The three vertices of a parallelogram taken in order are (−1,0), (3,1)and (2,2) respectively. Find the coordinate of the fourth vertex.

Let A(-1, 0), B(3, 1), C(2, 2) and D(x, y) be the vertices of a parallelogram ABCD taken in order. Since the diagonals of a parallelogram bisect each other.
Then, Coordinates of the mid-point of AC is the same as the coordinates of the mid-point of BD.

We know the mid-point formula is given by:

$\left(\mathrm{x},\mathrm{y}\right)=\left(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2}{2},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2}{2}\right)$
Now, mid-point of AC = mid-point of BD,

$\Rightarrow \left(\frac{-1+2}{2},\frac{0+2}{2}\right)=\left(\frac{3+\mathrm{x}}{2},\frac{1+\mathrm{y}}{2}\right)$
 $\Rightarrow \left(\frac{1}{2},1\right)=\left(\frac{3+\mathrm{x}}{2},\frac{1+\mathrm{y}}{2}\right)$
Comparing the LHS and the RHS we get,

 $\frac{3+\mathrm{x}}{2}=\frac{1}{2}$ and $\frac{1+\mathrm{y}}{2}=1$
$\Rightarrow \mathrm{x}=-3 \mathrm{and} \mathrm{y}=1$
Hence, the fourth vertex of the parallelogram is (-3, 1).