Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9)

We have to find a point on x-axis.

Hence, its y-coordinate will be 0

Let the point on x-axis be (x, 0)

Distance between (x, 0) and (2, -5) = [(x- 2)² + (0+ 5)²]^1/2

= [(x- 2)² + (5)²]^1/2

Distance between (x, 0) and (-2, 9) = [(x + 2)² + (0 + 9)²]^1/2

= [(x + 2)² + (9)²]^1/2

By the given condition, these distances are equal in measure

[(x - 2)² + (5)²]^1/2 = [(x + 2)² + (9)²]^1/2

(x – 2)² + 25 = (x + 2)² + 81

x² + 4 – 4x + 25 = x² + 4 + 4x + 81

8x = 25 – 81

8x = -56

x = - 7

Therefore, the point is (- 7, 0)