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

x cos 3 π 2 − y sin 3 π 2 = 3

x cos 5 π 6 + y sin 5 π 6 = 2

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

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x \cos π + y \sin π = 3

x \cos π - y \sin π = 4

x \cos \frac{3 π}{2} - y \sin \frac{3 π}{2} = 3

x \cos \frac{5 π}{6} + y \sin \frac{5 π}{6} = 2

answer is A.

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

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