Answer
The sides $\overline{AB}$ and $\overline{BC}$ are perpendicular and have the same length.
The fourth vertex is $D=(-2,-2)$
Work Step by Step
Apply the distance formula
$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
$|AB|=\sqrt{(x_{B}-x_{A})^{2}+(y_{B}-y_{A})^{2}}=\sqrt{(1-2)^{2}+(3-(-1))^{2}}=\sqrt{1+16} =\sqrt{17}$
$|BC|=\sqrt{(x_{C}-x_{B})^{2}+(y_{C}-y_{B})^{2}}=\sqrt{(-3-1)^{2}+(2-3)^{2}}=\sqrt{16+1} =\sqrt{17}$
So, $\overline{AB}$ and $\overline{BC}$ have the same length.
Testing slopes
$m_{AB}=\displaystyle \frac{y_{B}-y_{A}}{x_{B}-x_{A}}=\frac{3-(-1)}{1-2}=\frac{-4}{1}=-4$
$m_{BC}==\displaystyle \frac{y_{C}-y_{B}}{x_{C}-x_{B}}=\frac{2-3}{-3-1}=\frac{-1}{-4}=-\frac{1}{4}$
Since $m_{AB}=-\displaystyle \frac{1}{m_{BC}}$, $\overline{AB}$ and $\overline{BC}$ lie on perpencular lines.
Thus, A, B, and C are three vertices of a square.
The fourth vertex lies opposite to B.
In a square, the diagonals bisect each other (and are perpendicular, but we don't need that now).
The midpoint of $\overline{AC}$ is also the midpoint of $\overline{BD}$
$(\displaystyle \frac{x_{A}+x_{C}}{2},\frac{y_{A}+y_{C}}{2})=(\frac{x_{B}+x_{D}}{2},\frac{y_{B}+y_{D}}{2})$
$(\displaystyle \frac{2-3}{2},\frac{-1+2}{2})=(\frac{1+x_{D}}{2},\frac{3+y_{D}}{2})$
$\left\{\begin{array}{lll}
1+x_{D}=-1 & \Rightarrow & x_{D}=-2\\
3+y_{D}=1 & \Rightarrow & y_{D}=-2
\end{array}\right.$
$D=(-2,-2)$
