I can't seem to find a formula to find the perpendicular bisector of a line. Anyone know what it is?
Suppose each of $\displaystyle P: (x_1,y_1)~\&~Q: (x_2,y_2)$ is a point and $\displaystyle y_1\ne y_2$.
The perpendicular bisector of $\displaystyle \overline{PQ}$ is $\displaystyle y = \left( { - \frac{{{x_1} - {x_2}}}{{{y_1} - {y_2}}}} \right)\left( {x - \frac{{{x_1} + {x_2}}}{2}} \right) + \left( {\frac{{{y_1} + {y_2}}}{2}} \right)$
The slope of the line through A and B is $\displaystyle \displaystyle \begin{align*} \frac{1 - 2}{3 - (-2)} = -\frac{1}{5} \end{align*}$. So the slope of your perpendicular bisector is 5, since the two slopes will multiply to give -1.
The midpoint of AB is $\displaystyle \displaystyle \begin{align*} \left( \frac{-2 + 3}{2} , \frac{ 2 + 1}{2} \right) = \left( \frac{ 1}{2} , \frac{3}{2} \right) \end{align*}$
So the perpendicular bisector is of the form $\displaystyle \displaystyle \begin{align*} y = mx + c \end{align*}$ with $\displaystyle \displaystyle \begin{align*} m = 5 \end{align*}$ and a point it passes through is $\displaystyle \displaystyle \begin{align*} (x, y) = \left( \frac{1}{2} , \frac{3}{2} \right) \end{align*}$. Substituting these gives
$\displaystyle \displaystyle \begin{align*} \frac{3}{2} &= 5 \left( \frac{1}{2} \right) + c \\ \frac{3}{2} &= \frac{5}{2} + c \\ c &= -1 \end{align*}$
and so the equation of your perpendicular bisector is $\displaystyle \displaystyle \begin{align*} y = 5x - 1 \end{align*}$.