Question

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

Answer 1

Let the point A on x-axis equidistant from P(2, –5) and Q(–2, 9).

Point A lies on the X-axis.
∴ Its y co-ordinate is 0.

Let A =(x, 0)

By using distance formula

= \[\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\]

\[AP = \sqrt{\left( x - 2 \right)^2 + \left[ 0 - \left( - 5 \right) \right]^2} = \sqrt{\left(x - 2 \right)^2 + 25}\]

\[QA = \sqrt{\left[ x - \left( - 2 \right) \right]^2 + \left( 0 - 9 \right)^2} = \sqrt{\left( x + 2 \right)^2 + 81}\]

\[AP = QA\]

\[ \Rightarrow \sqrt{\left( x - 2 \right)^2 + 25} = \sqrt{\left( x + 2 \right)^2 + 81}\]

Squaring both sides

∴ (x − 2)² + 25 = (x + 2)² + 81

∴ x² − 4x + 4 + 25 = x² + 4x − 4 + 81

∴ −8x = 56

∴ x = −7

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