A line forms a triangle of area $54\sqrt{3}$ sq-units with the coordinate axes. If the perpendicular drawn from the origin to the line makes an angle of $60°$ with the $x-$axis, then the equation the line is 

Let $p$ be the length of the perpendicular drawn from the origin to the required line. The perpendicular makes an angle of $60°$ with the $x-$axis. Putting $a= 60°$ in $x\cos \alpha +y\sin \alpha =p$ the equation of the line is $x\cos {60}^{\circ }+y\sin {60}^{\circ }=p$

or, $x+\sqrt{3}y=2p$.

This cuts the coordinate axes at $A(2p,0)$ and $B\left(0,\frac{2p}{\sqrt{3}}\right)$

It is given that

           Area of $\mathrm{△}OAB=54\sqrt{3}$ sq. units.

$\begin{array}{l}\Rightarrow \frac{1}{2}\left(OA\right)\left(OB\right)=54\sqrt{3}\\ \Rightarrow \frac{1}{2}\times 2p\times \frac{2p}{\sqrt{3}}=54\sqrt{3}\Rightarrow p^2=81\Rightarrow p=9\end{array}$

Hence, the equation of the line is $x+\sqrt{3}y=18$