Question

Find the equation of the line whose perpendicular distance from the origin is $5$ units and the angle made by the perpendicular with the positive $x$-axis is ${30}^{\circ }$.

Answer 1

Given : perpendicular distance of line from origin is $5$ units.
Normal makes an angle of ${30}^{\circ }$ with the positive $x$-axis.
Normal form of equation of line is $x\cos \omega +y\sin \omega =p$
Here $p=5$ and $\omega ={30}^{\circ }$
$\Rightarrow x\cos {30}^{\circ }+y\sin {30}^{\circ }=5$
$\Rightarrow x\cdot \frac{\sqrt{3}}{2}+y\cdot \frac{1}{2}=5$
$\Rightarrow \frac{\sqrt{3}x+y}{2}=5$
$\Rightarrow \sqrt{3}x+y=10$
Thus, equation of line is $\sqrt{3}x+y=10$