Question

Reduce the following equation to the normal form and find p and α in \[x + \sqrt{3}y - 4 = 0\] .

Answer 1

\[x + \sqrt{3}y - 4 = 0\]

\[\Rightarrow x + \sqrt{3}y = 4\]

\[ \Rightarrow \frac{x}{\sqrt{1^2 + \left( \sqrt{3} \right)^2}} + \frac{\sqrt{3}y}{\sqrt{1^2 + \left( \sqrt{3} \right)^2}} = \frac{4}{\sqrt{1^2 + \left( \sqrt{3} \right)^2}} \left[ \text { Dividing both sides by } \sqrt{\left( \text { coefficient of x } \right)^2 + \left( \text { coefficient of y } \right)^2} \right]\]

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

This is the normal form of the given line, where p = 2,

\[cos\alpha = \frac{1}{2}\] and  \[sin\alpha = \frac{\sqrt{3}}{2} \Rightarrow \alpha = \frac{\pi}{3}\].