Official solution
\[x + \sqrt 3 y = \sqrt 3 \] or \[y = - {1 \over {\sqrt 3 }}x + 1\]
Let \[\theta\] be the angle which the line makes with the positive x-axis.
\[\therefore\] \[\tan \theta = - {1 \over {\sqrt 3 }} = \tan \left( {\pi - {\pi \over 6}} \right)\] or \[\theta = \pi - {\pi \over 6}\]
\[\therefore\] \[\angle ABC = {\pi \over 6}\]; \[\therefore\] \[\angle DBE = {\pi \over 6}\]
\[\therefore\] the equation of the line BD is,
\[y = \tan {\pi \over 6}x + c\] or \[y = {x \over {\sqrt 3 }} + c\] ..... (1)
The line \[x + \sqrt 3 y = \sqrt 3 \] intersects the x-axis at \[B(\sqrt 3 ,0)\] and, the line (1) passes through \[B(\sqrt 3 ,0)\].
\[\therefore\] \[0 = {{\sqrt 3 } \over {\sqrt 3 }} + c\] or, c = -1
Hence, the equation of the reflected ray is,
\[y = {x \over {\sqrt 3 }} - 1\] or \[y\sqrt 3 = x - \sqrt 3 \]