Question

The equation of the line which is at a distance of $3$ units from the origin and the perpendicular from the origin makes an angle of ${30}^{\circ }$ with positive $x-$axis is

• A. $\sqrt{3}x+y=6$
• B. $\sqrt{3}x-y=6$
• C. $x-\sqrt{3}y=6$
• D. $x+\sqrt{3}y=6$

Answer 1

The correct option is A $\sqrt{3}x+y=6$

Equation of line in normal form is
$x\cos \alpha +y\sin \alpha =p$
Given : $p=3,\alpha ={30}^{\circ }$
$\Rightarrow x\cos {30}^{\circ }+y\sin {30}^{\circ }=3$
$\Rightarrow \frac{x\sqrt{3}}{2}+\frac{y}{2}=3$
$\Rightarrow \sqrt{3}x+y=6$