Question

Reduce the equation $\sqrt{3}x-y+5=0$ to normal form and find the value of $p$ and $\alpha$

Answer 1

Given $\sqrt{3}x-y+5=0$

$⟹-\frac{\sqrt{3}}{2}x+\frac{1}{2}y=\frac{5}{2}$

$⟹x\cos \frac{5\pi }{6}+y\sin \frac{5\pi }{6}=\frac{5}{2}$

This is in the form of $x\cos \alpha +y\sin \alpha =p$

$⟹\alpha =\frac{5\pi }{6},p=\frac{5}{2}$