Find the equation of the line which is at a perpendicular distance of 5 units

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., $\sqrt{3} x+y=10$