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

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.

$\therefore$Coordinates of the mid point of AC

$\begin{array}{l}\Rightarrow \left(\frac{-1+2}{2},\frac{0+2}{2}\right)=\left(\frac{3+x}{2},\frac{1+y}{2}\right)\\ \Rightarrow \left(\frac{1}{2},1\right)=\left(\frac{3+x}{2},\frac{y+1}{2}\right)\\ \Rightarrow \frac{3+x}{2}=\frac{1}{2}\text{ and }\frac{y+1}{2}=1\\ \Rightarrow x=-2\text{ and }y=1.\end{array}$

Hence the fourth vertex of the parallelogram is (–2, 1)