If the points A (a, −11), B (5, b), C(2, 15) and D (1, 1)

Solution:

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (a,−11); B (5, b); C (2, 15) and D (1, 1).

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 $\mathrm{BD}$

Therefore,

$\left(\frac{a+2}{2}, \frac{15-11}{2}\right)=\left(\frac{5+1}{2}, \frac{b+1}{2}\right)$

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

$\frac{a+2}{2}=3$

$a=4$

Similarly,

$\frac{b+1}{2}=2$

$b=3$

Therefore,

$a=4$

 

$b=3$