The point P(x, y) is equidistant from the points A(5, 1) and B(– 1, 5), means PA = PB

By distance formula, as shown below:

PA = √{(5 – x)² + (1 – y)²}

= √{(25 + x² – 10x) + (1 + y² – 2y)}

⇒ PA = √{26 + x² – 10x + y² – 2y}

PB = √{(– 1 – x)² + (5 – y)²}

= √{(1 + x² + 2x + 25 + y² – 10y)}

⇒ PB = √{(26 + x² + 2x + y² – 10y)}

Now, PA = PB

Squaring both sides, we get

26 + x² – 10x + y² – 2y = 26 + x² + 2x + y² – 10y

⇒ 12x = 8y

⇒3x = 2y

Hence proved.