Since P (x, y) is equidistant from A (5, 1) and B (-1, 5), then
PA = PB
or PA2 = PB2
(5 – x)2 + (1 – y)2 = (– 1 – x)2 + (5 – y)2
(25 + x2 – 10x) + (1 + y2 – 2y) = (1 + x2 + 2x + 25 + y2 – 10y)
26 + x2 – 10x + y2 – 2y = (26 + x2 + 2x + y2 – 10y)
12x = 8y
3x = 2y
Hence proved.