Find the distance between the points P(−7, 2) and Q(5, −3).

Class 10Coordinate GeometryGeometry — Coordinates

A17
B13
C $29$
D5

Step-by-Step Solution

PQ =

(5 - (- 7))^{2} + (- 3 - 2)^{2} = 1 2^{2} + (- 5)^{2} = 144 + 25 = 169 = 13

. The value 17 comes from adding 12 + 5 without squaring and taking the root;

29

comes from writing Δ

x

as 5 − 7 = −2, losing the sign of the −7.

Answer: B) 13

Key idea: Distance formula

**Distance** between two points P

(x

₁,

y

₁) and Q

(x

₂,

y

₂):

**PQ =

(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

**

This is just the Pythagoras theorem applied to the right triangle formed with horizontal and vertical legs.

**Example**: Distance between (3, 4) and (7, 1).
=

(7 - 3)^{2} + (1 - 4)^{2} = 16 + 9 = 25 = 5

.

**Special distance from origin**: distance of

(x, y)

from (0, 0) is

x^{2} + y^{2}

Master this chapter in 30 minutes a day.

Don't just memorise the answer — build a daily practice habit.

Practice 5 Similar Questions Free →

All Coordinate Geometry practice questions

More Class 10 Coordinate Geometry practice

A ship's radar sits at the origin and shows a buoy at the point (−9, −12), with each…The point A(k, −5) is 10 units away from B(2, 3). Find the NEGATIVE value of k.Using distances only, decide whether P(−2, −1), Q(1, 3) and R(4, 7) lie on one straight…For the points A(0, 1), B(3, 5) and C(7, 8) a student finds AB = 5 and BC = 5 and…Find the value of p for which the points (−3, 4), (p, 0) and (6, −8) are collinear.