Question

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

Answer 1

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 $= p _1 p _2 \operatorname{cosec} \theta=\left(\frac{3}{\sqrt{2}}\right)\left(\frac{3}{2}\right)(5 \sqrt{2})=9$ (square units) where $p_1, p_2$ are distances between opposite parallel sides of parallelogram.