How would I find a point on the y-axis equidistant from two other points? The points are$$(5,-5) and (1,1)$$
I tried doing this visually and came up with (0,-5). This wasn't correct once I applied the distance formula to check the distance between that point and the two others. 

You want the point where the perpendicular bisector of the two points cuts the $y$-axis.  

The slope of the line between $(5,-5)$ and $(1,1)$ is $-\frac{3}{2}$, so the slope of the normal is $\frac23$. Hence the normal through the mid-point $(3,-2)$ cuts the $y$-axis at $(0,-4)$.
