Question

Find the values of $\theta$ and $p$, if the equation $x\cos \theta +y\sin \theta =p$ is the normal form of the line $\sqrt{3}x+y+2=0.$

Answer 1

The equation of the given line is $\sqrt{3}x+y+2=0$
this equation can be reduced as $-\sqrt{3}x-y=2$
on dividing both sides by $\sqrt{(-\sqrt{3}{)}^2+(-1{)}^2}=2,$
we obtain $-\frac{\sqrt{3}}{2}x-\frac{1}{2}y=\frac{2}{2}$
$\Rightarrow \{-\frac{\sqrt{3}}{2}\}x+\{-\frac{1}{2}\}y=1$....(1)
On comparing equation (1) to $x\cos \theta +y\sin \theta =p$,
we obtain $\cos \theta =-\frac{\sqrt{3}}{2},\sin \theta =-\frac{1}{2},$ and $p=1$
Since the value of $\sin \theta$ and $\cos \theta$ are both negative, $\theta$ is in the third quadrant
$\therefore \theta =\pi +\frac{\pi }{6}=\frac{7\pi }{6}$

Thus, the respective values of $\theta$ and $p$ are $\frac{7\pi }{6}$ and $1$