Three vertices of a parallelogram are (a+b, a−b),

Solution:

Let $\mathrm{ABCD}$ be a parallelogram in which the co-ordinates of the vertices are $\mathrm{A}(a+b, a-b) ; \mathrm{B}(2 a+b, 2 a-b)$ and $\mathrm{C}(a-b, a+b)$. 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.

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 mid-point of BD

Therefore,

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

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

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

$x=-b$

 

$y=b$

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