Since the point of intersection lies on fourth quadrant and equidistant from the two axes,
i.e., let the point be (k, $$-$$k) and this point satisfies the two equations of the given lines.
$$\therefore$$ 4ak $$-$$ 2ak + c = 0 ......... (1)
and 5bk $$-$$ 2bk + d = 0 ..... (2)
From (1) we get, $$k = {{ - c} \over {2a}}$$
Putting the value of k in (2) we get,
$$5b\left( { - {c \over {2a}}} \right) - 2b\left( { - {c \over {2a}}} \right) + d = 0$$
or, $$ - {{5bc} \over {2a}} + {{2bc} \over {2a}} + d = 0$$ or, $$ - {{3bc} \over {2a}} + d = 0$$
or, $$ - 3bc + 2ad = 0$$ or, $$3bc - 2ad = 0$$
$$x + \sqrt 3 y = \sqrt 3 $$ 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 $$y\sqrt 3 = x - \sqrt 3 $$