Question

Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Answer 1

Let (x, 0) be the points on the x-axis. 

Then the distance will be same;

(x – 3)² + 16 = (x – 7)² + 36

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

⇒ 14x – 6x = 49 + 36 – 9 – 16

⇒ 8x = 60 ⇒ x = \(\frac{15}{2}\)

Hence the point is (\(\frac{15}{2}\), 0).