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

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

1. A
$x\mathrm{cos}\frac{5\pi }{6}+y\mathrm{sin}\frac{5\pi }{6}=2$
2. B
$x\mathrm{cos}\pi +y\mathrm{sin}\pi =3$
3. C
$x\mathrm{cos}\pi -y\mathrm{sin}\pi =4$
4. D
$x\mathrm{cos}\frac{3\pi }{2}-y\mathrm{sin}\frac{3\pi }{2}=3$

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### Solution:

$\sqrt{3}x-y+4=0⇒-\sqrt{3}x+1y=4$

Divide with 2 an both sides we get

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

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

Hence $\alpha$  is in ${Q}_{2}⇒\alpha =\frac{5\pi }{6}$

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

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