If (−1, 2), (2, −1) and (3, 1) are any three vertices of a parallelogram, then

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (−1, 2);

B (2,−1) and C(3, 1). We have to find the co-ordinates of the forth vertex.

Let the forth vertex be $\mathrm{D}(a, b)$

Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide.

Now to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as,

$\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

The mid-point of the diagonals of the parallelogram will coincide.

So,

Co-ordinate of mid-point of $\mathrm{AC}=$ Co-ordinate of mid-point of $\mathrm{BD}$

Therefore,

$\left(\frac{3-1}{2}, \frac{2+1}{2}\right)=\left(\frac{a+2}{2}, \frac{b-1}{2}\right)$

$\left(\frac{a+2}{2}, \frac{b-1}{2}\right)=\left(1, \frac{3}{2}\right)$

Now equate the individual terms to get the unknown value. So,

$\frac{a+2}{2}=1$

$a=0$

Similarly,

$\frac{b-1}{2}=\frac{3}{2}$

$b=4$

So the forth vertex is $\mathrm{D}(0,4)$