Question

The length of the perpendicular from the origin to a line is 7 and the perpendicular makes an angle of ${150}^{\circ }$ with the positive direction of the x-axis. Find the equation of the line.

• A. $\sqrt{3x}+y-14=0$
• B. $\sqrt{3x}-y-14=0$
• C. $-\sqrt{3x}-y-14=0$
• D. $-\sqrt{3x}+y-14=0$

Answer 1

The correct option is D

$-\sqrt{3x}+y-14=0$

Equation of the line in the normal form is

$xcos\alpha +ysin\alpha =p$

Here, give p = 7

$\alpha ={150}^{\circ }$

$cos\alpha =cos{150}^{\circ }=cos({180}^{\circ }-{30}^{\circ })=-cos{30}^{\circ }=\frac{-\sqrt{3}}{2}$

$sin\alpha =sin{150}^{\circ }=sin({180}^{\circ }-{30}^{\circ })=sin{30}^{\circ }=\frac{1}{2}$

Equation of the line is

$x\left(\frac{-\sqrt{3}}{2}\right)+7\left(\frac{1}{2}\right)=7$

$-\sqrt{3x}+y=14$

$-\sqrt{3x}+y-14=0$