A(5,1)≡(x1,y1) B(-1,5)≡(x2,y2) P≡(x,y)PA=PB(given)∴By distance formula √(x-x1)²+(y-y1)² we have,⇒(x-5)²+(y-1)²=(x+1)²+(y-5)²⇒x²-10x+25+y²-2y+1=x²+2x+1+y²-10y+25∴-10x-2y=2x-10y∴-12x=-8y2y=3x. [hence proved]