The equation of the line such that the normal ray from the origin makes an angle of 30⁰ with the x-axis and the area of the triangle formed by the line with the coordinate axes is $50\sqrt{3}is$

$\begin{array}{l}Let\alpha ={30}^{0}EquationoflineinNormalformis\\ x\cos \alpha +y\sin \alpha =p\\ \Rightarrow x\cos {30}^0+y\sin {30}^0=p\\ \Rightarrow \sqrt{3}x+y-2p=0\\ GivenArea=50\sqrt{3}\\ \Rightarrow \frac{4p^2}{2\left|\sqrt{3}\times 1\right|}=50\sqrt{3}\\ \Rightarrow 2p^2=50\times 3\\ \Rightarrow p^2=25\times 3\\ \Rightarrow p=\pm 5\sqrt{3}\\ \therefore \mathrm{Re}quiredEquationis\sqrt{3}x+y\pm 10\sqrt{3}=0\end{array}$