Question

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

Answer 1

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

$\text{The normal form of line is}$

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

$-\sqrt{3}x-y=2...\left(1\right)$

$\left[\text{Dividing both sides by}\sqrt{(\text{coefficient of x}{)}^2+(\text{coefficient of y}{)}^2}\right]$

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

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

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

$\text{Comparing the equations x}cos\theta +ysin\theta =p$

$\text{and}\frac{-\sqrt{3}}{2}x-\frac{1}{2}y=1\text{we get,}$

$\text{So,}$

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

$\therefore \theta ={210}^{\circ }$

$p=2\left[\text{From equation (1)}\right]$