Answer
$(0, -4)$
Work Step by Step
Points on the y axis have x-coordinates equal to 0.
Let $P=(0, y)$ be such a point.
We want its distances from $(5,-5)$ and $(1,1)$ to be equal:
$\sqrt{(0-5)^{2}+(y-(-5))^{2}}=\sqrt{(0-1)^{2}+(y-1)^{2}}\quad\text{ ...expand squares}$
$\sqrt{25+y^{2}+10y+25}=\sqrt{1+y^{2}-2y+1} \quad\text{ ...square both sides}$
$y^{2}+10y+50=y^{2}-2y+2$
$12y=-48$
$y=-4$.
The point is $P=(0, -4)$.
