If the point P(x, y) is equidistant from the points A(5, 1)

As per the question, we have

$A P=B P$

$\Rightarrow \sqrt{(x-5)^{2}+(y-1)^{2}}=\sqrt{(x+1)^{2}+(y-5)^{2}}$

$\Rightarrow(x-5)^{2}+(y-1)^{2}=(x+1)^{2}+(y-5)^{2} \quad$ (Squaring both sides)

$\Rightarrow x^{2}-10 x+25+y^{2}-2 y+1=x^{2}+2 x+1+y^{2}-10 y+25$

$\Rightarrow-10 x-2 y=2 x-10 y$

$\Rightarrow 8 y=12 x$

$\Rightarrow 3 x=2 y$

Hence, 3x = 2y.