Question

Find the value of x which is equidistant from the points A (6, 5) and B(-4, 5).

Answer 1

Let co-ordinate of any point P on x-axis = (x, 0)

∴ PA = PB

⇒ PA² = PB²

⇒ (x – 6)² + (0 – 5)² = (x + 4)² + (0 – 5)²

⇒ x² + 36 – 12x + 25 = x² + 16 + 8x + 25

⇒ -12x + 36 = 8x + 16

⇒ -12x – 8x = 16 – 36

⇒ -20x = -20

⇒ x = -20/-20 = 1

Hence, the co-ordinate of required point = (1,0)