The points (3, −4) and (−6, 2) are the extremities of a diagonal

Solution:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (3,−4); B (−1,−3) and C (−6, 2). 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 BD

Therefore,

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

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

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

$x=-2$

 

$y=1$

So the forth vertex is $\mathrm{D}(-2,1)$