If A (1, 2) B (4, 3) and C (6, 6) are the three vertices

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

B (4, 3) and C (6, 6). We have to find the co-ordinates of the forth vertex.

Let the forth vertex be $\mathrm{D}(x, y)$

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,

$\left(\frac{1+6}{2}, \frac{2+6}{2}\right)=\left(\frac{x+4}{2}, \frac{y+3}{2}\right)$

$\left(\frac{x+4}{2}, \frac{y+3}{2}\right)=\left(\frac{7}{2}, 4\right)$

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

$\frac{x+4}{2}=\frac{7}{2}$

$x=3$

Similarly,

$\frac{y+3}{2}=4$

$y=5$

So the forth vertex is $D(3,5)$