Question

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

Answer 1

we have PA = PB 

∴ PA² = PB² 

x² – 4x + 29 = x² + 4x + 85 

-4x – 4x = 85 – 29 -8x = 56 

8x = -56 

∴ x = \(\frac{-56}{8}\)

∴ x = – 7 

∴ Coordinates of P are (x, 0), it means (-7, 0). 

∴ Required point (x, 0) = (-7, 0) 

∴ x = -7.