Question

The two adjacent sides of a parallelogram are represented by the lines $x-y+1=0$ and $4x-3y=2$. If one of the diagonal of the parallelogram is the line $y=\frac{3x}{2}$, then find the area of parallelogram.

Answer 1

Answer: 9

Here, $\tan \theta =\left|\frac{\frac{4}{3}-1}{1+\frac{4}{3}}\right|=\frac{1}{7}$, where $\theta$ is the angle between adjacent sides of given parallelogram.
So, area of parallelogram $=p_1{p}_2\mathrm{cosec}\theta =\left(\frac{3}{\sqrt{2}}\right)\left(\frac{3}{2}\right)(5\sqrt{2})=9$ (square units) where $p_1,p_2$ are distances between opposite parallel sides of parallelogram.