Question

Find equation of a straight line on which length of perpendicular from one the origin is four units and the line makes an angle of ${120}^{\circ }$ with the positive direction of x-axis.

• A. $\sqrt{3}x-y=0$
• B. $\sqrt{3}x+y=8$
• C. $x+\sqrt{3}y=8$
• D. $x-\sqrt{3}y=8$

Answer 1

The correct option is C $x+\sqrt{3}y=8$

Given $p=4$ units and $\alpha ={120}^{\circ }$

$\therefore$Equation of the line in the normal form is

$x\cos {60}^{\circ }+y\sin {60}^{\circ }=4$

or $x\times \frac{1}{2}+y\times \frac{\sqrt{3}}{2}=4$

or $x+\sqrt{3}y=8$ is the required equation.