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, $$3bc - 2ad = 0$$

Let $$P,Q,R,$$ be the vertices of $$\Delta PQR$$



Since $$PS$$ is the median, $$S$$ is mid-point of $$QR$$

So, $$S = \left( {{{7 + 6} \over 2},{{3 - 1} \over 2}} \right) = \left( {{{13} \over 2},1} \right)$$

Now, slope of $$PS$$ $$ = {{2 - 1} \over {2 - {{13} \over 2}}} = - {2 \over 9}$$

Since, required line is parallel to $$PS$$ therefore slope of required line $$=$$ slope of $$PS$$

Now, equation of line passing through $$(1, -1)$$ and having slope $$ - {2 \over 9}$$ is

$$y - \left( { - 1} \right) = - {2 \over 9}\left( {x - 1} \right)$$

$$9y + 9 = - 2x + 2$$

$$ \Rightarrow 2x + 9y + 7 = 0$$

From the figure, we have

$$a = 2,b = 2\sqrt 2 ,c = 2$$

$${x_1} = 0,\,{x^2} = 0,\,{x_3} = 2$$



Now, $$x$$-co-ordinate of incenter is given as

$${{a{x_1} + b{x_2} + c{x_3}} \over {a + b + c}}$$

$$ \Rightarrow x$$-coordinate of incentre

$$ = {{2 \times 0 + 2\sqrt 2 .0 + 2.2} \over {2 + 2 + 2\sqrt 2 }}$$

$$=$$ $${2 \over {2 + \sqrt 2 }} = 2 - \sqrt 2 $$

$$x + \sqrt 3 y = \sqrt 3 $$ 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 $$y\sqrt 3 = x - \sqrt 3 $$