Rbse Solutions for Class 10 Maths Chapter 7 Exercise 7.1 | Distance Formula

Get complete, step-by-step solutions for NCERT Class 10 Maths Chapter 7 Exercise 7.1.

Master the use of the Distance Formula to find the distance between pairs of points (Q.1, Q.2). Solutions cover essential coordinate geometry concepts: determining if three points are collinear (Q.3), checking if given coordinates form an isosceles triangle (Q.4), and classifying quadrilaterals (Square, Parallelogram, General Quadrilateral) by comparing side and diagonal lengths (Q.5, Q.6). Learn to solve for unknown coordinates or establish a relation between $x$ and $y$ for points equidistant from two given points (Q.7, Q.9, Q.10). Crucial practice for finding distances when coordinates contain variables (Q.8).

The problems in this exercise are solved using the Distance Formula for two points $P(x_1, y_1)$ and $Q(x_2, y_2)$:

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

1. Find the distance between the following pairs of points:

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

$$d = \sqrt{(4 – 2)^2 + (1 – 3)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = \mathbf{2\sqrt{2} \text{ units}}$$

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

$$d = \sqrt{(-1 – (-5))^2 + (3 – 7)^2} = \sqrt{(-1 + 5)^2 + (-4)^2} = \sqrt{(4)^2 + 16} = \sqrt{16 + 16} = \sqrt{32} = \mathbf{4\sqrt{2} \text{ units}}$$

(iii) $(a, b)$ and $(-a, -b)$

$$d = \sqrt{(-a – a)^2 + (-b – b)^2} = \sqrt{(-2a)^2 + (-2b)^2} = \sqrt{4a^2 + 4b^2} = \sqrt{4(a^2 + b^2)} = \mathbf{2\sqrt{a^2 + b^2} \text{ units}}$$

2. Distance between $(0, 0)$ and $(36, 15)$

$$d = \sqrt{(36 – 0)^2 + (15 – 0)^2} = \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = \mathbf{39 \text{ units}}$$

Distance between towns A and B (Section 7.2 context):

Section 7.2 often discusses a town A at $(0, 0)$ and a town B at $(36, 15)$.

Since the coordinates are the same as above, the distance between the two towns A and B is 39 km (assuming the units are km as is common in that example).

3. Determine if the points are collinear

Points $A(1, 5)$, $B(2, 3)$, and $C(-2, -11)$. Points are collinear if the sum of the distances between two pairs equals the distance of the third pair (e.g., $AB + BC = AC$).

1. Distance $AB$:$$AB = \sqrt{(2 – 1)^2 + (3 – 5)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$
2. Distance $BC$:$$BC = \sqrt{(-2 – 2)^2 + (-11 – 3)^2} = \sqrt{(-4)^2 + (-14)^2} = \sqrt{16 + 196} = \sqrt{212}$$
3. Distance $AC$:$$AC = \sqrt{(-2 – 1)^2 + (-11 – 5)^2} = \sqrt{(-3)^2 + (-16)^2} = \sqrt{9 + 256} = \sqrt{265}$$

Since $\sqrt{5} + \sqrt{212} \approx 2.23 + 14.56 = 16.79$, and $\sqrt{265} \approx 16.27$.

$AB + BC \neq AC$.

The points are not collinear.

4. Check for an Isosceles Triangle

Points $A(5, -2)$, $B(6, 4)$, and $C(7, -2)$. A triangle is isosceles if at least two sides have equal length.

1. Distance $AB$:$$AB = \sqrt{(6 – 5)^2 + (4 – (-2))^2} = \sqrt{1^2 + 6^2} = \sqrt{1 + 36} = \sqrt{37}$$
2. Distance $BC$:$$BC = \sqrt{(7 – 6)^2 + (-2 – 4)^2} = \sqrt{1^2 + (-6)^2} = \sqrt{1 + 36} = \sqrt{37}$$
3. Distance $AC$:$$AC = \sqrt{(7 – 5)^2 + (-2 – (-2))^2} = \sqrt{2^2 + 0^2} = \sqrt{4} = 2$$

Since $AB = BC = \sqrt{37}$, the triangle is an isosceles triangle.

5. Check if ABCD is a Square

The coordinates from the figure are typically:

• $A(3, 4)$
• $B(6, 7)$
• $C(9, 4)$
• $D(6, 1)$

For ABCD to be a square, it must satisfy:

1. All four sides are equal ($AB = BC = CD = DA$).
2. The diagonals are equal ($AC = BD$).
3. Side Lengths:
   • $AB = \sqrt{(6-3)^2 + (7-4)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18}$
   • $BC = \sqrt{(9-6)^2 + (4-7)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18}$
   • $CD = \sqrt{(6-9)^2 + (1-4)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18}$
   • $DA = \sqrt{(3-6)^2 + (4-1)^2} = \sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18}$(All sides are equal: $AB = BC = CD = DA = \sqrt{18}$)
4. Diagonal Lengths:
   • $AC = \sqrt{(9-3)^2 + (4-4)^2} = \sqrt{6^2 + 0^2} = \sqrt{36} = 6$
   • $BD = \sqrt{(6-6)^2 + (1-7)^2} = \sqrt{0^2 + (-6)^2} = \sqrt{36} = 6$(The diagonals are equal: $AC = BD = 6$)

Since all sides are equal and the diagonals are equal, ABCD is a square.

Champa is correct.

6. Name the type of quadrilateral formed

(i) $A(-1, -2), B(1, 0), C(-1, 2), D(-3, 0)$

1. Side Lengths:
   • $AB = \sqrt{(1 – (-1))^2 + (0 – (-2))^2} = \sqrt{2^2 + 2^2} = \sqrt{8}$
   • $BC = \sqrt{(-1 – 1)^2 + (2 – 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{8}$
   • $CD = \sqrt{(-3 – (-1))^2 + (0 – 2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{8}$
   • $DA = \sqrt{(-1 – (-3))^2 + (-2 – 0)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{8}$(All sides are equal)
2. Diagonal Lengths:
   • $AC = \sqrt{(-1 – (-1))^2 + (2 – (-2))^2} = \sqrt{0^2 + 4^2} = 4$
   • $BD = \sqrt{(-3 – 1)^2 + (0 – 0)^2} = \sqrt{(-4)^2 + 0^2} = 4$(Diagonals are equal)

Since all sides are equal and the diagonals are equal, it is a Square.

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

1. Side Lengths:
   • $AB = \sqrt{(3 – (-3))^2 + (1 – 5)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52}$
   • $BC = \sqrt{(0 – 3)^2 + (3 – 1)^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$
   • $CD = \sqrt{(-1 – 0)^2 + (-4 – 3)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50}$
   • $DA = \sqrt{(-3 – (-1))^2 + (5 – (-4))^2} = \sqrt{(-2)^2 + 9^2} = \sqrt{4 + 81} = \sqrt{85}$(All four sides are different)

Since all four side lengths are different, it is a General Quadrilateral (or no specific type of named quadrilateral).

(iii) $A(4, 5), B(7, 6), C(4, 3), D(1, 2)$

1. Side Lengths:
   • $AB = \sqrt{(7 – 4)^2 + (6 – 5)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$
   • $BC = \sqrt{(4 – 7)^2 + (3 – 6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18}$
   • $CD = \sqrt{(1 – 4)^2 + (2 – 3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$
   • $DA = \sqrt{(4 – 1)^2 + (5 – 2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18}$(Opposite sides are equal: $AB = CD = \sqrt{10}$ and $BC = DA = \sqrt{18}$)
2. Diagonal Lengths:
   • $AC = \sqrt{(4 – 4)^2 + (3 – 5)^2} = \sqrt{0^2 + (-2)^2} = 2$
   • $BD = \sqrt{(1 – 7)^2 + (2 – 6)^2} = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52}$(Diagonals are unequal: $AC \neq BD$)

Since opposite sides are equal and the diagonals are unequal, it is a Parallelogram.

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

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

We require $PA = PB$, or $PA^2 = PB^2$.

$$PA^2 = (x – 2)^2 + (0 – (-5))^2 = (x – 2)^2 + 25$$

$$PB^2 = (x – (-2))^2 + (0 – 9)^2 = (x + 2)^2 + 81$$

Set $PA^2 = PB^2$:

$$(x – 2)^2 + 25 = (x + 2)^2 + 81$$

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

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

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

The point on the x-axis is $\mathbf{(-7, 0)}$.

8. Find the values of $y$

Given $P(2, -3)$, $Q(10, y)$, and $PQ = 10$.

$$PQ^2 = (10 – 2)^2 + (y – (-3))^2$$

$$10^2 = 8^2 + (y + 3)^2$$

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

$$(y + 3)^2 = 100 – 64 = 36$$

Take the square root of both sides:

$$y + 3 = \pm \sqrt{36}$$

$$y + 3 = 6 \quad \text{or} \quad y + 3 = -6$$

$$y = 3 \quad \text{or} \quad y = -9$$

The values of $y$ are $\mathbf{3}$ and $\mathbf{-9}$.

9. Find $x$, and distances $QR$ and $PR$

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

$PQ = QR$, or $PQ^2 = QR^2$.

1. Find $x$:$$PQ^2 = (5 – 0)^2 + (-3 – 1)^2 = 5^2 + (-4)^2 = 25 + 16 = 41$$$$QR^2 = (x – 0)^2 + (6 – 1)^2 = x^2 + 5^2 = x^2 + 25$$Set $PQ^2 = QR^2$:$$41 = x^2 + 25$$$$x^2 = 41 – 25 = 16$$$$x = \pm 4$$The values of $x$ are $\mathbf{4}$ and $\mathbf{-4}$.
2. Find distances $QR$ and $PR$ (Two cases):
   • Case 1: $x = 4$ (Point $R$ is $(4, 6)$)
     • $QR = \sqrt{QR^2} = \sqrt{41}$.
     • $PR = \sqrt{(4 – 5)^2 + (6 – (-3))^2} = \sqrt{(-1)^2 + 9^2} = \sqrt{1 + 81} = \sqrt{82}$.
     • Distances: $QR = \mathbf{\sqrt{41}}$ and $PR = \mathbf{\sqrt{82}}$.
   • Case 2: $x = -4$ (Point $R$ is $(-4, 6)$)
     • $QR = \sqrt{QR^2} = \sqrt{41}$.
     • $PR = \sqrt{(-4 – 5)^2 + (6 – (-3))^2} = \sqrt{(-9)^2 + 9^2} = \sqrt{81 + 81} = \sqrt{162} = \mathbf{9\sqrt{2}}$.
     • Distances: $QR = \mathbf{\sqrt{41}}$ and $PR = \mathbf{9\sqrt{2}}$.

10. Find a relation between $x$ and $y$

Let $P(x, y)$ be the equidistant point, and $A(3, 6)$ and $B(-3, 4)$ be the two given points.

$PA = PB$, or $PA^2 = PB^2$.

$$PA^2 = (x – 3)^2 + (y – 6)^2$$

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

Set $PA^2 = PB^2$:

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

Expand the squares:

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

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

$$-6x – 12y + 36 = 6x – 8y + 16$$

Group $x$ and $y$ terms:

$$36 – 16 = 6x + 6x – 8y + 12y$$

$$20 = 12x + 4y$$

Divide the entire equation by 4:

$$5 = 3x + y \quad \text{or} \quad \mathbf{3x + y – 5 = 0}$$

The required relation is $\mathbf{3x + y – 5 = 0}$.

Last Updated on November 28, 2025 by Aman Singh