If two vertices of a parallelogram are (3, 2)

Solution:

We have a parallelogram ABCD in which A (3, 2) and B (−1, 0) and the co-ordinate of the intersection of diagonals is M (2,−5).

We have to find the co-ordinates of vertices C and D.

So let the co-ordinates be $\mathrm{C}\left(x_{1}, y_{1}\right)$ and $\mathrm{D}\left(x_{2}, y_{2}\right)$

In general 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 $\mathrm{M}$

Therefore,

$\left(\frac{3+x_{1}}{2}, \frac{2+y_{1}}{2}\right)=(2,-5)$

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

$x=1$

 

$y=-12$

So the co-ordinate of vertex C is (1,−12) 

Similarly,

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

Therefore,

$\left(\frac{-1+x_{2}}{2}, \frac{0+y_{2}}{2}\right)=(2,-5)$

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

$x=5$

 

$y=-10$

So the co-ordinate of vertex C is (5,−10)