JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Straight Lines

The incident ray strikes the vertical line $x=1$ at $30^{\circ }$ to it, so it makes $60^{\circ }$ with the horizontal and has slope $\tan 60^{\circ }=\sqrt{3}$. From $(2,2\sqrt{3})$ the ray $y-2\sqrt{3}=\sqrt{3}(x-2)$ meets $x=1$ at $y=2\sqrt{3}-\sqrt{3}=\sqrt{3}$, so $A=(1,\sqrt{3})$. Reflection in the vertical line keeps the angle but reverses the horizontal sense, so the reflected ray has slope $-\sqrt{3}$: $y-\sqrt{3}=-\sqrt{3}(x-1)$, i.e. $\sqrt{3}x+y=2\sqrt{3}$. Testing $(3,-\sqrt{3})$ gives $3\sqrt{3}-\sqrt{3}=2\sqrt{3}$, which holds, so the reflected line passes through $(3,-\sqrt{3})$.