NCERT Class 10 Maths – Exercise 7.1 Solutions

NCERT Class 10 Maths

Chapter 7 – Coordinate Geometry | Exercise 7.1

(Rationalized Syllabus 2025-26)

💡 Distance Formula

The distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is given by:

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

Find the distance between the following pairs of points:

(i) (2, 3), (4, 1)

$$d = \sqrt{(4 – 2)^2 + (1 – 3)^2}$$

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

✔️ $2\sqrt{2}$ units

(ii) (-5, 7), (-1, 3)

$$d = \sqrt{(-1 – (-5))^2 + (3 – 7)^2}$$

$$= \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32}$$

✔️ $4\sqrt{2}$ units

(iii) (a, b), (-a, -b)

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

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

✔️ $2\sqrt{a^2 + b^2}$ units

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?

Let points be $A(0, 0)$ and $B(36, 15)$.

$$AB = \sqrt{(36 – 0)^2 + (15 – 0)^2}$$

$$= \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521}$$

$\sqrt{1521} = 39$.

✔️ 39 units

Yes, the distance between the two towns A and B is 39 km (assuming coordinates represent km).

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

Let $A(1, 5), B(2, 3), C(-2, -11)$.

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

$$BC = \sqrt{(-2 – 2)^2 + (-11 – 3)^2} = \sqrt{(-4)^2 + (-14)^2} = \sqrt{16 + 196} = \sqrt{212} = 2\sqrt{53}$$

$$AC = \sqrt{(-2 – 1)^2 + (-11 – 5)^2} = \sqrt{(-3)^2 + (-16)^2} = \sqrt{9 + 256} = \sqrt{265}$$

For collinearity, sum of two smaller distances must equal the largest distance.

$AB + BC = \sqrt{5} + 2\sqrt{53} \approx 2.23 + 14.56 = 16.79$.

$AC = \sqrt{265} \approx 16.27$.

Since $AB + BC \neq AC$ (and no other combination works), they are not collinear.

✖️ No, points are not collinear

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

Let $A(5, -2), B(6, 4), C(7, -2)$.

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

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

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

Since $AB = BC$, two sides are equal.

✔️ Yes, it is an isosceles triangle

In a classroom, 4 friends are seated at the points A, B, C and D. Determine if ABCD is a square.

Coordinates from figure: $A(3, 4), B(6, 7), C(9, 4), D(6, 1)$.

Sides:

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

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

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

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

All sides are equal. Now check diagonals.

Diagonals:

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

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

Sides are equal and diagonals are equal.

✔️ Yes, ABCD is a square

Name the type of quadrilateral formed, if any, by the following points:

(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)

Let points be A, B, C, D.

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

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

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

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

All sides equal. Check diagonals:

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

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

✔️ Square

(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)

Calculating distances:

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

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

$AC = \sqrt{(0+3)^2 + (3-5)^2} = \sqrt{9+4} = \sqrt{13}$.

Since $AC + BC = \sqrt{13} + \sqrt{13} = 2\sqrt{13} = AB$, points A, B, C are collinear.

✔️ No quadrilateral formed (Points are collinear)

(iii) (4, 5), (7, 6), (4, 3), (1, 2)

$AB = \sqrt{3^2 + 1^2} = \sqrt{10}$.

$BC = \sqrt{(-3)^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$.

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

$DA = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$.

Opposite sides are equal ($AB=CD, BC=DA$). It’s a parallelogram.

Check diagonals: $AC = 2$, $BD = \sqrt{52}$. Not equal.

✔️ Parallelogram

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

Let point on x-axis be $P(x, 0)$. Let $A(2, -5)$ and $B(-2, 9)$.

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

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

$$(x – 2)^2 + 25 = (x + 2)^2 + 81$$

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

$$-4x + 29 = 4x + 85$$

$$-8x = 56 \Rightarrow x = -7$$

✔️ Point is (-7, 0)

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

Given $PQ = 10 \Rightarrow PQ^2 = 100$.

$$(10 – 2)^2 + (y – (-3))^2 = 100$$

$$8^2 + (y + 3)^2 = 100$$

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

$$(y + 3)^2 = 36$$

Taking square root: $y + 3 = \pm 6$.

Case 1: $y + 3 = 6 \Rightarrow y = 3$.

Case 2: $y + 3 = -6 \Rightarrow y = -9$.

✔️ y = 3 or y = -9

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.

Given $QP = QR \Rightarrow QP^2 = QR^2$.

$$(5 – 0)^2 + (-3 – 1)^2 = (x – 0)^2 + (6 – 1)^2$$

$$25 + 16 = x^2 + 25$$

$$x^2 = 16 \Rightarrow x = \pm 4$$

Case 1 ($x=4$): $R(4, 6)$.

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

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

Case 2 ($x=-4$): $R(-4, 6)$.

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

$PR = \sqrt{(-4-5)^2 + (6-(-3))^2} = \sqrt{81+81} = \sqrt{162} = 9\sqrt{2}$.

✔️ x = ±4; QR = $\sqrt{41}$; PR = $\sqrt{82}$ or $9\sqrt{2}$

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 $P(x, y)$ be equidistant from $A(3, 6)$ and $B(-3, 4)$.

$$PA^2 = PB^2$$

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

$$x^2 – 6x + 9 + y^2 – 12y + 36 = x^2 + 6x + 9 + y^2 – 8y + 16$$

Cancel $x^2, y^2, 9$ from both sides:

$$-6x – 12y + 36 = 6x – 8y + 16$$

$$-12x – 4y + 20 = 0$$

Divide by -4:

$$3x + y – 5 = 0$$

✔️ Relation: 3x + y – 5 = 0

🎉 Exercise 7.1 Completed