Question

Reduce the equation \[\sqrt{3}\] x + y + 2 = 0 to the normal form and find p and α.

Answer 1

\[\sqrt{3}\] x + y + 2 = 0

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

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

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

This is the normal form of the given line.
Here,  p = 1,

\[cos\alpha = - \frac{\sqrt{3}}{2}\] and \[sin\alpha = - \frac{1}{2}\]

\[\Rightarrow \alpha = {210}^\circ\]