Question

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

• A. $30^{\circ },1$
• B. $210^{\circ },1$
• C. $30^{\circ },4$
• D. $210^{\circ },4$

Answer 1

Answer: (B)

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 $p=1$