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

\[- \sqrt{3}x - y = 2\]

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

\[ \Rightarrow \frac{- \sqrt{3}}{2}x - \frac{1}{2}y = 1\]

Comparing the equations x cos θ + y sin θ = p  and \[\frac{- \sqrt{3}}{2}x - \frac{1}{2}y = 1\] we get, 

\[\cos\theta = - \frac{\sqrt{3}}{2}, \sin\theta = - \frac{1}{2} \text { and }p = 1\]

∴ \[\theta = {210}^\circ \text { and }p = 1\]