Question

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

Answer 1

Here $\sqrt{3}x+y+2=0$

$\Rightarrow \sqrt{3}x+y=-2\Rightarrow -\sqrt{3}x-y=2$

Dividing both sides by $\sqrt{(-\sqrt{3}{)}^2+(-1{)}^2}=2$, we have

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

put $cos\alpha =\frac{-\sqrt{3}}{2}$ and $sin\alpha =\frac{-1}{2}$

$\Rightarrow \alpha$ lies in IIIrd quadrant

$\therefore cos\alpha =\frac{-\sqrt{3}}{2}=-cos{30}^{\circ }$

$=cos({180}^{\circ }+{30}^{\circ })$

$\Rightarrow \alpha ={210}^{\circ }$

\(\therefore) Equation of line in normal form is

$xcos\frac{7\pi }{6}+ysin\frac{7\pi }{6}=1$

Comparing it with $xcos\alpha +ysin\alpha =p$, we have

$\alpha =\frac{7\pi }{6}$ and p = 1