Question

Which point on the -axis is equidistant from (7, 6) and (– 3, 4).

Answer 1

Let the point on x-axis be P(x, 0) and the given points be A(7, 6) and B(– 3, 4). 

∴ AP = BP 

⇒ \(\sqrt{(7-x)^2 +(6-0)^2}\) 

= \(\sqrt{(-3-x)^2+(4-0)^2}\)

⇒ (7 − x)² + 36 = (3 + x)² + 16 

⇒ 49 − 14x + x² + 36 = 9 + 6x + x² + 16 

⇒ 20x = 85 − 25 = 60 

⇒ x = 3 

∴ The required point is P(3, 0).