Question:
Find the equation of the line which is at a perpendicular distance of 5 units from the origin and the angle made by the perpendicular with the positive x-axis is $30^{\circ}$
Solution:
If $p$ is the length of the normal from the origin to a line and $\omega$ is the angle made by the normal with the positive direction of the $x$-axis, then the equation of the line is given by $x \cos \omega+y \sin \omega=p$.
Here, $p=5$ units and $\omega=30^{\circ}$
Thus, the required equation of the given line is
$x \cos 30^{\circ}+y \sin 30^{\circ}=5$
$x \frac{\sqrt{3}}{2}+y \cdot \frac{1}{2}=5$
i.e., $\sqrt{3} x+y=10$