# NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

NCERT solutions for class 10 maths chapter 7 Coordinate Geometry: A system of mathematics in which the points on a plane are described using an ordered pair of numbers is called coordinate geometry. In solutions of NCERT class 10 maths chapter 7 coordinate geometry, you will get a detailed explanation to each and every question present in the chapter including the optional exercises. Coordinate Geometry is one of the important chapters for higher mathematics in upcoming classes. You will use at least some part of coordinate geometry in the field whether you are going to study commerce, engineering, maths, medical, arts, etc.

CBSE NCERT solutions for class 10 maths chapter 7 Coordinate Geometry is designed to give you a step by step solution for a particular question. In Coordinate Geometry, you will study the algebraic forms of geometric figures. NCERT solutions for class 10 maths chapter 7 Coordinate Geometry will be helpful for doing homework as well as while preparing for the board examinations. Apart from this chapter, if you want NCERT solutions for other classes and subjects then you can download these solutions by clicking on the above link.

## NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Excercise: 7.1

Given points:  (2, 3), (4, 1)

Distance between the points will be: $(x_{1},y_{1})\ and\ (x_{2},y_{2})$

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

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

Given points:  (– 5, 7), (– 1, 3)

Distance between the points will be: $(x_{1},y_{1})\ and\ (x_{2},y_{2})$

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

$D= \sqrt{(-1+5)^2+(3-7)^2} = \sqrt{16+16} = 4\sqrt{2}$

Given points:  (a, b), (– a, – b)

Distance between the points will be: $(x_{1},y_{1})\ and\ (x_{2},y_{2})$

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

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

Given points:  (0, 0) and (36, 15)

Distance between the points will be: $(x_{1},y_{1})\ and\ (x_{2},y_{2})$

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

$D= \sqrt{(36-0)^2+(15-0)^2} =\sqrt{1296+225} = \sqrt{1521} = 39$

The distance between the two towns A and B is, thus 39 km for given towns location

$(0,0)$  and  $(36,15)$.

Let the points (1, 5), (2, 3) and (– 2, – 11) be representing the vertices A, B, and C of the given triangle respectively.

$A = (1,5),\ B = (2,3),\ C = (-2,-11)$

Therefore,

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

$BC = \sqrt{(2-(-2))^2+(3-(-11))^2} = \sqrt{4^2+14^2} = \sqrt{16+196} = \sqrt{212}$$CA = \sqrt{(1-(-2))^2+(5-(-11))^2} = \sqrt{3^2+16^2} = \sqrt{9+256} = \sqrt{265}$Since these are not satisfied.

$AB+BC \neq CA$

$BA+AC \neq BC$

$BC+CA \neq BA$

As these cases are not satisfied.

Hence the points are not collinear.

The distance between two points $A(x_{1},y_{1})\ and\ B(x_{2},y_{2})$ is given by:

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

So, we have the following points: (5, – 2), (6, 4) and (7, – 2) assuming it to be the vertices of triangle A, B, and C respectively.

$AB = \sqrt{(5-6)^2+(-2-4)^2} = \sqrt{1+36} = \sqrt{37}$

$BC = \sqrt{(6-7)^2+(4+2)^2} = \sqrt{1+36} = \sqrt{37}$

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

Therefore, AB = BC

Here two sides are equal in length.

Therefore, ABC is an isosceles triangle.

The coordinates of the points:

$A(3,4),\ B(6,7),\ C(9,4),\ and\ D(6,1)$ are the positions of 4 friends.

The distance between two points $A(x_{1},y_{1}), \ and\ B(x_{2},y_{2})$ is given by:

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

Hence,

$AB = \sqrt{(3-6)^2+(4-7)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$

$BC = \sqrt{(6-9)^2+(7-4)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$

$CD = \sqrt{(9-6)^2+(4-1)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$

$AD= \sqrt{(3-6)^2+(4-1)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$

And the lengths of diagonals:

$AC = \sqrt{(3-9)^2+(4-4)^2} =\sqrt{36+0} = 6$

$BD = \sqrt{(6-6)^2+(7-1)^2} =\sqrt{36+0} = 6$

So, here it can be seen that all sides of quadrilateral ABCD are of the same lengths and diagonals are also having the same length.

Therefore, quadrilateral ABCD is a square and Champa is saying right.

Let the given points $(-1,-2),\ (1,0),\ (-1,2),\ and\ (-3,0)$ be representing the vertices A, B, C, and D of the given quadrilateral respectively.

The distance formula:

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

$AB= \sqrt{(-1-1)^2+(-2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$

$BC= \sqrt{(1+1)^2+(0-2)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$

$CD= \sqrt{(-1+3)^2+(2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$

$AD= \sqrt{(-1+3)^2+(-2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$

Finding the length of the diagonals:

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

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

It is clear that all sides are of the same lengths and also the diagonals have the same lengths.

Hence, the given quadrilateral is a square.

Let the given points $(-3,5),\ (3,1),\ (0,3),\ and\ (-1,-4)$ be representing the vertices A, B, C, and D of the given quadrilateral respectively.

The distance formula:

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

$AB= \sqrt{(-3-3)^2+(5-1)^2} =\sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$

$BC= \sqrt{(3-0)^2+(1-3)^2} =\sqrt{9+4} = \sqrt{13}$

$CD= \sqrt{(0+1)^2+(3+4)^2} =\sqrt{1+49} = \sqrt{50} = 5\sqrt2$

$AD= \sqrt{(-3+1)^2+(5+4)^2} =\sqrt{4+81} = \sqrt{85}$

All the sides of the given quadrilateral have different lengths.

Therefore, it is only a general quadrilateral and not a specific one like square, rectangle, etc.

Let the given points $(4,5),\ (7,6),\ (4,3),\ (1,2)$ be representing the vertices A, B, C, and D of the given quadrilateral respectively.

The distance formula:

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

$AB= \sqrt{(4-7)^2+(5-6)^2} =\sqrt{9+1} = \sqrt{10}$

$BC= \sqrt{(7-4)^2+(6-3)^2} =\sqrt{9+9} = \sqrt{18}$

$CD= \sqrt{(4-1)^2+(3-2)^2} =\sqrt{9+1} = \sqrt{10}$

$AD= \sqrt{(4-1)^2+(5-2)^2} =\sqrt{9+9} = \sqrt{18}$

And the diagonals:

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

$BD =\sqrt{(7-1)^2+(6-2)^2} = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$

Here we can observe that the opposite sides of this quadrilateral are of the same length.

However, the diagonals are of different lengths.

Therefore, the given points are the vertices of a parallelogram.

Let the point which is equidistant from $A(2,-5)\ and \ B(-2,9)$ be $X(x,0)$ as it lies on X-axis.

Then, we have

Distance AX: $= \sqrt{(x-2)^2+(0+5)^2}$

and Distance BX $= \sqrt{(x+2)^2+(0+9)^2}$

According to the question, these distances are equal length.

Hence we have,

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

Solving this to get the required coordinates.

Squaring both sides we get,

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

$\Rightarrow (x-2+x+2)(x-2-x-2)= 81 - 25 = 56$

$\Rightarrow (-8x)= 56$

Or, $\Rightarrow x = -7$

Hence the point is $X(-7,0)$.

Given the distance between the points $P(2,-3)$ and $Q(10,y)$ is 10 units.

The distance formula :

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

So, given $PQ = 10\ units$

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

After squaring both sides

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

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

$\Rightarrow y+3 = \pm 6$

$\Rightarrow y = 6 - 3\ or\ y = -6-3$

Therefore, the values are  $y = 3\ or -9$.

Given $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x,6)$.

Then, the distances $PQ = RQ$.

Distance $PQ = \sqrt{(5-0)^2+(-3-1)^2} = \sqrt{25+16} = \sqrt{41}$

Distance $RQ = \sqrt{(x-0)^2+(6-1)^2} = \sqrt{x^2+25}$

$\Rightarrow \sqrt{x^2+25} = \sqrt{41}$

Squaring both sides, we get

$\Rightarrow x^2 = 16$

$\Rightarrow x = \pm 4$

The points are: $R(4,6)\ or\ R(-4,6.)$

CASE I: when R is  $(4,6)$

The distances QR and PR.

$QR = \sqrt{(0-4)^2+(1-6)^2} = \sqrt{16+25} = \sqrt{41}$

$PR = \sqrt{(5-4)^2+(-3-6)^2} = \sqrt{1^2+(-9)^2} = \sqrt{1+81} = \sqrt{82}$

CASE II: when R is  $(-4,6)$

The distances QR and PR.

$QR = \sqrt{(0-(-4))^2+(1-6)^2} = \sqrt{16+25} = \sqrt{41}$

$PR = \sqrt{(5-(-4))^2+(-3-6)^2} = \sqrt{9^2+(-9)^2} = \sqrt{81+81} = 9\sqrt{2}$

## Q10 Find a relation between x and y such that the point (x, y) is equidistant from the point  (3, 6) and (– 3, 4).

Let the point $P(x,y )$ is equidistant from $A(3,6)$ and $B(-3,4)$.

Then, the distances $AP =BP$

$AP = \sqrt{(x-3)^2+(y-6)^2}$  and  $BP = \sqrt{(x-(-3))^2+(y-4)^2}$

$\Rightarrow \sqrt{(x-3)^2+(y-6)^2} = \sqrt{(x-(-3))^2+(y-4)^2}$

Squaring both sides: we obtain

$\Rightarrow (x-3)^2+(y-6)^2= (x+3)^2+(y-4)^2$

$\Rightarrow (2x)(-6)+(2y-10)(-2)= 0$        $\left [\because a^2-b^2 = (a+b)(a-b) \right ]$

$\Rightarrow -12x-4y+20 = 0$

$\Rightarrow 3x+y-5 = 0$

Thus, the relation is  $3x+y-5 = 0$ between x and y.

## NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Excercise: 7.2

Let the coordinates of point $P(x,y)$ which divides the line segment joining the points $A(-1,7)$ and $B(4,-3)$, internally, in the ratio $m_{1}:m_{2}$ then,

Section formula: $\left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

Substituting the values in the formula:

Here, $m_{1}:m_{2} = 2:3$

$\Rightarrow \left (\frac{2(4)+3(-1)}{2+3} , \frac{2(-3)+3(7)}{2+3} \right )$

$\Rightarrow \left (\frac{5}{5} , \frac{15}{5} \right )$

Hence the coordinate is $P \left (1 , 3 \right )$.

Let the trisection of the line segment $A(4,-1)$ and  $B(-2,-3)$ have the points $P(x_{1},y_{1})$  and $Q(x_{2},y_{2})$

Then,

Section formula: $\left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

By observation point, P divides AB internally in the ratio $1:2$.

Hence, $m:n = 1:2$

Substituting the values in the equation we get;

$\Rightarrow P\left (\frac{1(-2)+2(4)}{1+2} , \frac{1(-3)+2(-1)}{1+2} \right )$

$\Rightarrow P \left (\frac{-2+8}{3} , \frac{-3-2}{3} \right )$

$\Rightarrow P \left (2 , \frac{-5}{3} \right )$

And by observation point Q, divides AB internally in the ratio $2:1$

Hence, $m:n = 2:1$

Substituting the values in the equation above, we get

$\Rightarrow Q\left (\frac{2(-2)+1(4)}{2+1} , \frac{2(-3)+1(-1)}{2+1} \right )$

$\Rightarrow Q \left (\frac{-4+4}{3} , \frac{-6-1}{3} \right )$

$\Rightarrow Q\left (0 , \frac{-7}{3} \right )$

Hence, the points of trisections are $\dpi{100} P \left (2 , \frac{-5}{3} \right )$  and $\dpi{100} Q\left (0 , \frac{-7}{3} \right )$

Niharika posted the green flag at the distance P, i.e.,

$\frac{1}{4}\times100\ m = 25\ m$  from the starting point of $2^{nd}$ line.

Therefore, the coordinates of this point $P$are $(2,25).$

Similarly, Preet posted red flag at $\frac{1}{5}$ of the distance Q i.e.,

$\frac{1}{5}\times100\ m = 20\ m$  from the starting point of $8^{th}$ line.

Therefore, the coordinates of this point Q are $(8,20)$.

The distance $PQ$ is given by,

$PQ = \sqrt{(8-2)^2+(25-20)^2} = \sqrt{36+25} = \sqrt{61} m$

and the point at which Rashmi should post her Blue Flag is the mid-point of the line joining these points. Let this point be $R(x,y)$.

Then, by Section Formula, 

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

$x = \frac{2+8}{2},\ y = \frac{25+20}{2}$

$x = 5,\ y = 22.5$

Therefore, Rashmi should post her Blue Flag at 22.5 m on the 5th line.

Let the ratio be : $k:1$

Then, By section formula:

$P(x,y) = \left (\frac{kx_{2}+x_{1}}{k+1} , \frac{ky_{2}+y_{1}}{k+1} \right )$

Given point $P(x,y) = (-1,6)$

$-1 = \frac{6k-3}{k+1}$

$\Rightarrow -k-1 = 6k-3$

$\Rightarrow k = \frac{7}{2}$

Hence, the point $P$ divides the line AB in the ratio $2:7$.

Let the point on the x-axis be $P(x,0)$ and it divides it in the ratio $k:1$.

Then, we have

Section formula:

$P(x,y) = \left (\frac{kx_{2}+x_{1}}{k+1} , \frac{ky_{2}+y_{1}}{k+1} \right )$

$\implies \frac{ky_{2}+y_{1}}{k+1} = 0$

$k =-\frac{y_{1}}{y_{2}}$

Hence, the value of k will be: $k =-\frac{-5}{5}= 1$

Therefore, the x-axis divides the line in the ratio $1:1$ and the point will be,

Putting the value of $k=1$ in section formula.

$P(x,0) = \left ( \frac{x_{2}+x_{1}}{2}, 0 \right )$

$P(x,0) = \left ( \frac{1-4}{2}, 0 \right ) = \left ( \frac{-3}{2}, 0 \right )$

Let the given points $A(1,2),\ B(4,y),\ C(x,6),\ D(3,5)$.

Since the diagonals of a parallelogram bisect each other. Intersection point O of diagonal AC and BD also divides these diagonals.

Therefore, O is the mid-point of AC and BD.

The coordinates of the point O when it is mid-point of AC.

$\left ( \frac{1+x}{2}, \frac{2+6}{2} \right ) \Rightarrow \left ( \frac{x+1}{2}, 4 \right )$

The coordinates of the point O when it is mid-point of BD.

$\left ( \frac{4+3}{2}, \frac{5+y}{2} \right ) \Rightarrow \left ( \frac{7}{2}, \frac{5+y}{2} \right )$

Since both coordinates are of same point O.

Therefore,

$\frac{x+1}{2} =\frac{7}{2}$  and  $4 = \frac{5+y}{2}$

Or,

$x = 6\ and\ y = 3$

As the centre point $C(2,-3)$ will be the mid-point of the diameter AB.

Then, the coordinates of point A will be $A(x,y)$.

Given point $B(1,4)$.

Therefore,

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

$\frac{x+1}{2} = 2\ and\ \frac{y+4}{2} = -3$

$\Rightarrow x = 3\ and\ y = -10$.

Therefore, the coordinates of A are $(3,-10).$

From the figure:

As $AP = \frac{3}{7}AB$

$\Rightarrow PB = \frac{4}{7}AB$ hence the ratio is  3:4,

Now, from the section formula, we can find the coordinates of Point P.

Section Formula:

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

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

$P(x,y)= \left (\frac{6-8}{7} , \frac{-12-8}{7} \right )$

$P(x,y)= \left (\frac{-2}{7} , \frac{-20}{7} \right )$

From the figure:

Points C, D, and E divide the line segment AB into four equal parts.

Now, from the section formula, we can find the coordinates of Point C, D, and E.

Section Formula:

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

Here point D divides the line segment AB into two equal parts hence

$D(x_{2},y_{2})= \left (\frac{-2+2}{2} , \frac{2+8}{2} \right )$

$D(x_{2},y_{2})= \left (0 , 5 \right )$

Now, point C divides the line segment AD into two equal parts hence

$C(x_{1},y_{1})= \left (\frac{-2+0}{2} , \frac{2+5}{2} \right )$

$C(x_{2},y_{2})= \left (-1 , \frac{7}{2} \right )$

Also, point E divides the line segment DB into two equal parts hence

$E(x_{1},y_{1})= \left (\frac{2+0}{2} , \frac{8+5}{2} \right )$

$E(x_{2},y_{2})= \left (1 , \frac{13}{2} \right )$

From the figure:

Let the vertices of the rhombus are:

$A(3,0),\ B(4,5),\ C(-1,4),\ D(-2,-1)$

Area of the rhombus ABCD is given by;

$= \frac{1}{2}\times(Product\ of\ lengths\ of\ diagonals)$

Hence we have to find the lengths of the diagonals AC and BD of the rhombus.

The distance formula:

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

Length of the diagonal AC:

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

Length of the diagonal BD:

$BD = \sqrt{(4-(-2))^2+(5-(-1))^2} = \sqrt{36+36} = 6\sqrt{2}$

Thus, the area will be,

$= \frac{1}{2}\times (AC)\times(BD)$

$= \frac{1}{2}\times (4\sqrt{2})\times(6\sqrt{2}) = 24\ square\ units.$

## Q1 (i) Find the area of the triangle whose vertices are : (2, 3), (–1, 0), (2, – 4)

As we know, the area of a triangle with vertices (x1,x2) ,(y1, y2) and (z1 z2 ) is given by :

$A= \frac{1}{2}(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))$

So Area of a triangle  whose vertices are(2, 3), (–1, 0)and (2, – 4) is

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

$A= \frac{1}{2}[8+7+6]$

$A= \frac{1}{2}[21]$

$A= \frac{21}{2}$

$A= 10.5\:unit^2$

Hence, the area of the triangle is 10.5 per unit square.

From the figure:

Area of the triangle is given by:

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

Substituting the values in the above equation, we obtain

$Area = \frac{1}{2}\left [ (-5)((-5)-(-2))+3(2-(1))+5(-1-(-5)) \right ]$

$= \frac{1}{2}\left [ 35+9+20 \right ] = 32 \ square\ units.$

The points (7, –2), (5, 1), (3, k) are collinear if the area of the triangle formed by the points will be zero.

Area of the triangle is given by:

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

Substituting the values in the above equation, we obtain

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

$\left [ 7-7k+5k+10-9 \right ] = 0$

$\Rightarrow -2k+8 = 0$

$\Rightarrow k = 4$

Hence, the points are collinear for k=4.

The points (8,1), (k, -4), (2,-5) are collinear if the area of the triangle formed by these points will be zero.

Area of the triangle is given by:

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

Substituting the values in the above equation, we obtain

$\frac{1}{2}\left [ 8(-4-(-5))+k((-5)-1)+2(1-(-4)) \right ] = 0$

$\Rightarrow 8-6k+10 = 0$

$\Rightarrow 6k = 18$

$\Rightarrow k = 3$

Hence, the points are collinear for k = 3.

From the figure:

The coordinates of the point P, Q, and R are:

Point P is the midpoint of side AB, hence the coordinates of P are :

$P(x_{1},y_{1}) = \left (\frac{0+2}{2}, \frac{3+1}{2} \right ) = \left (1, 2 \right )$

Point Q is the midpoint of side AC, hence the coordinates of Q are :

$Q(x_{2},y_{2}) = \left (\frac{2+0}{2}, \frac{1-1}{2} \right ) = \left (1, 0 \right )$

Point R is the midpoint of side BC, hence the coordinates of R are :

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

Hence, the area of the triangle formed by the midpoints PQR will be,

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

$= \frac{1}{2}\left [ (2-1)+1(1-0)+0(0-2) \right ]$

$=\frac{1}{2}(1+1) = 1\ square\ units.$

And the area formed by the triangle ABC will be:

$Area_{(ABC)} = \frac{1}{2}\left [ 0(1-3)+2(3-(-1))+0(-1-1) \right ]$

$= \frac{1}{2}\left [ 8 \right ] = 4\ square\ units.$

Thus, the ratio of Area of $\triangle PQR$ to the Area of $\triangle ABC$  will be $1:4$.

From the figure:

The coordinates are  $A(-4,-2),\ B(-3,-5),\ C(3,-2)\ and\ D(2,3)$

Divide the quadrilateral into 2 parts of triangles.

Then the area will be, $ABC + ADC$

Area of the triangle formed by ABC will be,

$Area_{(ABC)} = \frac{1}{2}\left [ (-4)((-5)-(-2))+(-3)((-2)-(-2))+3((-2)-(-2)) \right ]$                       $= \frac{1}{2}\left [ 12+0+9 \right ] = \frac{21}{2}\ Square\ units.$

Area of the triangle formed by ADC will be,

$Area_{(ADC)} = \frac{1}{2}\left [ (-4)((-2)-(-3))+3(3-(-2))+2((-2)-(-2)) \right ]$                        $= \frac{1}{2}\left [ 20+15+0 \right ] = \frac{35}{2}\ Square\ units.$

Therefore, the area of the quadrilateral will be:

$= \frac{21}{2}+\frac{35}{2} = 28\ square\ units.$

Alternatively,

The points A and C are in the same ordinates.

Hence, the length of base AC will be $(3-(-4)) = 7\ units.$

Therefore,

Area of triangle ABC:

$= \frac{1}{2} \times (Base) \times (Height) = \frac{1}{2}\times(7)(3)$

$= \frac{1}{2} \times (Base) \times (Height) = \frac{1}{2}\times(7)(5)$

Therefore, the area will be, $\frac{1}{2}\times(7)\times(5+3) =28\ square\ units.$

From the figure:

The coordinates of midpoint M of side BC is:

$M = \left ( \frac{3+5}{2}, \frac{-2+2}{2} \right ) = \left ( 4,0 \right )$

Now, calculating the areas of the triangle ABM and ACM :

Area of triangle, ABM:

$Area_{(ABM)} = \frac{1}{2}\left [ 4((-2)-0)+3(0-(-6))+4((-6)-(-2)) \right ]$

$= \frac{1}{2}\left [ -8+18-16 \right ] = 3\ Square\ units.$

Area of triangle, ACM:

$Area_{(ACM)} = \frac{1}{2}\left [ 4(0-(-2))+4(2-(-6))+5((-6)-0) \right ]$

$= \frac{1}{2}\left [ -8+32-30 \right ] = -3\ Square\ units.$

However, the area cannot be negative, Therefore, area of $\triangle ACM$ is 3 square units.

Clearly, the median AM divided the $\triangle ABC$ in two equal areas.

## Q1 Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

Let the line divide the line segment AB in the ratio  $k:1$ at point C.

Then, the coordinates of point C will be:

$C(x,y) = \left ( \frac{3k+2}{k+1},\frac{7k-2}{k+1} \right )$

Point C will also satisfy the given line equation $2x + y - 4 = 0$, hence we have

$\Rightarrow 2\left ( \frac{3k+2}{k+1} \right )+\left (\frac{7k-2}{k+1} \right ) - 4 = 0$

$\Rightarrow \frac{6k+4+7k-2-4k-4}{k+1} = 0$

$\Rightarrow 9k-2 = 0$

$\Rightarrow k=\frac{2}{9}$

Therefore, the ratio in which the line $2x + y - 4 = 0$ divides the line segment joining the points $A(2,-2)$ and $B(3,7)$ is $2:9$ internally.

If the points $(x, y), (1, 2)\ and\ (7, 0)$ are collinear then, the area formed by these points will be zero.

The area of the triangle is given by,

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

Substituting the values in the above equation, we have

$Area = \frac{1}{2}\left [ x(2-0)+1(0-y)+7(y-2) \right ]= 0$

$\Rightarrow 2x-y+7y-14= 0$

Or,

$\Rightarrow x+3y-7= 0$

Hence, the required relation between x and y is $x+3y-7= 0$.

From the figure:

Let the center point be $O(x,y)$.

Then the radii of the circle $OA,\ OB,\ and\ OC$  are equal.

The distance OA:

$OA = \sqrt{(x-6)^2+(y+6)^2}$

The distance OB:

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

The distance OC:

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

Equating the radii of the same circle.

When equating, $OA = OB$

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

Squaring both sides and applying $a^2-b^2 = (a+b)(a-b)$

$\Rightarrow (x-6+x-3)(x-6-x+3)+(y+6+y+7)(y+6-y-7) = 0$

$\Rightarrow (2x-9)(-3) + (2y+13)(-1) = 0$

$\Rightarrow -6x+27-2y-13 = 0$  or

$\Rightarrow 3x+y -7= 0$                                 ...................................(1)

When equating, $OA = OC$

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

Squaring both sides and applying $a^2-b^2 = (a+b)(a-b)$

$\Rightarrow (x-6+x-3)(x-6-x+3)+(y+6+y-3)(y+6-y+3) = 0$

$\Rightarrow (2x-9)(-3) + (2y+3)(9) = 0$

$\Rightarrow -3x+9y+27 = 0$                 ...................................(2)

Now, adding the equations (1) and (2), we get

$\Rightarrow 10y = -20$

$\Rightarrow y = -2$.

From equation (1), we get

$\Rightarrow 3x-2 = 7$

$\Rightarrow 3x =9$

$\Rightarrow x =3$

Therefore, the centre of the circle is $(3,-2)$.

From the figure:

We know that the sides of a square are equal to each other.

Therefore, AB = BC

So,

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

Squaring both sides, we obtain

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

Now, doing $\left ( a^2-b^2 = (a+b)(a-b) \right )$

We get

$\implies (x-1+x-3)(x-1-x+3) = 0$

Hence $x = 2$.

Applying the Pythagoras theorem to find out the value of y.

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

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

$\Rightarrow \left (\sqrt{1+(y-2)^2} \right )^2 + \left (\sqrt{1+(y-2)^2} \right )^2 = \left (\sqrt{16} \right )^2$

$\Rightarrow \left ({1+(y-2)^2 \right ) + \left (1+(y-2)^2 \right ) = 16$

$\Rightarrow (y-2)^2 = 7$

Taking A as origin then, the coordinates of P, Q, and R can be found by observation:

Coordinates of point P is $(4,6).$

Coordinates of point Q is $(3,2).$

Coordinates of point R is $(6,5).$

The area of the triangle, in this case, will be:

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

$=\frac{1}{2}\left [ 4(2-5)+3(5-6)+6(6-2) \right ]$

$=\frac{1}{2}\left [ -12-3+24\right ] = \frac{9}{2}\ Square\ units.$

Taking C as origin, then CB will be x-axis and CD be y-axis.

The coordinates fo the vertices P, Q, and R are: $(12,2),\ (13,6),\ (10,3).$ respectively.

The area of the triangle, in this case, will be:

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

$=\frac{1}{2}\left [ 12(6-3)+13(3-2)+10(2-6) \right ]$

$=\frac{1}{2}\left [ 36-13+40 \right ] = \frac{9}{2}\ Square\ units.$

It can be observed that in both cases the area is the same so, it means that the area of any figure does not depend on the reference which you have taken.

From the figure:

Given ratio:

$\frac{AD }{AB} = \frac{AE }{AC } = \frac{1}{4}$

Therefore, D and E are two points on side AB and AC respectively, such that they divide side AB an AC in the ratio of $1:3$.

Section formula:

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

Then, coordinates of point D:

$D(x_{1},y_{1})= \left (\frac{1\times1+3\times 4}{1+3} , \frac{1\times 5+3\times 6}{1+3} \right )$

Coordinates of point E:

$E(x_{2},y_{2})= \left (\frac{1\times7+3\times 4}{1+3} , \frac{1\times 2+3\times 6}{1+3} \right )$

$= \left ( \frac{19}{4}, \frac{20}{4} \right )$

Then, the area of a triangle:

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

Substituting the values in the above equation,

$Area\ of\ \triangle ADE = \frac{1}{2}\left [ 4\left ( \frac{23}{4} - \frac{20}{4}\right )+\frac{13}{4}\left ( \frac{20}{4} - 6 \right )+\frac{19}{4}\left (6-\frac{23}{4} \right )\right ]$$= \frac{1}{2}\left [ 3-\frac{13}{4} +\frac{19}{16}\right ] = \frac{1}{2}\left [ \frac{48-52+19}{16} \right ] = \frac{15}{32}\ square\ units.$

$Area\ of\ \triangle ABC = \frac{1}{2}\left [ 4(5-2)+1(2-6)+7(6-5) \right ]$

$= \frac{1}{2}\left [ 12-4+7 \right ] = \frac{15}{2}\ Square\ units.$

Hence the ratio between the areas of $\triangle ADE$ and $\triangle ABC$ is $1:16.$

From the figure:

Let AD be the median of the triangle

Then, D is the mid-point of BC

Coordinates of Point D:

$\left ( \frac{6+1}{2},\frac{5+4}{2} \right ) = \left ( \frac{7}{2}, \frac{9}{2} \right )$

From the figure,

The point P divides the median AD in the ratio, AP: PD = 2: 1

Hence using the section formula,

$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

$P(x,y)= \left (\frac{2\times\frac{7}{2}+1\times4}{2+1} , \frac{2\times\frac{9}{2}+1\times2}{2+1} \right ) = \left ( \frac{11}{3}, \frac{11}{3} \right )$

From the figure,

$\Rightarrow$ The point Q divides the median BE in the ratio, BQ : QE = 2 : 1

Hence using the section formula,

$Q(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

$Q(x,y)= \left (\frac{2\times\frac{5}{2}+1\times6}{2+1} , \frac{2\times3+1\times5}{2+1} \right ) = \left ( \frac{11}{3}, \frac{11}{3} \right )$

$\Rightarrow$ The point R divides the median CF in the ratio, CR: RF = 2: 1

Hence using the section formula,

$R(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$

$R(x,y)= \left (\frac{2\times 5+1\times1}{2+1} , \frac{2\times\frac{7}{2}+1\times4}{2+1} \right ) = \left ( \frac{11}{3}, \frac{11}{3} \right )$

We observed that the coordinates of P, Q, and R are the same. Therefore, all these are representing the same point on the plane. i.e., the centroid of the triangle.

From the figure,

Let the median be AD which divides the side BC into two equal parts.

Therefore, D is the mid-point of side BC.

Coordinates of D:

$= \left ( \frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2} \right )$

Let the centroid of this triangle be O.

Then, point O divides the side AD in a ratio 2:1.

Coordinates of O:

$= \left ( \frac{2\times\frac{x_{2}+x_{3}}{2}+1\times x_{1}}{2+1}, \frac{2\times\frac{y_{2}+y_{3}}{2}+1\times y_{1} }{2+1} \right )$

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

From the figure:

P is the mid-point of side AB.

Therefore, the coordinates of P are, $\left ( \frac{-1-1}{2}, \frac{-1+4}{2} \right ) = \left ( -1, \frac{3}{2} \right )$

Similarly, the coordinates of Q, R and S are:$\left ( 2,4 \right ),\ \left ( 5, \frac{3}{2} \right ),\ and\ \left ( 2,-1 \right )$ respectively.

The distance between the points P and Q:

$PQ = \sqrt{(-1-2)^2+\left ( \frac{3}{2} -4 \right )^2} = \sqrt{9+\frac{25}{4}} = \sqrt{\frac{61}{4}}$

and the distance between the points Q and R:

$QR = \sqrt{(2-5)^2+\left ( 4-\frac{3}{2} \right )^2} = \sqrt{9+\frac{25}{4}} = \sqrt{\frac{61}{4}}$

Distance between points R and S:

$RS = \sqrt{(5-2)^2+\left ( \frac{3}{2}+1 \right )^2} = \sqrt{9+\frac{25}{4}} = \sqrt{\frac{61}{4}}$

Distance between points S and P:

$SP = \sqrt{(2+1)^2+\left ( -1-\frac{3}{2} \right )^2} = \sqrt{9+\frac{25}{4}} = \sqrt{\frac{61}{4}}$

Distance between points P and R the diagonal length:

$PR = \sqrt{(-1-5)^2+\left ( \frac{3}{2}-\frac{3}{2} \right )^2} = 6$

Distance between points Q and S the diagonal length:

$QS = \sqrt{(2-2)^2+\left ( 4+1 \right )^2} = 5$

Hence, it can be observed that all sides have equal lengths. However, the diagonals are of different lengths.

Therefore, PQRS is a rhombus.

## NCERT solutions for class 10 maths chapter wise

 Chapter No. Chapter Name Chapter 1 CBSE NCERT solutions for class 10 maths chapter 1 Real Numbers Chapter 2 NCERT solutions for class 10 maths chapter 2 Polynomials Chapter 3 Solutions of NCERT class 10 maths chapter 3 Pair of Linear Equations in Two Variables Chapter 4 CBSE NCERT solutions for class 10 maths chapter 4 Quadratic Equations Chapter 5 NCERT solutions for class 10 chapter 5 Arithmetic Progressions Chapter 6 Solutions of NCERT class 10 maths chapter 6 Triangles Chapter 7 CBSE NCERT solutions for class 10 maths chapter 7 Coordinate Geometry Chapter 8 NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry Chapter 9 Solutions of NCERT class 10 maths chapter 9 Some Applications of Trigonometry Chapter 10 CBSE NCERT solutions class 10 maths chapter 10 Circles Chapter 11 NCERT solutions  for class 10 maths chapter 11 Constructions Chapter 12 Solutions of NCERT class 10 chapter maths chapter 12 Areas Related to Circles Chapter 13 CBSE NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes Chapter 14 NCERT solutions for class 10 maths chapter 14 Statistics Chapter 15 Solutions of NCERT class 10 maths chapter 15 Probability

## How to use NCERT solutions for class 10 maths chapter 7 Coordinate Geometry?

• First of all, go through the concepts and examples given in the textbook to understand the basics of the chapter.

• Memorize the formulae to find out some directly asked questions.

• After doing the above two activities you can come to the practice exercises.

• During the practice of exercises, you can take the help of NCERT solutions for class 10 maths chapter 7 Coordinate Geometry.

• After doing all the above-mentioned things, you can come to practice the previous year's question to boost your preparation.

Keep working hard & happy learning!