Question

Prove that 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

∴ d = `sqrt((x_2 - x_1)^2 = (y_2 - y_1)^2)`

Distance PA = Distance PB

`sqrt ((7 - x)^2 + (1 - y)^2) = sqrt ((3 - x)^2 + (5 - y)^2)`

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

⇒ −14x − 2y + 6x + 10y = 34 − 50

⇒ −8x + 8y = −16

⇒ −8 (x − y) = −16

⇒ x − y = 2

⇒ x = 2 + y