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

Let P(x,0) be any point on x – axis, which is equidistant from A(2,-5) and B(-2,9).
Therefore, PA = PB
⇒ √((2-x)²+(-5-0)²) = √((-2-x)² + (9-0)²)
⇒ √(4+x²-4x +25) = √(4+x²+4x+81)
Squaring both the sides
4+x²-4x + 25 = 4+x² + 4x + 81
⇒ -8x = 81- 25 = 56
⇒ X = -56/8 = -7
Hence, P(-7,0) is the point on the x-axis which is equidistant from (2,-5) and (-2,9).

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