NCERT Solutions
Class 10 - Mathematics - Chapter 7: COORDINATE GEOMETRY - Exercise 7.1

Question 8

We are given two points: \(P(2, -3)\) and \(Q(10, y)\), and their distance is 10 units.

Use the distance formula: \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

Substitute the values: \( 10 = \sqrt{(10 - 2)^2 + (y + 3)^2} \).

Compute the square: \( (10 - 2)^2 = 8^2 = 64 \).

So the equation becomes: \( 10 = \sqrt{64 + (y + 3)^2} \).

Square both sides to remove the square root: \( 100 = 64 + (y + 3)^2 \).

Subtract 64 from both sides: \( (y + 3)^2 = 36 \).

Take square roots: \( y + 3 = 6 \) or \( y + 3 = -6 \).

So the solutions are: \( y = 3 \) or \( y = -9 \).