Answer
Three different parallelograms with the fourth vertex:
$D=(-1,-2),\ D=(-1,4),\ D=(5,2)$
.
Work Step by Step
The diagonals of a parallelogram bisect each other (their intersection point is the midpoint of each diagonal).
Name the points $A(-1,1),\ B(2,0),\ C(2,3)$
1. If we take $\overline{AB} $ as a diagonal, its midpoint is equal to the midpoint of $\overline{CD}$
$(\displaystyle \frac{x_{A}+x_{B}}{2},\frac{y_{A}+y_{B}}{2})=(\frac{x_{C}+x_{D}}{2},\frac{y_{C}+y_{D}}{2})$
$(\displaystyle \frac{-1+2}{2},\frac{1+0}{2})=(\frac{2+x_{D}}{2},\frac{3+y_{D}}{2}) \Rightarrow\left\{\begin{array}{lll}
2+x_{D}=1 & \Rightarrow & x_{D}=-1\\
3+y_{D}=1 & \Rightarrow & y_{D}=-2
\end{array}\right.$
$D=(-1,-2)$
$2$. If we take $\overline{AC} $as a diagonal, its midpoint is equal to the midpoint of $\overline{BD}$
$(\displaystyle \frac{x_{A}+x_{C}}{2},\frac{y_{A}+y_{C}}{2})=(\frac{x_{B}+x_{D}}{2},\frac{y_{B}+y_{D}}{2})$
$(\displaystyle \frac{-1+2}{2},\frac{1+3}{2})=(\frac{2+x_{D}}{2},\frac{0+y_{D}}{2}) \Rightarrow\left\{\begin{array}{lll}
2+x_{D}=1 & \Rightarrow & x_{D}=-1\\
y_{D}=4 & \Rightarrow & y_{D}=4
\end{array}\right.$
$D=(-1,4)$
$3$. If we take $\overline{BC} $as a diagonal, its midpoint is equal to the midpoint of $\overline{AD}$
$(\displaystyle \frac{x_{B}+x_{C}}{2},\frac{y_{B}+y_{C}}{2})=(\frac{x_{A}+x_{D}}{2},\frac{y_{A}+y_{D}}{2})$
$(\displaystyle \frac{2+2}{2},\frac{0+3}{2})=(\frac{-1+x_{D}}{2},\frac{1+y_{D}}{2}) \Rightarrow\left\{\begin{array}{lll}
-1+x_{D}=4 & \Rightarrow & x_{D}=5\\
1+y_{D}=3 & \Rightarrow & y_{D}=2
\end{array}\right.$
$D=(5,2)$
