The area of the parallelogram formed by the lines $x\cos \alpha +y\sin \alpha =p, x\cos \alpha +y\sin \alpha =q,$

$x\cos \beta +y\sin \beta =r$ and $x\cos \beta +y\sin \beta =s$ for given value of $p,q,r$and $s$ is least, if $(\alpha -\beta )=$

The equations of the sides of the parallelogram are:

$x\cos \alpha +y\sin \alpha -p=0$ $x\cos \alpha +y\sin \alpha -q=0$

$x\cos \beta +y\sin \beta -r=0$ $x\cos \beta +y\sin \beta -s=0$

$\therefore$ Area of the parallelogram

$=\left|\frac{\left\{\right(-p)-(-q\left)\right\}\left\{\right(-r)-(-s\left)\right\}}{\left|\begin{array}{c}\cos \alpha \sin \alpha \\ \cos \beta \sin \beta \end{array}\right|}\right|=\left|\frac{(p-q)(r-s)}{\sin (\alpha -\beta )}\right|$

Clearly, it is minimum when $\alpha -\beta =\pm \frac{\pi }{2}$