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,

$-\frac{\sqrt{3}}{2}x-\frac{1}{2}y=\frac{2}{2}$

$(-\frac{\sqrt{3}}{2})x+(-\frac{1}{2})y=1$

On comparing equation to $x\cos \theta +y\sin \theta =p$, we get,

$\cos \theta =-\frac{\sqrt{3}}{2},\sin \theta =-\frac{1}{2},p=1$

Since, the values of $\sin \theta$ and $\cos \theta$ are negative, $\theta =\pi +\frac{\pi }{6}=\frac{7\pi }{6}$

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