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Q.
If the equation $x \cos \theta+y \sin \theta=p$ is the normal form of the line $\sqrt{3} x+y+2=0$, then the values of $\theta$ and $p$ respectively are
Straight Lines
Solution:
Given, equation of line is $\sqrt{3} x+y+2=0$
$\Rightarrow \sqrt{3} x+y =-2$
$\Rightarrow -\sqrt{3} x-y =2$
On dividing above equation by
$\sqrt{(\text { coefficient of } x)^2+(\text { coefficient of } y)^2}$
i.e., $\sqrt{(-\sqrt{3})^2+(-1)^2}=\sqrt{3+1}=\sqrt{4}=2$, we get
$\Rightarrow \frac{-\sqrt{3}}{2} x-\frac{1}{2} y=\frac{2}{2} $
$ \Rightarrow ($ for convert in the form of $x \cos \theta+y \sin \theta=p)$
$\Rightarrow - \cos \,30^{\circ} x - \sin\,30^{\circ}y = 1$
$ \Rightarrow \cos \left(180^{\circ}+30^{\circ}\right) x+\sin \left(180^{\circ}+30^{\circ}\right) y=1 $
$ ( \because \cos \theta $ and $ \sin \theta $ both are negative in third quadrant $)$
$ \Rightarrow \left(\cos 210^{\circ}\right) x+\left(\sin 210^{\circ}\right) y=1$
On comparing with $x \cos \theta+y \sin \theta=p$, we get
$\theta=210^{\circ}$
and $p=1$