The perpendicular form of the line 3x−y+4=0 is

\sqrt{3} x - y + 4 = 0 \Rightarrow - \sqrt{3} x + 1 y = 4

Divide with 2 an both sides we get

$\frac{- \sqrt{3} x}{2} + \frac{1}{2} y = \frac{4}{2} \Rightarrow x (\frac{- \sqrt{3}}{2}) + y (\frac{1}{2}) = 2$

Here $\cos α = \frac{- \sqrt{3}}{2}, \sin α = \frac{1}{2}, p = 2$

Hence $α$ is in $Q_{2} \Rightarrow α = \frac{5 π}{6}$

$∴$ The equation n of the line in perpendicular form is $x \cos \frac{5 π}{6} + y \sin \frac{5 π}{6} = 2$

The perpendicular form of the line $\sqrt{3} x - y + 4 = 0$ is