#### If the points A (1, –2), B (2, 3) C (a, 2) and D (– 4, –3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.

Solution

We know that diagonals bisect each other
Hence, mid-point of AC = mid-point of BD
$\left ( \frac{1+a}{2},\frac{-2+2}{2} \right )= \left ( \frac{2-4}{2},\frac{3-3}{2} \right )$
$\left ( \frac{1+a}{2},0 \right )= \left ( -1,0 \right )$
$\frac{1+a}{2}= -1$
1 + a = –2
a = –3
C(–3, 2)
$Area \, of \, \triangle ABC = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]$
$= \frac{1}{2}\left [ 1\left ( 3-2 \right ) +2\left ( 2+2 \right )+\left ( -3 \right )\left ( -2-3 \right )\right ]$
$= \frac{1}{2}\left [ 1+2\left ( 4 \right )+15\right ]$
$= \frac{1}{2}\left [ 24 \right ]= 12sq\cdot units$
Area of parallelogram = 2 × Area of $\triangle$ABC
Area of parallelogram = 2 × 12 = 24sq.units
$Length\, of\, AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
$= \sqrt{\left ( 2-1 \right )^{2}+\left ( 3+2 \right )^{2}}$
$AB=\sqrt{1+25}= \sqrt{26}units$
Area of parallelogram = Base × height
$\frac{24}{Base}= Height$
$Height= \frac{24}{AB}$
$Height= \frac{24}{\sqrt{26}}\times\frac{\sqrt{26}}{\sqrt{26}}$
$\frac{24\sqrt{26}}{13}units$