JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Straight Lines

Write $L$ in normal form $x\cos \alpha +y\sin \alpha =4$, since its distance from the origin is $4$. The perpendicular from the origin lies at angle $\alpha$, and it must make $60^{\circ }$ with $x+y=0$, which sits at $135^{\circ }$. Taking $\alpha =15^{\circ }$ gives $\cos 15^{\circ }=\frac{\sqrt{3}+1}{2\sqrt{2}}$ and $\sin 15^{\circ }=\frac{\sqrt{3}-1}{2\sqrt{2}}$, both intercepts then positive. Multiplying $x\cos 15^{\circ }+y\sin 15^{\circ }=4$ through by $2\sqrt{2}$ gives $(\sqrt{3}+1)x+(\sqrt{3}-1)y=8\sqrt{2}$.