let the fourth vertex

$D$ be

$\left(x,y,z\right)$

Midpoint of

$AC=\left(\dfrac{-2+3}{2},\dfrac{3+5}{2},\dfrac{5+1}{2}\right)=\left(\dfrac{1}{2},\dfrac{8}{2},\dfrac{6}{2}\right)$

Midpoint of

$BD=\left(\dfrac{-3+x}{2},\dfrac{2+y}{2},\dfrac{1+z}{2}\right)$

Since

$ABCD$ is a parallelogram and in a parallelogram, the diagonals bisect each other, so the mid-points of

$AC$ and

$BD$ are same,

$\therefore\,\left(\dfrac{-3+x}{2},\dfrac{2+y}{2},\dfrac{1+z}{2}\right)=\left(\dfrac{1}{2},\dfrac{8}{2},\dfrac{6}{2}\right)$

$\Rightarrow\,\dfrac{-3+x}{2}=\dfrac{1}{2},\dfrac{2+y}{2}=\dfrac{8}{2},\dfrac{1+z}{2}=\dfrac{6}{2}$

$\Rightarrow\,x-3=1,\,y+2=8,\,z+1=6$

$\Rightarrow\,x=1+3=4,\,y=8-2=6,\,z=6-1=5$

Hence, the fourth vertex

$D$ be

$\left(4,6,5\right)$

Three consecutive vertices of a parallelogram $ABCD$ are $A\left(5,2\right),\,B\left(3,5\right)$ and $C\left(2,3\right)$ .Find the fourth vertex $D$

In a parallelogram

$ABCD$ co-ordinates are

$A\left(5,2\right),\,B\left(3,5,\right)$

and

$C\left(2,3\right)$ .Let the fourth vertex

$D$ be

$\left(x,y\right)$

Midpoint of

$AC=\left(\dfrac{5+2}{2},\dfrac{2+3}{2}\right)=\left(\dfrac{7}{2},\dfrac{5}{2}\right)$

Midpoint of

$BD=\left(\dfrac{3+x}{2},\dfrac{5+y}{2}\right)$

Since

$ABCD$ is a parallelogram and in a parallelogram, the diagonals bisect each other, so the mid-points of

$AC$ and

$BD$ are same,

$\therefore\,\left(\dfrac{3+x}{2},\dfrac{5+y}{2}\right)=\left(\dfrac{7}{2},\dfrac{5}{2}\right)$

$\Rightarrow\,\dfrac{3+x}{2}=\dfrac{7}{2},\dfrac{5+y}{2}=\dfrac{5}{2}$

$\Rightarrow\,x+3=7,\,y+5=5$

$\Rightarrow\,x=7-3=4,\,\\y=5-5=0$

Hence, the fourth vertex

$D$ be

$\left(4,0\right)$

Show that the perpendicular from the origin upon the straight line joining the points
$(a \ cos \ \alpha, a \ sin \ \alpha)$ and $(a \ cos \ \beta, a \ sin \ \beta)$
bisects the distance between them.

1759109_1740695_ans_39ebff20f587474b9d1324767684ae7d.jpg

The mid-points of the sides of a triangle are $(3, 4), (4, 6)$ and $(5, 7).$ Find the coordinates of the vertices of the triangle.

1961877_1816125_ans_b9048ed38c9a427c9887ea88e54de571.jpg

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

1897432_1865429_ans_97fcc3d082af47328df7ac2ed6040539.jpeg

A point $P(2,-1)$ is equidistant from the points $(a,7)$ and $(-3,a)$ . Find $a$ .

Point

$P(2,-1)$ is equidistant from the points

$A(a,7)$ and

$B(-3,a)$

$PA=PB$

${ \left( PA \right) }^{ 2 }={ \left( PB \right) }^{ 2 }$

$\quad { \left( a-2 \right) }^{ 2 }+{ \left( 7+1 \right) }^{ 2 }={ \left( -3-2 \right) }^{ 2 }+{ \left( a+1 \right) }^{ 2 }\quad$

${ a }^{ 2 }+4-4a+64=25+{ a }^{ 2 }+1+2a\Rightarrow 42=6a\Rightarrow a=7$

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

1) By distance formula, we can write,

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

$\therefore BD=\sqrt { { \left( -6 \right) }^{ 2 }+{ \left( -6 \right) }^{ 2 } }$

$\therefore BD=\sqrt { 36+36 }$

$\therefore BD=\sqrt { 2\times 36 }$

$\therefore BD=6\sqrt { 2 }$

Similarly,

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

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

$\therefore AC=\sqrt { 16+16 }$

$\therefore AC=\sqrt { 2\times 16 }$

$\therefore AC=4\sqrt { 2 }$

2) Area of the rhombus is given by,

$A=\frac { 1 }{ 2 } \left( product\quad of\quad diagonals \right)$

$\therefore A=\frac { 1 }{ 2 } \left( BD\times AC \right)$

$\therefore A=\frac { 1 }{ 2 } \left( 6\sqrt { 2 } \times 4\sqrt { 2 } \right)$

$\therefore A=\frac { 1 }{ 2 } \left( 24\times 2 \right)$

$\therefore A=24\quad { cm }^{ 2 }$

Coordinate Geometry - Class 10 Maths - Extra Questions

If the points(-3, 6), (-9, a) and (0, 15) are colliner then find a

Find the coordinates of the point which divides the line segment joining the points (5,−2)(5,-2) and (9,6)(9,6) in the ratio 3:13:1.

Find the midpoint of the segment connecting the points (a,b) and (3a, c)

Find the coordinates of the mid-point of the line segment joining the points (2,3)(2, 3) and (4,7)(4, 7).

Find the distance between the points (3,0)(3,0) and (0,4)(0,4) without using the distance formula.

Find the co-ordinates of a point AA, where ABAB is the diameter of a circle, whose centre is (2,−3)(2, - 3) and BB is (1,4)(1, 4).

Find the midpoint of the line segment joining the points (3,0)(3,0) and (-1,4)(-1,4).

Prove that the coordinates, x and y, of the middle point of the line joining the point (2,3) to the point (3, 4) satisfy the equation x-y + 1=0x-y + 1=0.

Find the coordinates of the point which divides the line joining the points \left(1, 3\right)\left(1, 3\right) and \left(2, 7\right)\left(2, 7\right) in the ratio 3 : 43 : 4.

If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.

Find the co ordinates of the midpoint line joining of A\left(3,2\right)A\left(3,2\right) and B\left(5,4\right)B\left(5,4\right).

Find the sum of 'x' co-ordinates of the vertices of a triangle if the mid-points of its sides be the points (6, -1), (-1,-2)and(1,-4)(6, -1), (-1,-2)and(1,-4)

Find the co-ordinates of the mid-point of the line joining the points (2, 3)(2, 3) and (4, 7)(4, 7).

Find the coordinates of the point which divides the line segment joining the points(6,3) and (-4,5) in the ratio 3:2 internally.

Find the coordinates of the point which divides internally the join of the pointsa) (8,9)(8,9) and (-7,4)(-7,4) in the ratio 2:32:3b) (1,-2)(1,-2) and (4,7)(4,7) in the ratio 1:21:2.

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

If the points A(-2, -1), B(1, 0), C(a, 3)A(-2, -1), B(1, 0), C(a, 3) and D(1, b)D(1, b) form a parallelogram ABCDABCD, find the value of aa and bb.

Given A\left( {4, - 3} \right)A\left( {4, - 3} \right), B\left( {8,5} \right)B\left( {8,5} \right). Find the coordinates of the point that divides segment \text AB\text AB in two equal parts.

A(2,3,7), B (-1,3,2) and C (p,5,r) are vertices of a triangle. If the median through A is equally inclined to the co-ordinate axes then find the co-ordinates of the midpoint of BC in terms of p and r.

Find the mid point of line joining (2,3)(2,3) and (-6,5)(-6,5).

Find the value of aa, for which point P( a , 2 )P( a , 2 ) is the midpoint of the lines segment joining the points Q ( - 5,4 )Q ( - 5,4 ) and R ( - 1,0 )R ( - 1,0 ) .

Find the other point if the origin is mid point of them and one vertex is (4,3)(4,3)

Find the co-ordinates of points which divides the line segment joining the points (-2,5)(-2,5) and (6,8)(6,8) in the ratio 3:23:2 internally.

The shape of a garden is rectangular in the middle and semicircular at the ends.Find the axes of the garden?

AA and BB are the points (2,0)(2,0) and (0,2)(0,2) respectively. Find the coordinates of point PP if it is the mid point.

Find the ratio in which the point PP whose abscissa is 33 divides the line joining A(6,5)A(6,5) and B(-1,4)B(-1,4) and hence find the coordinates of PP.

Find the ratio in which the line joining A(6,5) & B(4, -3) is divided by the y =2, Also find the coordinate of the point of divisions.

Find the mid-points of the line segment joining the points.\left( {a,b} \right)\left( {a,b} \right) and \left( {a + 2b,2a - b} \right)\left( {a + 2b,2a - b} \right)

The mid point of (2,3)(2,3) and (3,4)(3,4) is

Find the area of the triangle whose vertices are (3,8),(-4,2) and (5,-1). If the area is \left ( \dfrac a2 \right )\left ( \dfrac a2 \right ) sq. units, then what will be the value of aa?

If A and B are (-2, -2)(-2, -2) and (2, -4)(2, -4), respectively, find the coordinates of P such that AP = \dfrac{3}{7} ABAP = \dfrac{3}{7} AB and P lies on the line segment AB.

If the midpoints of the line segment joining the points P ( 6 , b - 2 )P ( 6 , b - 2 ) and ( -2 , 4)( -2 , 4) is ( 2 , - 3 ) ,( 2 , - 3 ) , find the value of b .b .

Find the mid point of (9,0)(9,0) and (5,4)(5,4)

Solve the following :Find the co-ordinates of midpoint of the segment joining the points (22, 20) (22, 20) and (0,16)(0,16).

If the area of the triangle with vertices at the points: (2,7),(1,1) \ and \ (10,8)(2,7),(1,1) \ and \ (10,8)is \left (\dfrac {b}{2}\right )\left (\dfrac {b}{2}\right ) sq. uts.Then what is the value of b?

If the points A(a, -11), B(5, b), C(2, 15)A(a, -11), B(5, b), C(2, 15) and D(1, 1)D(1, 1) and the vertices of a parallelogram ABCDABCD, find the values of aa and bb.

Draw a quadrilateral in the Cartesian plane, whose vertices are (-4, 5), (0, 7), (5, -5) and (-4, -2). Also, find its area

What point on the x-axis is equidistant from the points (7,6)(7,6) and (-3,4)(-3,4)?

Determine whether the point is collinear.L(-2,3),M(1,-3),N(5,4)L(-2,3),M(1,-3),N(5,4)

Given MM is the mid-point of ABAB , find the co-ordinates of :BB ; if A = (3 , -1) A = (3 , -1) and M = (-1 , 3) M = (-1 , 3) .

Find a point on the y-axis which is equidistant from the points (5,2)(5,2) and (-4,3)(-4,3)

Verify whether the following points are collinear or not:A (1, -3), B (2, -5), C (-4, 7)A (1, -3), B (2, -5), C (-4, 7)

The line segment joining the points A (3, 2)A (3, 2) and B (5,1)B (5,1) is divided at the point P in the ratio 1:21:2 and it lies on the line 3x -18y + k = 0.3x -18y + k = 0. Find the value of kk.

Column II gives the coordinates of the point P that divides the line segment joining the points given in column I, match them correctly

The co-ordinates of the midpoint P of segment AB with A \equiv\equiv (3.5, 9.5), B \equiv\equiv (-1.5, 0.5)(-1.5, 0.5) is (1,a)(1,a), then the value of a is _____

A (-3, 2)A (-3, 2) and B (5, 4) are the end points of a line segment, find the sum of coordinates of the mid points of the line segment.

A line is such that its segment between the axes is bisected at the point (23, 27),(23, 27), the product of the intercepts made by the line on the coordinate axes is

Also, find the value of mm if, coordinates of the point of intersection of the diagonals of the square ABCDABCD is (1,m)(1,m)

The co-ordinates of the mid-point of CD is (1,m)(1,m).Value of mm is

Find the abscissa of the point which divides the line segment joining the points (6, 3)(6, 3) and (-4, 5)(-4, 5) in the ratio 3 : 23 : 2 internally.

The area of a triangle is 55. Two of its vertices are (2, 1)(2, 1) and (3, -2)(3, -2). The third vertex lies on y = x + 3y = x + 3. Find the third vertex.

In the adjoining figure, D,\,E,\,FD,\,E,\,F are the midpoints of the sides BC,\,CA\;and\;ABBC,\,CA\;and\;AB of \Delta ABC\Delta ABC. If BE\;and\;DFBE\;and\;DF intersect at XX while CF\;and\;DECF\;and\;DE intersect at YY, prove that XY=\displaystyle\frac{1}{4}BCXY=\displaystyle\frac{1}{4}BC.

Find the midpoint of the segment connecting the points (6,4)(6,4) and (3,-4)(3,-4).

The base of an equilateral triangle with side 2a2a lies along the yy-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

The two opposite vertices of a square are (-1, 2)(-1, 2) and (3, 2)(3, 2). Find the coordinates of the other two vertices.

Prove that the diagonals of a rectangle ABCD, with vertices A(2, -1)(2, -1), B(5, -1)(5, -1), C (5, 6)(5, 6) and D (2, 6)(2, 6), are equal and bisect each other.

Find the ratio in which the line segment joining the points A(3,- 3)A(3,- 3) and B(- 2, 7)B(- 2, 7) is divided by the x-axis. Also find the coordinates of the point of division.

Let P and Q be the points of trisection of the line segment joining the points A(2, -2)(2, -2) and B(-7, 4)(-7, 4) such that P is nearer to A. Find the coordinates of P and Q.

Find the ratio in which the y-y-axis divides the line segment joining the points (-4, -6)(-4, -6) and (10, 12)(10, 12). Also, find the coordinates of the point of division.

If the coordinates of points AA and BB are (-2, -2)(-2, -2) and (2, -4)(2, -4) respectively, find the coordinates of PP such that AP = \dfrac{3}{7} ABAP = \dfrac{3}{7} AB, where PP lies on the line segment ABAB.

The coordinates of the midpoint of the line segment joining the points \left( 2a+2,3 \right) \left( 2a+2,3 \right) and \left( 4,2b+1 \right) \left( 4,2b+1 \right) are \left( 2a,2b \right) \left( 2a,2b \right) . Find the values of aa and bb.

Prove that the points (-7, -3), (5, 10), (15, 8)(-7, -3), (5, 10), (15, 8) and (3, -5)(3, -5) taken in order are the vertices of a parallelogram.

If M (4, 5)M (4, 5) is the mid-point of the line segment ABAB and co-ordinates of AA are (3, 4)(3, 4), then find the co-ordinates of point BB.

A line intersects the y-axis and x-axis at points P and Q respectively. If (2, -5) is the mid point of PQ, then find the coordinates of P and Q.

If AA and BB are two points having coordinates (-2,-2)(-2,-2) and (2,-4)(2,-4) respectively, find the coordinates of PP such that AP=\cfrac{3}{7}ABAP=\cfrac{3}{7}AB.

A(4,2),B(6,5)A(4,2),B(6,5) and C(1,4)C(1,4) are the vertices of \triangle {ABC}\triangle {ABC}.(i) The median from AA meets BCBC at DD. Find the coordinates of the point DD.(ii) Find the coordinates of point PP on ADAD such that AP:PD=2:1AP:PD=2:1

If the coordinates of the mid-points of the sides of a triangle are (1,1), (2,-3)(1,1), (2,-3) and (3,4)(3,4), find its vertices.

If the coordinates of the mid-points of the sides of a triangle be (3,-2),(-3,1)(3,-2),(-3,1) and (4,-3)(4,-3), then find the coordinates of its vertices.

Find the distance of the point (1, 2) from the midpoint of the line segment joining the points (6, 8) and (2, 4).

If AA and BB are (1,4)(1,4) and (5,2)(5,2) respectively, find the coordinates of PP when \dfrac{AP}{BP}= \dfrac{3}{4}\dfrac{AP}{BP}= \dfrac{3}{4}.

If 3 \cos\cos \theta\theta = 1, find the value of \dfrac{6 \sin^2 \theta \, + \, \tan^2 \theta}{4 \cos \theta}\dfrac{6 \sin^2 \theta \, + \, \tan^2 \theta}{4 \cos \theta}

Show that the mid-point of the line segment joining the points (5,7)(5,7) and (3,9)(3,9) is also the mid-point of the line segment joining the points (8,6)(8,6) and (0,10)(0,10).

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), then prove that x+y=a+bx+y=a+b

Determine the equation of the right bisector of the line joining the points A(-1,2)A(-1,2) & B(-3,4)B(-3,4)

The sides of a triangle ABCABC are as under : AB : 2x+3y=29: AB : 2x+3y=29 and AC : x+2y=16AC : x+2y=16. If the mid-point of BCBC is (5,6)(5,6) then find the equation of BCBC.

Find the length of the medians of a \triangle \triangle ABC whose vertices are A(7, -3), B(5, 3) and C(3, -1).

If AA and BB are (-2, -2)(-2, -2) and (2, -4)(2, -4), respectively, find the coordination of PP such that AP = \dfrac{3}{7}ABAP = \dfrac{3}{7}AB and PP lies on the line segment ABAB.

Find the co-ordinate of the midpoint of the line segment joining points (4, 7)(4, 7) and (2, -3)(2, -3).

Show that A(6,4),B(5,-2)A(6,4),B(5,-2) and C(7,-2)C(7,-2) are the vertices of an isosceles triangle. Also, find the length of the median through AA.

Find the coordinates of the point which divides the line segment joining the points $(5,-2)$ and $(9,6)$ in the ratio $3:1$ .

Find the coordinates of the mid-point of the line segment joining the points $(2, 3)$ and $(4, 7)$ .

Find the distance between the points $(3,0)$ and $(0,4)$ without using the distance formula.

Find the co-ordinates of a point $A$ , where $AB$ is the diameter of a circle, whose centre is $(2, - 3)$ and $B$ is $(1, 4)$ .

Find the midpoint of the line segment joining the points $(3,0)$ and $(-1,4)$ .

Prove that the coordinates, x and y, of the middle point of the line joining the point (2,3) to the point (3, 4) satisfy the equation
$x-y + 1=0$ .

Find the coordinates of the point which divides the line joining the points $\left(1, 3\right)$ and $\left(2, 7\right)$ in the ratio $3 : 4$ .

Find the co ordinates of the midpoint line joining of $A\left(3,2\right)$ and $B\left(5,4\right)$ .

Find the sum of 'x' co-ordinates of the vertices of a triangle if the mid-points of its sides be the points $(6, -1), (-1,-2)and(1,-4)$

Find the co-ordinates of the mid-point of the line joining the points $(2, 3)$ and $(4, 7)$ .

Find the coordinates of the point which divides internally the join of the points
a) $(8,9)$ and $(-7,4)$ in the ratio $2:3$
b) $(1,-2)$ and $(4,7)$ in the ratio $1:2$ .

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

If the points $A(-2, -1), B(1, 0), C(a, 3)$ and $D(1, b)$ form a parallelogram $ABCD$ , find the value of $a$ and $b$ .

Given $A\left( {4, - 3} \right)$ , $B\left( {8,5} \right)$ . Find the coordinates of the point that divides segment $\text AB$ in two equal parts.

Find the mid point of line joining $(2,3)$ and $(-6,5)$ .

Find the value of $a$ , for which point $P( a , 2 )$ is the midpoint of the lines segment joining the points $Q ( - 5,4 )$ and $R ( - 1,0 )$ .

Find the other point if the origin is mid point of them and one vertex is $(4,3)$

Find the co-ordinates of points which divides the line segment joining the points $(-2,5)$ and $(6,8)$ in the ratio $3:2$ internally.

$A$ and $B$ are the points $(2,0)$ and $(0,2)$ respectively. Find the coordinates of point $P$ if it is the mid point.

Find the ratio in which the point $P$ whose abscissa is $3$ divides the line joining $A(6,5)$ and $B(-1,4)$ and hence find the coordinates of $P$ .

Find the mid-points of the line segment joining the points.
$\left( {a,b} \right)$ and $\left( {a + 2b,2a - b} \right)$

The mid point of $(2,3)$ and $(3,4)$ is

Find the area of the triangle whose vertices are (3,8),(-4,2) and (5,-1). If the area is $\left ( \dfrac a2 \right )$ sq. units, then what will be the value of $a$ ?

If A and B are $(-2, -2)$ and $(2, -4)$ , respectively, find the coordinates of P such that $AP = \dfrac{3}{7} AB$ and P lies on the line segment AB.

If the midpoints of the line segment joining the points $P ( 6 , b - 2 )$ and $( -2 , 4)$ is $( 2 , - 3 ) ,$ find the value of $b .$

Find the mid point of $(9,0)$ and $(5,4)$

Solve the following :
Find the co-ordinates of midpoint of the segment joining the points $(22, 20)$ and $(0,16)$ .

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?

If the points $A(a, -11), B(5, b), C(2, 15)$ and $D(1, 1)$ and the vertices of a parallelogram $ABCD$ , find the values of $a$ and $b$ .

What point on the x-axis is equidistant from the points $(7,6)$ and $(-3,4)$ ?

Determine whether the point is collinear.
$L(-2,3),M(1,-3),N(5,4)$

Given $M$ is the mid-point of $AB$ , find the co-ordinates of :
$B$ ; if $A = (3 , -1)$ and $M = (-1 , 3)$ .

Find a point on the y-axis which is equidistant from the points $(5,2)$ and $(-4,3)$

Verify whether the following points are collinear or not:
$A (1, -3), B (2, -5), C (-4, 7)$

The line segment joining the points $A (3, 2)$ and $B (5,1)$ is divided at the point P in the ratio $1:2$ and it lies on the line $3x -18y + k = 0.$ Find the value of $k$ .

The co-ordinates of the midpoint P of segment AB with A $\equiv$ (3.5, 9.5), B $\equiv$ $(-1.5, 0.5)$ is $(1,a)$ , then the value of a is _____

$A (-3, 2)$ and B (5, 4) are the end points of a line segment, find the sum of coordinates of the mid points of the line segment.

A line is such that its segment between the axes is bisected at the point $(23, 27),$ the product of the intercepts made by the line on the coordinate axes is

Also, find the value of $m$ if, coordinates of the point of intersection of the diagonals of the square $ABCD$ is $(1,m)$

The co-ordinates of the mid-point of CD is $(1,m)$ .
Value of $m$ is

Find the abscissa of the point which divides the line segment joining the points $(6, 3)$ and $(-4, 5)$ in the ratio $3 : 2$ internally.

The area of a triangle is $5$ . Two of its vertices are $(2, 1)$ and $(3, -2)$ . The third vertex lies on $y = x + 3$ . Find the third vertex.

In the adjoining figure, $D,\,E,\,F$ are the midpoints of the sides $BC,\,CA\;and\;AB$ of $\Delta ABC$ . If $BE\;and\;DF$ intersect at $X$ while $CF\;and\;DE$ intersect at $Y$ , prove that $XY=\displaystyle\frac{1}{4}BC$ .

Find the midpoint of the segment connecting the points $(6,4)$ and $(3,-4)$ .

The base of an equilateral triangle with side $2a$ lies along the $y$ -axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

The two opposite vertices of a square are $(-1, 2)$ and $(3, 2)$ . Find the coordinates of the other two vertices.

Prove that the diagonals of a rectangle ABCD, with vertices A $(2, -1)$ , B $(5, -1)$ , C $(5, 6)$ and D $(2, 6)$ , are equal and bisect each other.

Find the ratio in which the line segment joining the points $A(3,- 3)$ and $B(- 2, 7)$ is divided by the x-axis. Also find the coordinates of the point of division.

Let P and Q be the points of trisection of the line segment joining the points A $(2, -2)$ and B $(-7, 4)$ such that P is nearer to A. Find the coordinates of P and Q.

Find the ratio in which the $y-$ axis divides the line segment joining the points $(-4, -6)$ and $(10, 12)$ . Also, find the coordinates of the point of division.

If the coordinates of points $A$ and $B$ are $(-2, -2)$ and $(2, -4)$ respectively, find the coordinates of $P$ such that $AP = \dfrac{3}{7} AB$ , where $P$ lies on the line segment $AB$ .

The coordinates of the midpoint of the line segment joining the points $\left( 2a+2,3 \right)$ and $\left( 4,2b+1 \right)$ are $\left( 2a,2b \right)$ . Find the values of $a$ and $b$ .

Prove that the points $(-7, -3), (5, 10), (15, 8)$ and $(3, -5)$ taken in order are the vertices of a parallelogram.

If $M (4, 5)$ is the mid-point of the line segment $AB$ and co-ordinates of $A$ are $(3, 4)$ , then find the co-ordinates of point $B$ .

If $A$ and $B$ are two points having coordinates $(-2,-2)$ and $(2,-4)$ respectively, find the coordinates of $P$ such that $AP=\cfrac{3}{7}AB$ .

$A(4,2),B(6,5)$ and $C(1,4)$ are the vertices of $\triangle {ABC}$ .
(i) The median from $A$ meets $BC$ at $D$ . Find the coordinates of the point $D$ .
(ii) Find the coordinates of point $P$ on $AD$ such that $AP:PD=2:1$

If the coordinates of the mid-points of the sides of a triangle are $(1,1), (2,-3)$ and $(3,4)$ , find its vertices.

If the coordinates of the mid-points of the sides of a triangle be $(3,-2),(-3,1)$ and $(4,-3)$ , then find the coordinates of its vertices.

If $A$ and $B$ are $(1,4)$ and $(5,2)$ respectively, find the coordinates of $P$ when $\dfrac{AP}{BP}= \dfrac{3}{4}$ .

If 3 $\cos$ $\theta$ = 1, find the value of $\dfrac{6 \sin^2 \theta \, + \, \tan^2 \theta}{4 \cos \theta}$

Show that the mid-point of the line segment joining the points $(5,7)$ and $(3,9)$ is also the mid-point of the line segment joining the points $(8,6)$ and $(0,10)$ .

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

Determine the equation of the right bisector of the line joining the points $A(-1,2)$ & $B(-3,4)$

The sides of a triangle $ABC$ are as under $: AB : 2x+3y=29$ and $AC : x+2y=16$ . If the mid-point of $BC$ is $(5,6)$ then find the equation of $BC$ .

Find the length of the medians of a $\triangle$ ABC whose vertices are A(7, -3), B(5, 3) and C(3, -1).

If $A$ and $B$ are $(-2, -2)$ and $(2, -4)$ , respectively, find the coordination of $P$ such that $AP = \dfrac{3}{7}AB$ and $P$ lies on the line segment $AB$ .

Find the co-ordinate of the midpoint of the line segment joining points $(4, 7)$ and $(2, -3)$ .

Show that $A(6,4),B(5,-2)$ and $C(7,-2)$ are the vertices of an isosceles triangle. Also, find the length of the median through $A$ .

Find the image of the point (3,8) with respect to the line $x+3y=7$ assuming the line to be a plane mirror.

If $A(3,y)$ is equidistant from points $P(8,-3)$ and $Q(7,6)$ , find the vaue of $y$ and find the distance $AQ$

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 $BM$ .

The point which divides the line joining the points $(1, 3)$ and $(2, 7)$ in the ratio $3:4$ is

Find the coordinates of the point which divides the line joining the points $(4, -3)$ and $(8, 5)$ in the ratio $3:1$ internally.

The coordinates of the midpoint of the sides of a $\triangle ABC$ are $D(2,1),\ E(5,3),\ F(3,7)$ . Find the length and equation of its sides.

If $Q$ is mid-point of line segment $AB$ and $P$ is mid-point of $QB$ , then show that $PB=\dfrac14AB$

$(-2.4),(4.8),(10.7)$ 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.

Find the area of triangle formed by joining the mid point sides of the whose vertices are A (0,-1) B (2,1) C(0,3) Find the ratio of this area to the area of $\bigtriangleup ABC$ .

If the mid-point of the line joining $(3,4)$ and $(k,7)$ is $(x,y)$ and $2x+2y+1=0$ , find the value of $k$ .

The coordinates of $A$ and $B$ are $(-3,\alpha)$ and $(1,\alpha+4)$ and the mid point of $\overline {AB}$ is $P(-1,1)$ . Find the value of $\alpha$

Find the ratio in which $(-8,3)$ divides the join of points $(2,-2)$ and $(-4,1).$