Three consecutive vertices of a parallelogram are

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (−2,−1); B (1, 0) and C (4, 3). We have to find the co-ordinates of the forth vertex.

Let the forth vertex be

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 of two pointsand 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,

$\left(\frac{x+1}{2}, \frac{y}{2}\right)=\left(\frac{4-2}{2}, \frac{3-1}{2}\right)$

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

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

$x=1$

$y=2$

So the forth vertex is