Question

Prove that the abscissa of a point P, which is equidistant from points with coordinates A(7, 1) and B(3, 5) is 2 more than its ordinate.

Answer 1

Let point P(x, y) be given, with A(7, 1) and B(3, 5).

According to question,

AP = BP

AP² = BP²

By using distance formula,

(x − 7)² + (y − 1)² = (x – 3)² + (y – 5)²

⇒ x² + 49 − 14x + y² − 2y + 1 = x² + 9 − 6x + y² + 25 − 10y

⇒ 50 − 14x − 2y = 34 − 6x − 10y

⇒ −14x − 2y + 6x + 10y = 34 − 50

⇒ −8x + 8y = −16

⇒ −8(x − y) = −16

⇒ x − y = 2

⇒ x = 2 + y

Hence proved.