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NCERT Solutions · Class 10 Maths · Chapter 7

Chapter 7: Coordinate Geometry (Coordinate Geometry)

Step-by-step NCERT solutions for Class 10 Maths Chapter 7, Coordinate Geometry (2026-27 reprint): Exercise 7.1 on the distance formula (collinearity, isosceles triangles, squares and parallelograms) and Exercise 7.2 on the section formula, mid-points, trisection points and the area of a rhombus. All 20 questions are answered, with the key answer highlighted.

Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-10-maths/chapter-7-coordinate-geometry

Distance formula: $P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ . Section formula: the point dividing the join of $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ in the ratio $m_{1} : m_{2}$ is $(\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}, \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}})$ . The mid-point is $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$ .

Exercise 7.1

Find the distance between: (i) (2, 3), (4, 1) (ii) (–5, 7), (–1, 3) (iii) (a, b), (–a, –b)

Solution

$2^{2} + (- 2)^{2} = 8 = 22$
$4^{2} + (- 4)^{2} = 32 = 42$
$(- 2 a)^{2} + (- 2 b)^{2} = 2 a^{2} + b^{2}$

(i) 2√2 (ii) 4√2 (iii) 2√(a² + b²)

Find the distance between (0, 0) and (36, 15). Can you now find the distance between towns A and B of Section 7.2?

Solution

$3 6^{2} + 1 5^{2} = 1296 + 225 = 1521 = 39$

Town A is at (0, 0) and town B at (36, 15) (in km), so they are 39 km apart.

39 units; the towns are 39 km apart.

Determine whether (1, 5), (2, 3) and (–2, –11) are collinear.

Solution

Let A(1, 5), B(2, 3), C(–2, –11):

$A B = 1 + 4 = 5 \approx 2.24$
$B C = 16 + 196 = 212 \approx 14.56$
$A C = 9 + 256 = 265 \approx 16.28$

For collinear points, the sum of the two smaller distances must equal the largest. Here $A B + B C \approx 16.80 \neq = A C$ .

Not collinear.

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

Solution

$A B = 1 + 36 = 37$
$B C = 1 + 36 = 37$
$A C = 4 + 0 = 2$

Two sides are equal.

Yes, it is an isosceles triangle (AB = BC = √37).

Four friends sit at A, B, C and D in Fig. 7.8. Champa says ABCD is a square; Chameli disagrees. Who is correct?

Solution

From the figure, A(3, 4), B(6, 7), C(9, 4), D(6, 1).

ABCD: four equal sides and equal diagonals

Sides: $A B = B C = C D = D A = 9 + 9 = 32$
Diagonals: $A C = 6$ , $B D = 6$

All four sides are equal and the diagonals are equal, so ABCD is a square.

Champa is correct: ABCD is a square.

Name the type of quadrilateral formed, if any: (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)

Solution

(i) Each side $= 4 + 4 = 22$ , and the diagonals are $0 + 16 = 4$ and $16 + 0 = 4$ . Equal sides and equal diagonals: a square.

(ii) The points (–3, 5), (0, 3) and (3, 1) are collinear: from (–3, 5) to (0, 3) is $13$ , from (0, 3) to (3, 1) is $13$ , and from (–3, 5) to (3, 1) is $52 = 213$ . So no quadrilateral is formed.

(iii) $A B = 10$ , $B C = 18$ , $C D = 10$ , $D A = 18$ : opposite sides equal. The diagonals are $A C = 2$ and $B D = 52$ , unequal. So it is a parallelogram (not a rectangle).

(i) Square (ii) No quadrilateral (three points are collinear) (iii) Parallelogram

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

Solution

Let it be $(x, 0)$ : $(x - 2)^{2} + 25 = (x + 2)^{2} + 81 \Rightarrow - 8 x = 56 \Rightarrow x = - 7$

(–7, 0)

Find y if the distance between P(2, –3) and Q(10, y) is 10 units.

Solution

$64 + (y + 3)^{2} = 100 \Rightarrow (y + 3)^{2} = 36 \Rightarrow y + 3 = \pm 6$

y = 3 or y = –9

Q(0, 1) is equidistant from P(5, –3) and R(x, 6). Find x, and the distances QR and PR.

Solution

$Q P^{2} = 25 + 16 = 41$ and $Q R^{2} = x^{2} + 25$ . So $x^{2} = 16$ , $x = \pm 4$ , and $QR = 41$ .

If $x = 4$ : $P R = 1 + 81 = 82$
If $x = - 4$ : $P R = 81 + 81 = 92$

x = ±4; QR = √41; PR = √82 (x = 4) or 9√2 (x = –4)

Find a relation between x and y so that (x, y) is equidistant from (3, 6) and (–3, 4).

Solution

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

$- 6 x - 12 y + 45 = 6 x - 8 y + 25 \Rightarrow 12 x + 4 y = 20$

3x + y – 5 = 0

Exercise 7.2

Find the point dividing the join of (–1, 7) and (4, –3) in the ratio 2 : 3.

Solution

$(\frac{2 ( 4 ) + 3 ( - 1 )}{5}, \frac{2 ( - 3 ) + 3 ( 7 )}{5}) = (1, 3)$

(1, 3)

Find the points of trisection of the segment joining (4, –1) and (–2, –3).

Solution

The points divide it in the ratios 1 : 2 and 2 : 1:

1 : 2: $(\frac{- 2 + 8}{3}, \frac{- 3 - 2}{3}) = (2, - \frac{5}{3})$
2 : 1: $(\frac{- 4 + 4}{3}, \frac{- 6 - 1}{3}) = (0, - \frac{7}{3})$

(2, –5/3) and (0, –7/3)

Lines are drawn 1 m apart in a rectangular ground ABCD, with 100 flower pots 1 m apart along AD. Niharika runs 1/4 of AD on the 2nd line and posts a green flag; Preet runs 1/5 of AD on the 8th line and posts a red flag. Find the distance between the flags. Where should Rashmi post a blue flag halfway between them?

Solution

Take A as the origin, AB along the $x$ -axis and AD (100 m) along the $y$ -axis.

Green flag: 2nd line, $\frac{1}{4} \times 100 = 25$ m, so $(2, 25)$ .
Red flag: 8th line, $\frac{1}{5} \times 100 = 20$ m, so $(8, 20)$ .

Distance $= 6^{2} + 5^{2} = 61$ m. Mid-point $= (\frac{2 + 8}{2}, \frac{25 + 20}{2}) = (5, 22.5)$ .

The flags are √61 m apart; Rashmi should post the blue flag on the 5th line, 22.5 m from AD's starting end.

Find the ratio in which (–1, 6) divides the join of (–3, 10) and (6, –8).

Solution

Let the ratio be $k : 1$ : $- 1 = \frac{6 k - 3}{k + 1} \Rightarrow - k - 1 = 6 k - 3 \Rightarrow k = \frac{2}{7}$ . (Check with $y$ : $\frac{- 8 ( 2/7 ) + 10}{9/7} = \frac{54/7}{9/7} = 6$ ✓)

2 : 7

Find the ratio in which the x-axis divides the join of A(1, –5) and B(–4, 5), and the point of division.

Solution

Let the ratio be $k : 1$ . On the $x$ -axis $y = 0$ : $\frac{5 k - 5}{k + 1} = 0 \Rightarrow k = 1$ .

Point $= (\frac{- 4 + 1}{2}, 0) = (- \frac{3}{2}, 0)$

1 : 1, at (–3/2, 0)

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

Solution

The diagonals of a parallelogram bisect each other, so the mid-points of (1, 2)–(x, 6) and (4, y)–(3, 5) coincide:

$\frac{1 + x}{2} = \frac{7}{2} \Rightarrow x = 6$ ; $\frac{8}{2} = \frac{y + 5}{2} \Rightarrow y = 3$

x = 6, y = 3

AB is a diameter of a circle with centre (2, –3), and B is (1, 4). Find A.

Solution

The centre is the mid-point of AB: $A = (2 \times 2 - 1, 2 \times (- 3) - 4) = (3, - 10)$

A(3, –10)

A is (–2, –2) and B is (2, –4). Find P on AB such that AP = (3/7) AB.

Solution

$A P : P B = 3 : 4$ :

$P = (\frac{3 ( 2 ) + 4 ( - 2 )}{7}, \frac{3 ( - 4 ) + 4 ( - 2 )}{7}) = (- \frac{2}{7}, - \frac{20}{7})$

P(–2/7, –20/7)

Find the points that divide the join of A(–2, 2) and B(2, 8) into four equal parts.

Solution

The middle point is the mid-point of AB, $(0, 5)$ . The other two are the mid-points of each half:

between $(- 2, 2)$ and $(0, 5)$ : $(- 1, \frac{7}{2})$
between $(0, 5)$ and $(2, 8)$ : $(1, \frac{13}{2})$

(–1, 7/2), (0, 5) and (1, 13/2)

Find the area of the rhombus with vertices (3, 0), (4, 5), (–1, 4) and (–2, –1) in order.

Solution

The diagonals join (3, 0)–(–1, 4) and (4, 5)–(–2, –1):

$d_{1} = 16 + 16 = 42$
$d_{2} = 36 + 36 = 62$

Area $= \frac{1}{2} d_{1} d_{2} = \frac{1}{2} \times 42 \times 62 = 24$

24 square units

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