Answer
The point on the y-axis is $(0,-4)$
Work Step by Step
Let the point on the y-axis be $(0,y)$
We can find an expression for the distance from $(0,y)$ to the point $(1,1)$:
$d_1 = \sqrt{(0-1)^2+(y-1)^2}$
$d_1 = \sqrt{1+(y^2-2y+1)}$
$d_1 = \sqrt{y^2-2y+2}$
We can find an expression for the distance from $(0,y)$ to the point $(5,-5)$:
$d_2 = \sqrt{(0-5)^2+(y-(-5))^2}$
$d_2 = \sqrt{25+(y^2+10y+25)}$
$d_2 = \sqrt{y^2+10y+50}$
We can equate the two distances to find $y$:
$d_1 = d_2$
$\sqrt{y^2-2y+2}= \sqrt{y^2+10y+50}$
$y^2-2y+2= y^2+10y+50$
$-2y+2= 10y+50$
$12y = -48$
$y = -4$
The point on the y-axis is $(0,-4)$
