Question

Find the point on X-axis which is equidistant from P(2, –5) and Q(–2, 9).

Answer 1

P(2, –5) and Q(–2, 9)

Let point A(x, y) on the X-axis be equidistant from the points P and Q.

Point A lies on the X-axis.

∴ Its y-coordinate is 0.

Let A = (x, 0)

Now, AP = AQ    ...(Given)

By using distance formula

`sqrt((x - 2)^2 + [0 - (-5)]^2) = sqrt([x - (-2)]^2 + (0 - 9)^2)`

⇒ `sqrt((x - 2)^2 + 5^2) = sqrt((x + 2)^2 + (-9)^2)`

⇒ `sqrt((x - 2)^2 + 25) = sqrt((x + 2)^2 + 81)`

Squaring both sides

∴ (x − 2)² + 25 = (x + 2)² + 81

∴ x² − 4x + 4 + 25 = x² + 4x + 4 + 81

∴ −4x − 4x = 81 − 25

∴ −8x = 56

∴ x = `56/-8`

∴ x = −7

∴ The point on the X-axis, equidistant from points P and Q, is (−7,0).