Question

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

Answer 1

The normal form of the line $\sqrt{3}x+y+2=0$ is

$$\begin{gathered} -\sqrt{3}x-y=2 \\ \Rightarrow \frac{-\sqrt{3}}{\sqrt{{\left(-\sqrt{3}\right)}^2+{\left(-1\right)}^2}}x-\frac{1}{\sqrt{{\left(-\sqrt{3}\right)}^2+{\left(-1\right)}^2}}y=\frac{2}{\sqrt{{\left(-\sqrt{3}\right)}^2+{\left(-1\right)}^2}}\left[\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}\sqrt{{\left(\mathrm{coefficient}\mathrm{of}x\right)}^2+{\left(\mathrm{coefficient}\mathrm{of}y\right)}^2}\right] \\ \Rightarrow \frac{-\sqrt{3}}{2}x-\frac{1}{2}y=1 \end{gathered}$$

Comparing the equations xcos θ + ysin θ = p and $\frac{-\sqrt{3}}{2}x-\frac{1}{2}y=1$ we get,

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

∴ $\theta ={210}^{\circ }\mathrm{and}p=1$