Question

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

Answer 1

Solution:
Let the point on x-axis be P(x, 0)

Let A be (7, 6) and B be (3, 4)

Now PA = 

PA² = (x – 7)² + 36

and PB = 

PB² = (x – 3)² + 16

Given PA = PB

or PA² = PB²

or (x – 7)² + 36 = (x – 3)² + 16

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

or -14x + 6x = 25 – 85

or -8x = -60

or 8x = 60

or 2x = 15 ⇒ x = 15/2

Hence the required point is (15/2, 0)