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

Let M be the point on X-axis which is equidistant from P(2,-5) and Q(-2,9).
Since the point M is on X-axis, its y co-ordinate is zero.
M = (x,0)
Since M is equidistant from P and Q,
PM = QM ………..(i)
by Distance formula, PM = √[(x₂-x₁)²+(y₂-y₁)²]
= √[(x-2)²+(0-(-5))²]
= √[(x-2)²+(5)²]
= √(x²-4x+4+25)
= √(x²-4x+29)
by Distance formula, QM = √[(x₂-x₁)²+(y₂-y₁)²]
= √[(x-(-2))²+(0-9)²]
= √[(x+2)²+(9)²]
= √(x²+4x+85)
From (i)
√(x²-4x+29) = √(x²+4x+85)
Squaring both sides
x²-4x+29 = x²+4x+85
-8x = 85-29
-8x = 56
x = 56/-8
x = -7
Hence the point on X axis equidistant from P(2,-5) and Q(-2,9) is (-7,0).