# NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Written by Team Trustudies
Updated at 2021-05-07

## NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1

Q1 ) Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (-5, 7), (-1, 3)
(iii) (a, b), (- a, – b)

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Distance formula =$$\sqrt { (x_2 - x_1)^2 + (y_2 - y_1)^2}$$
(i) Applying Distance Formula to find distance between points (2, 3) and (4,1), we get
d = $$\sqrt { (4 - 2)^2 + (1 - 3)^2}$$
$$= \sqrt{ 2^2 + (-2) ^2} = \sqrt{4 + 4}$$
$$= \sqrt{8}$$
$$= 2\sqrt{2}$$

(ii) Applying Distance Formula to find distance between points (–5, 7) and (–1, 3), we get
d = $$\sqrt { (-5 + 1)^2 + (7 - 3)^2}$$
$$= \sqrt{ ((-4)^2 + (4) ^2} = \sqrt{16 + 16}$$
$$= \sqrt{32}$$
$$= 4\sqrt{2}$$

(iii) Applying Distance Formula to find distance between points (a, b) and (–a, –b), we get
d = $$\sqrt { (-a - a)^2 + (-b - b)^2}$$
$$= \sqrt{ ((-2a)^2 + (-2b) ^2}$$
$$= \sqrt{4a^2 + 4b^2}$$
$$= 2\sqrt{a^2 + b^2}$$

Q2 ) Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

The coordinates of point A are (0, 0) and coordinates of point B are (36, 15).
To find the distance between them, we use Distance formula:
d = $$\sqrt{(36-0)^2 + (15 - 0)^2}$$
$$= \sqrt{36^2 - 15^2}$$
$$= \sqrt{1296 + 225}$$
$$= \sqrt{1521}$$
$$= 39$$km

Q3 ) Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let A = (1, 5), B = (2, 3) and C = (–2, –11)
Using Distance Formula to find distance AB, BC and CA.
AB = $$\sqrt{(2-1)^2 + (3 - 5)^2}$$
$$= \sqrt{1^2 - (-2)^2}$$
$$= \sqrt{1 + 5}$$
$$= \sqrt{5}$$
BC = $$\sqrt{(-2-2)^2 + (-11 - 3)^2}$$
$$= \sqrt{(-4)^2 - (-14)^2}$$
$$= \sqrt{16 + 196}$$
$$= \sqrt{212}$$
CA = $$\sqrt{(-2-1)^2 + (-11 - 5)^2}$$
$$= \sqrt{(-3)^2 - (-15)^2}$$
$$= \sqrt{9 + 256}$$
$$= \sqrt{265}$$
Since AB + AC $$\ne$$ BC, BC + AC $$\ne$$ AB and AC $$\ne$$ BC.
Therefore, the points A, B and C are not collinear.

Q4 ) Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let A = (5, –2), B = (6, 4) and C = (7, –2)
Using Distance Formula to find distances AB, BC and CA.

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

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

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

Therefore, A, B and C are vertices of an isosceles triangle.

Q5 ) In a classroom, 4 friends are seated at the points A (3, 4), B (6, 7), C (9, 4) and D (6, 1). Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli. “Don’t you think ABCD is a square?”Chameli disagrees. Using distance formula, find which of them is correct.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

We have A = (3, 4), B = (6, 7), C = (9, 4) and D = (6, 1)
Using Distance Formula to find distances AB, BC, CD and DA, we get

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

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

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

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

Therefore, All the sides of ABCD are equal here. … (1)

Now, we will check the length of its diagonals.

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

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

So, Diagonals of ABCD are also equal. … (2)

From (1) and (2), we can definitely say that ABCD is a square.

Therefore, Champa is correct.

Q6 ) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0)
(ii) (- 3, 5), (3, 1), (0, 3), (- 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

i) Let A = (–1, –2), B = (1, 0), C= (–1, 2) and D = (–3, 0)
Using Distance Formula to find distances AB, BC, CD and DA, we get

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

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

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

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

Therefore, all four sides of quadrilateral are equal. … (1)
Now, we will check the length of diagonals.

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

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

Therefore, diagonals of quadrilateral ABCD are also equal. … (2)

From (1) and (2), we can say that ABCD is a square.

(ii) Let A = (–3, 5), B= (3, 1), C= (0, 3) and D= (–1, –4)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = $$\sqrt{(-3 - 3)^2 + (1 - 5)^2}$$
$$= \sqrt{36 + 16}$$
$$= 2 \sqrt{13}$$

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

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

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

We cannot find any relation between the lengths of different sides.

Therefore, we cannot give any name to the figure ABCD.

(iii) Let A = (4, 5), B= (7, 6), C= (4, 3) and D= (1, 2)
Using Distance Formula to find distances AB, BC, CD and DA, we get

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

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

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

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

Here opposite sides of quadrilateral ABCD are equal. … (1)

We can now find out the lengths of diagonals.

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

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

Here diagonals of ABCD are not equal. … (2)

From (1) and (2), we can say that ABCD is not a rectangle therefore it is a parallelogram.

Q7 ) Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let the point be (x, 0) on x–axis which is equidistant from (2, –5) and (–2, 9).

Using Distance Formula and according to given conditions we have:

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

=>$$\sqrt {x^2 + 4 - 4x + 25}$$
$$= \sqrt {x^2 + 4 + 4x + 81}$$

Squaring both sides, we get
$$\Rightarrow x^2 + 4 - 4x + 25 = x^2 + 4 + 4x + 81$$
$$\Rightarrow-4x + 29 = 4x + 85$$
$$\Rightarrow 8x = -56$$
$$\Rightarrow x = -7$$

Therefore, point on the x–axis which is equidistant from (2, –5) and (–2, 9) is (–7, 0)

Q8 ) Find the values of y for which the distance between the points P (2, –3) and Q (10, y) is 10 units.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Using Distance formula, we have
$$\Rightarrow 10 = \sqrt{(2 -10)^2 + (-3-y)^2}$$
$$\Rightarrow 10 = \sqrt{(-8)^2 + 9 + y^2 + 6y}$$
$$\Rightarrow 10 = \sqrt{64 + 9 + y^2 + 6y}$$
Squaring both sides, we get
$$\Rightarrow 100 = 73 + y^2 + 6y$$
$$\Rightarrow y^2 + 6y + 27 = 0$$

Solving this Quadratic equation by factorization, we can write
$$\Rightarrow y^2 + 9y - 3y - 27 = 0$$
$$\Rightarrow y (y + 9) – 3 (y + 9) = 0$$
$$\Rightarrow (y + 9) (y - 3) = 0$$
$$\Rightarrow y = 3, -9$$

Q9 ) If, Q (0, 1) is equidistant from P (5, –3) and R (x, 6), find the values of x. Also, find the distances QR and PR.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

It is given that Q is equidistant from P and R. Using Distance Formula, we get
PQ = RQ

PQ=$$\sqrt{(5-0)^2 + (-3 - 1)^2}$$
$$= \sqrt{(-5)^2 + (-4)^2}$$
$$= \sqrt{25 + 16}$$
$$= \sqrt{41}$$
QR= $$\sqrt{(0 - x)^2 + (1 - 6)^2}$$
$$= \sqrt{(-x)^2 + (-5)^2}$$
$$= \sqrt{x^2 + 25}$$
$$\therefore \sqrt{41} = \sqrt{x^2 + 25}$$
On squaring both sides,
$$\Rightarrow 41 = x^2 + 25$$
$$\Rightarrow x^2 = 16$$
$$\Rightarrow x = \pm4$$
Coordinates of point R will be (4,6) or (-4,6),
If R is (4,6),
QR = $$\sqrt{(0 - 4)^2 + (1 - 6)^2}$$
$$= \sqrt{4^2 + 5^2}$$
$$= \sqrt{16 + 25}$$
$$= \sqrt{41}$$
PR = $$\sqrt{(5 - 4)^2 + (-3 - 6)^2}$$
$$= \sqrt{1^2 + 9^2}$$
$$= \sqrt{1 + 81}$$
$$= \sqrt{82}$$
If R is (-4,6),
QR = $$\sqrt{(0 + 4)^2 + (1 - 6)^2}$$
$$= \sqrt{4^2 + 5^2}$$
$$= \sqrt{16 + 25}$$
$$= \sqrt{41}$$
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).

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

It is given that (x, y) is equidistant from (3, 6) and (–3, 4).
Using Distance formula, we can write

$$\Rightarrow \sqrt{(x-3)^2 + (y-6)^2} = \sqrt{(x-(-3))^2 + (y-4)^2}$$
$$\Rightarrow \sqrt{x^2 + 9 - 6x + y^2 + 36 -12y } = \sqrt{x^2 + 9 + 6x + y^2 + 16 -8y }$$
Squaring both sides, we get
$$\Rightarrow x^2 + 9 - 6x + y^2 + 36 -12y = x^2 + 9 + 6x + y^2 + 16 -8y$$
$$\Rightarrow -6x - 12y + 45 = 6x - 8y + 25$$
$$\Rightarrow 12x + 4y = 20$$
$$\Rightarrow 3x + y = 5$$

## NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2

Q1 ) Find the coordinates of the point which divides the join of (- 1, 7) and (4, – 3) in the ratio 2:3.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let the points be P(x,y), Using the section formula

$$P(x, y) = [\frac{mx_2 + nx_1}{m + n} , \frac{my_2 + ny_1}{m + n}]$$

$$x = {{2(4) + 3(-1)} \over {2 + 3}} = {{8 - 3} \over {5}} = 1$$
$$y = {{2(-3) + 3(7)} \over {2 + 3}} = {{-6 + 21} \over {5}} = 3$$
Therefore the point is(1,3)

Q2 ) Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let P (x1, y1) and Q (x2, y2) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB

Therefore, point P divides AB internally in the ratio 1:2.

x1 = $$\frac{(1×(-2)+2×4)}{1+2} = \frac{(-2+8)}{3} =\frac{6}{3} = 2$$

y1 = $$\frac{(1×(-3)+2×-1)}{1+2} = \frac{(-3-2)}{3} =\frac{-5}{3}$$

$$\therefore$$P = (x1, y1) = P(2, $$\frac{-5}{3}$$)

Point Q divides AB internally in the ratio 2:1.

x2 = $$\frac{(2×(-2)+1×4)}{2+1} = \frac{(-4+4)}{3} = 0$$

y2 = $$\frac{(2×(-3)+1×-1)}{2+1} = \frac{(-6-1)}{3} =\frac{-7}{3}$$

The coordinates of the point Q(x2 ,y2 ) = (0, $$\frac{-7}{3}$$ )

Q3 ) To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs $$\frac{1}{4}$$ th the distance AD on the 2nd line and posts a green flag. Preet runs $$\frac{1}{5}$$th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

The green flag is $$\frac{1}{4} th$$ of total distance =
$$\frac{1}{4} × 100 = 25$$ m in second line

$$\therefore$$ the coordinate of green flag
= (2 , 25 )
Similarly coordinate of red flag
= (8 , 20)

Distance between red and green flag
= $$\sqrt{(8-2)^2 + (20-25)^2}$$
$$= \sqrt{6^2+(-5)^2}$$
$$= \sqrt{36+25}$$
$$= \sqrt{61} m$$

Now blue flag is posted at the mid point of two flag
Then let, the coordinate of blue flag = (x , y )

$$\therefore( x , y ) =( \frac{(8+2)}{2} , \frac{20+25}{2} )$$
$$=( 5 , \frac{45}{2} )= ( 5, 22.5 )$$

$$\therefore$$ the blue flag is posted in fifth line at a distance of 22.5 m

Q4 ) Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6).

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let the required ratio be k : 1
By section formula ,

$$x = \frac{m1x2 + m2x1}{m1+m2}$$
$$\Rightarrow -1 = \frac{k×6 + 1× -3 )}{k+1}$$
$$\Rightarrow -k-1 = 6k -3$$
$$\Rightarrow 7k = 2$$
$$\Rightarrow k = \frac{2}{7}$$

The required ratio = k : 1
= $$\frac{2}{7}$$ : 1 = $$\frac{2}{7} ×7 : 1×7$$ = 2 : 7

Q5 ) Find the ratio in which the line segment joining A (1, – 5) and B (- 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let p(x , 0) be the point which devides the line segment joining A(1, -5 ) and B(-4,5 ) in the ratio m : 1

Then the using section formula

(x , 0 ) = $$( \frac{m×(-4) + 1 × 1 }{m+1} , \frac{m×5+1×(-5)}{m+1} )$$
$$\Rightarrow 0 = \frac{m×5+1×(-5)}{m+1}$$
$$\Rightarrow 0 = 5m - 5$$
$$\Rightarrow m = 5 /5 = 1$$

Hence the required ratio is 1 : 1

Since the ratio is 1 : 1 , so P is the mid point

$$\therefore$$ x = $$\frac{m×(-4) + 1 × 1 }{m+1}$$
x = $$\frac{1×(-4) + 1 × 1 }{1+1} = \frac{-4+1}{2} = \frac{-3}{2}$$

$$\therefore ( \frac{-3}{2} , 0 )$$ is the required point

Q6 ) If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Mid point of AC = Mid point BD
$$\Rightarrow \frac{x+1}{2} , \frac{6+2}{2} = \frac{4+3}{2} , \frac{y+5}{2}$$
$$\Rightarrow \frac{x+1}{2} = \frac{7}{2}$$
$$\Rightarrow x +1 = 7$$
$$\Rightarrow x = 7-1= 6$$
$$\Rightarrow \frac{y+5}{2} = \frac{8}{2}$$
$$\Rightarrow y+5 = 8$$
$$\Rightarrow y = 8 -5 = 3$$

Q7 ) Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, – 3) and B is (1,4).

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let the coordinates of point A be (x, y).

Mid-point of AB is (2, – 3), which is the centre of the circle.

Coordinate of B = (1, 4)
(2, -3) =$$(\frac{x+1}{2} , \frac{y+4}{2} )$$
$$\frac{x+1}{2} = 2$$
$$\Rightarrow x+1= 4$$
$$\Rightarrow x = 4-1 = 3$$
and$$\frac{y+4}{2} = -3$$
$$\Rightarrow y+4 = -6$$
$$\Rightarrow y = -6-4 = -10$$
$$\therefore$$ x = 3 and y = -10
The coordinates of A(3,-10).

Q8 ) If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = $$\frac{3}{7}$$ AB and P lies on the line segment AB.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

AP = $$\frac{3}{7}$$ AB
7 AP = 3 AB
AB : AP = 7: 3
Let AB = 7x
$$\therefore$$ AP = 3x
AB = AP + BP
7x = 3x + BP
BP = 7x - 3x = 4x

$$\therefore \frac{AP}{BP} = \frac{3x}{4x}$$
$$\therefore$$ AP : BP = 3:4
by section formula ,

(x, y ) = $$(\frac{m1x2 + m2x1}{m1+m2} , \frac{m1y2 + m2y1}{m1+m2} )$$
(x, y) = $$(\frac{3×2 +4×(-2) }{3+4} , \frac{3×(-4)+ 4(-2)}{3+4} )$$
(x ,y ) = $$(\frac{6 -8 }{7} , \frac{-12+ -8}{7} )$$
(x ,y ) = $$(\frac{-2}{7} , \frac{-20}{7} )$$

Hence the coordinates of P = $$(\frac{-2}{7} , \frac{-20}{7} )$$

Q9 ) Find the coordinates of the points which divide the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

From the figure, it can be observed that points X, Y, Z are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively.

Coordinate of X = $$( \frac{1×2+3×(-2)}{1+3} , \frac{1×8+3×2}{1+3} )$$
$$=( \frac{2-6}{4} , \frac{8+6}{4} )$$
$$= ( \frac{-4}{4} , \frac{14}{4} )$$
$$= ( -2 , \frac{7}{2} )$$
Coordinate of Y = $$( \frac{1×2+1×(-2)}{1+1} , \frac{1×8+1×2}{1+1} )$$
$$=( \frac{2-2}{2} , \frac{8+2}{2} )$$
$$= ( \frac{0}{2} , \frac{10}{2} )$$
$$= ( 0 , 5)$$
Coordinate of Y = $$( \frac{3×2+1×(-2)}{3+1} , \frac{3×8+1×2}{3+1} )$$
$$=( \frac{6-2}{4} , \frac{24+2}{4} )$$
$$= ( \frac{4}{4} , \frac{26}{4} )$$
$$= ( 1 , \frac{13}{2} )$$

Q10 ) 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 = $$\frac{1}{2}$$ (product of its diagonals)

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let A(3, 0), B (4, 5), C( – 1, 4) and D ( – 2, – 1) are the vertices of a rhombus ABCD.

Length of diagonal AC
$$=\sqrt{[3-(-1)]^2 + (0-4)^2}$$
$$= \sqrt{16+16}$$
$$= \sqrt{32}$$
$$= 4\sqrt{2}$$
Length of diagonal BD
$$=\sqrt{[(-2)-4]^2+ (-1-5)^2}$$
$$= \sqrt{(-6)^2+(-6)^2}$$
$$= \sqrt{36+36}$$
$$= \sqrt{72}$$
$$= 6\sqrt{2}$$

Area of rhombus = $$\frac{1}{2}$$ × product of diagonals
= $$\frac{1}{2} × 4\sqrt{2}×6\sqrt{2} = 24$$sq unit

## NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.3

Q1 ) Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

We know that formula for area of a triangle whose vertices are $$(x_1,y_1) , (x_2,y_2) , (x_3,y_3)$$ is,
= $$\frac{1}{2} \ | \ [x_1(y_2 - y_3) \ + \ x_2(y_3 - y_1) \ + \ x_3(y_1 - y_2)] \ |$$

(i) So, here $$x_1 \ = \ 2 \ , \ y_1 \ = \ 3 \ , \ x_2 \ = \ -1 \ , \ y_2 \ = \ 0 \ x_3 \ = \ 2 \ , \ y_3 \ = \ -4$$

So, area of triangle = $$\frac{1}{2} \ | \ [ 2( 0 + 4 ) \ - \ 1(-4 - 3) \ + \ 2(3 - 0)] \ | \ = \ \frac{1}{2} \ | \ [ 8 + 7 + 6 ] \ | \ = \ \frac{21}{2}$$

∴ Area of triangle is $$\frac{21}{2}$$ sq. units.

(ii) Similarly, here $$x_1 \ = \ -5 \ , \ y_1 \ = \ -1 \ , \ x_2 \ = \ 3 \ , \ y_2 \ = \ -5 \ x_3 \ = \ 5 \ , \ y_3 \ = \ 2$$
So, area of triangle = $$\frac{1}{2} \ | \ [ (-5)( -5 - 2 ) \ - \ 3(2 + 1) \ + \ 5(-1 + 5)] \ | \ = \ \frac{1}{2} \ | \ [ 35 + 9 + 20 ] \ | \ = \ 32$$

∴ Area of triangle is $$32$$ sq. units.

Q2 ) In each of the following find the value of ‘k’, for which the points are collinear.
(i) (7, -2), (5, 1), (3, -k)
(ii) (8, 1), (k, -4), (2, -5)

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

We know that for collinear points area of triangle = 0 ,i.e.,
$$0 \ = \frac{1}{2}[x_1(y_2 - y_3) \ + \ x_2(y_3 - y_1) \ + \ x_3(y_1 - y_2)]$$

(i) $$7(1-k) + 5(k+2) + 3(-2-1) \ = \ 0$$
=> $$7 - 7k + 5k + 10 - 9 \ = \ 0$$
=> $$2k \ = \ 8$$
=> $$k \ = \ 4$$

(ii) $$8(-4+5) + k(-5-1) + 2(1+4) \ = \ 0$$
=> $$8 - 6k + 10 \ = \ 0$$
=> $$6k \ = \ 18$$
=> $$k \ = \ 3$$

Q3 ) Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

The vertices of the triangle ABC are A (0, -1), B (2, 1), C (0, 3).
Let D, E, F be the mid-points of the sides AB,BC,CA of this triangle ABC respectively.
Coordinates of D, E, and F are given by :

D = $$(\frac{0+2}{2} \ , \ \frac{-1+1}{2})$$ = (1,0)
E = ( $$\frac{0+2}{2} \ , \ \frac{1+3}{2}$$ ) = (1,2)
F = ( $$\frac{0+0}{2} \ , \ \frac{3-1}{2}$$ ) = (0,1)

So, Area of $$∆ \ DEF \ = \ \frac{1}{2} \ | \ [ 1(2-1) + 1(1-0) + 0(0-2)] \ | \ = \ 1$$

∴ Area of ∆ DEF = 1 sq. units

Area of ∆ ABC = $$\frac{1}{2} \ | \ [0(1-3) \ + \ 2(3+1) \ + \ 0(-1-1) ] \ | \ = \ 4$$

$$\therefore$$ Area of $$\triangle$$ ABC = 4 sq. units

So, $$∆ \ DEF \ : \ ∆ \ ABC \ = \ 1 \ : \ 4$$.

Q4 ) Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Draw a line from B to D

Now we have,
Area of quad. ABCD = | Area of ∆ DAB | + | Area of ∆ BCD |

So,
Area of ∆ DAB = $$| \ \frac{1}{2} [2(-2+5) \ - \ 4(-5-3) \ - \ 3(3+2)] \ |$$ = $$\ | \frac{1}{2} [6 + 32 - 15 ] \ |$$ = $$\frac{23}{2}$$ sq. units

Similarly, Area of ∆ BCD = $$| \ \frac{1}{2} [-3(-2-3) \ + \ 3(3+5) \ + \ 2(-5+2)] \ | \ = \ | \ \frac{1}{2} [15 + 24 - 6 ] \ |$$ = $$\frac{33}{2}$$ sq. units

$$\therefore$$ Area of quad. ABCD = $$\frac{23}{2} \ + \ \frac{33}{2}$$ = $$\frac{56}{2} \ = \ 28$$ sq. units

Q5 ) You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for $$\triangle$$ ABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

The vertices of triangle ABC are A(4,-6), B (3, – 2) and C (5, 2).

Let D be the mid-point of side BC of ∆ ABC. Therefore, AD is the median in ∆ ABC.

So, coordinates of D are ( $$\frac{3+5}{2} \ , \ \frac{-2+2}{2}$$ )
Therefore, coordinates of D = (4,0)

Now,
Area of ∆ ABD = $$\frac{1}{2} \ | \ [4(-2-0) \ + \ 3(0+6) \ + \ 4(-6+2)] |$$
= $$\frac{1}{2} \ | \ [ -8 + 18 - 16] \ |$$
= $$|-3| \ = \ 3$$ sq. units

Similarly,
Area of ∆ ACD = $$\frac{1}{2} \ | \ [4(0-2) \ + \ 4(2+6) \ + \ 5(-6-0)] \ |$$
= $$\frac{1}{2} \ | \ [ -8 + 32 - 30] \ |$$
= $$|-3| \ = \ 3$$ sq. units

$$\therefore$$ Clearly, Area of ∆ ABD = Area of ∆ ACD = 3 sq. units.

Hence, median of a triangle divides it into two triangles of equal areas.

## NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.4

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).

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let the line 2x + y - 4 = 0 divides AB in 1 : k ratio, then the coordinates of the point of division is

$$x \ = \ \frac{2+3k}{k+1} \ and \ y \ = \ \frac{-2+7k}{k+1}$$

The point of intersection lie on both lines

$$\therefore x \ = \ \frac{2+3k}{k+1} \ and \ y \ = \ \frac{-2+7k}{k+1}$$ should satisfy 2x + y - 4 = 0
$$\Rightarrow 2( \frac{2+3k}{k+1}) \ + \ (\frac{-2+7k}{k+1}) \ - \ 4 \ = \ 0$$
$$\Rightarrow 4 \ + \ 6k \ - \ 2 \ + \ 7k \ = \ 4k \ + \ 4$$
$$\Rightarrow 9k - 2 = 0$$
$$\Rightarrow k \ = \ \frac{2}{9}$$

$$\therefore$$ The line 2x + y – 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7) in ratio 2 : 9 internally.

Q2 ) Find the relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

If given points are collinear then area of triangle formed by them must be zero.

Let (x, y), (1, 2) and (7, 0) are vertices of a triangle,

Area of triangle = $$\frac{1}{2} \ | \ [ (x_1(y_2 – y_3 ) + x_2(y_3 – y_1) + x_3(y_1 – y_2) ) ] \ | \ = \ 0$$

$$\Rightarrow \frac{1}{2} \ | \ [x(2 – 0) + 1 (0 – y) + 7( y – 2)] \ | \ = \ 0$$

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

$$\Rightarrow 2x \ + \ 6y \ – \ 14 \ = \ 0$$

$$\Rightarrow x \ + \ 3y \ – \ 7 \ = \ 0$$ .

Hence this is the required relation between x and y.

Q3 ) Find the centre of a circle passing through points (6, -6), (3, -7) and (3, 3).

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let the coordinate of centre O be (x,y)

Then, OA = OB = OC {$$\because$$ radius is equal}

Now,
OA = $$\sqrt{ (x-6)^2+(y+6)^2 }$$
OB = $$\sqrt{(x-3)^2 + (y+7)^2 }$$
OC = $$\sqrt{(x-3)^2 + (y-3)^2 }$$

Now, OA = OB
$$\Rightarrow \sqrt{(x-6)^2 + (y+6)^2 } = \sqrt{(x-3)^2 + (y+7)^2 }$$
$$\Rightarrow x^2 \ + \ 36 \ - \ 12x \ + \ y^2 \ + \ 36 \ - \ 12 \ = \ x^2 \ + \ 9 \ - \ 6x \ + \ y^2 \ + \ 9 \ - \ 6y$$
$$\Rightarrow 6x \ + 2y \ - \ 14 \ = \ 0$$ ..............(i)

Similarly, OB = OC
$$\Rightarrow \sqrt{(x-3)^2 + (y+7)^2 } = \sqrt{(x-3)^2 + (y-3)^2 }$$
$$\Rightarrow y^2 \ + \ 14y \ + \ 49 \ = \ y^2 \ - \ 6y \ + \ 9$$
$$\Rightarrow y \ = \ -2$$

Putting this value of y in eq. (i) , we get

$$\Rightarrow 6x \ + \ 2(-2) \ - \ 14 \ = \ 0$$
$$\Rightarrow x \ = \ 3$$

$$\therefore$$ The centre of circle is at (3,-2)

Q4 ) The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Let ABCD be a square with vertices A( $$x_1, y_1) , B(3 , 2) , C(x_2 , y_2) , D(-1 , 2)$$

Let each side of square be 'a'.
So applying Pythagoras theorem in ∆ABD, we get

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

Now, DA = AB {$$\because$$ ABCD is a square }

$$\Rightarrow \sqrt{(x_1 +1)^2 + (y_1 - 2)^2 } \ = \ \sqrt{(3-x_1)^2 + (2 - y_1)^2}$$ {by distance formula}
$$\Rightarrow (x_1 +1)^2 \ = \ (3-x_1)^2$$
$$\Rightarrow x_1^2 \ + \ 1 \ + \ 2x_1 \ = \ x_1^2 \ + \ 9 \ - \ 6x_1$$
$$\Rightarrow x_1 \ = \ 1$$

Now since each side is $$2 \sqrt{2}$$

$$\therefore$$ AB $$= \ 2 \sqrt{2}$$
$$\Rightarrow \sqrt{(3-x_1)^2 + (2 - y_1)^2} \ = \ 2 \sqrt{2}$$
$$\Rightarrow(2-y_1)^2 = \ 4$$
$$\Rightarrow y \ = \ 4$$

Now, point O is id point of BD . SO coordinates of O are

x = $$\frac{3-1}{2}$$ , y = $$\frac{2+2}{2}$$
$$\Rightarrow$$ x = 1 , y = 2

Now, since ABCD is a square $$\therefore$$ O is mid point of AC also

$$\frac{1+x_2}{2} \ = \ 1$$ , $$\frac{4+ y_2}{2} \ = \ 2$$
$$\Rightarrow x_2 \ = \ 1$$ , $$y_2 \ = \ 0$$

$$\therefore$$ Coordinates of vertices square ABCD are A(1, 4) , B(3,2) , C(1,0) , D(-1,2)

Q5 ) The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular lawn in the plot as shown in the fig. 7.14. The students are to sow the seeds of flowering plants on the remaining area of the plot.

(i) Taking A as origin, find the coordinates of the vertices of the triangle.

(ii) What will be the coordinates of the vertices of triangle PQR if C is the origin?

Also calculate the areas of the triangles in these cases. What do you observe?

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

(i) Taking A as origin, coordinates of the vertices P, Q and R are,

From figure:
P = (4, 6)
Q = (3, 2)
R = (6, 5)

Here AD is the x-axis and AB is the y-axis.

Area of triangle PQR in case of origin A:

Area of a triangle = $$\frac{1}{2} \ | \ [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)] \ | \$$
$$= \ \frac{1}{2} \ | \ [– 12 – 3 + 24] \ | \$$
$$= \ \frac{9}{2}$$ sq. units

(ii) Taking C as origin, coordinates of vertices P, Q and R are,

From figure:
P = (12, 2)
Q = (13, 6)
R = (10, 3)

Here CB is the x-axis and CD is the y-axis.

Area of triangle PQR in case of origin C:

Area of a triangle = $$\frac{1}{2} \ | \ [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)] \ | \$$
$$= \frac{1}{2} \ | \ [36 + 13 – 40] \ | \$$
$$= \ \frac{9}{2}$$ sq unit

This implies, Area of triangle PQR at origin A = Area of triangle PQR at origin C

Area is same in both case because triangle remains the same no matter which point is considered as origin.

Q6 ) The vertices of a $$∆ \ ABC$$ are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that $$\frac{AD}{AB} \ = \ \frac{AE}{AC} \ = \ \frac{1}{4}$$ . Calculate the area of the $$∆ \ ADE$$ and compare it with area of $$∆ \ ABC$$ . (Recall Theorem 6.2 and Theorem 6.6)

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Given, the vertices of a $$∆ \ ABC$$ are A (4, 6), B (1, 5) and C (7, 2) and,
$$\frac{AD}{AB} \ = \ \frac{AE}{AC} \ = \ \frac{1}{4}$$
$$\Rightarrow \frac{AB}{AD} \ = \ \frac{AC}{AE} \ = \ 4$$
$$\Rightarrow \frac{AD + DB}{AD} \ = \ \frac{AE + EC}{AE} \ = \ 4$$
$$\Rightarrow 1 + \frac{DB}{AD} \ = \ 1 + \frac{EC}{AE} \ = \ 4$$
$$\Rightarrow \frac{DB}{AD} \ = \ \frac{EC}{AE} \ = \ 3$$

$$\therefore$$ Point D and E divides AB and AC in ratio 1 : 3

Now coordinates of D calculated by section formula are,
$$x \ = \ \frac{3(4) + 1(1)}{4} \ = \ \frac{13}{4}$$ and,
$$y \ = \ \frac{3(6) + 1(5)}{4} \ = \ \frac{23}{4}$$

Similarly, coordinates of E are,
$$x \ = \ \frac{1(7) + 3(4)}{4} \ = \ \frac{19}{4}$$ and,
$$y \ = \ \frac{1(2) + 3(6)}{4} \ = \ \frac{20}{4} \ = \ 5$$

Now area of triangle ADE is,

$$\frac{1}{2} \ | \ [4(5-2) \ + \ 1(2-6) \ + \ 7(6-5)] \ |$$
= $$\frac{1}{2} \ | \ [(12-4+7)] \ |$$
= $$\frac{15}{2}$$ sq. units

Similarly, area of triangle ABC is,

$$\frac{1}{2} \ | \ [4( \frac{23}{4} -5) \ + \ \frac{13}{4}(5-6) \ + \ \frac{19}{4}(6- \frac{23}{4})] \ |$$
= $$\frac{1}{2} \ | \ [(3- \frac{13}{4} + \frac{19}{4})] \ |$$
= $$\frac{15}{32}$$ sq. units

$$\therefore$$ ratio of $$area \ of \ ∆ \ ADE \ to \ area \ of \ ∆ \ ABC \ = \ 1:16$$

Q7 ) Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of $$∆ \ ABC$$ .

(i) The median from A meets BC at D. Find the coordinates of point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.

(iii) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2 : 1.

(iv) What do you observe?

[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.]

(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

(i) Given, D is the median to BC. So, coordinates of D using distance formula are

$$x \ = \ \frac{6+1}{2} \ = \ \frac{7}{2}$$ and $$y \ = \ \frac{5+4}{2} \ = \ \frac{9}{2}$$

(ii) Coordinates of P can be calculated using section formula

$$x \ = \ \frac{2( \frac{7}{2}) + 1(4)}{3} \ = \ \frac{11}{3}$$ and $$y \ = \ \frac{2( \frac{9}{2}) + 1(2)}{3} \ = \ \frac{11}{3}$$

(iii) Coordinates of E
= ( $$\frac{4+1}{2} \ , \ \frac{2+4}{2}$$ )
= ( $$\frac{5}{2} , 3$$ )

So, coordinates of Q are,
( $$\frac{2( \frac{5}{2}) + 1(6)}{3} \ , \ \frac{2(3) + 1(5)}{3}$$ )
= ( $$\frac{11}{3} , \frac{11}{3}$$ )

Coordinates of F
= ( $$\frac{4+6}{2} , \frac{2+5}{2}$$ )
= ( $$5, \frac{7}{2}$$ )

So, coordinates of R are,
( $$\frac{2(5) + 1(1)}{3} , \frac{2 ( \frac{7}{2}) + 1(4)}{3}$$ )
= ( $$\frac{11}{3} , \frac{11}{3}$$ )

(iv) We observed that coordinates of P , Q, R are same i.e., ( $$\frac{11}{3} , \frac{11}{3}$$ ) which shows that medians intersect each other at a common point which is called centroid of the triangle.

(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, then the coordinates of the centroid of the triangle is

$$x \ = \ \frac{x_1 + x_2 + x_3}{3} \ and \ \frac{y_1 + y_2 + y_3}{3}$$

Q8 ) 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 midpoints of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

P, Q, R and S are the midpoints of AB, BC, CD and DA respectively

So, coordinate of P
= ( $$\frac{-1-1}{2} , \frac{-1+4}{2}$$ )
= ( $$-1 , \frac{3}{2}$$ )

Similarly, coordinate of Q
= ( $$\frac{5-1}{2} , \frac{4+4}{2}$$ )
= ( 2,4)

R = ( $$\frac{5+5}{2} , \frac{4-1}{2}$$ ) = ( $$5 , \frac{3}{2}$$ )

S = ( $$\frac{5-1}{2} , \frac{-1-1}{2}$$ ) = ( 2,-1)

Now, PQ = $$\sqrt{(-1-2)^2 + ( \frac{3}{2} - 4)^2} \$$
$$= \ \sqrt{ \frac{61}{4} } \$$
$$= \ \frac{ \sqrt{61}}{2}$$

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

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

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

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

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

Clearly, PQ = QR = RS = SP

But, $$PR \ne QS$$

Hence the lengths of all the sides are equal, while the lengths of the diagonals are different.

$$\therefore$$ The PQRS is a rhombus.

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