$$x + \sqrt 3 y = \sqrt 3 $$ 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 $$y\sqrt 3 = x - \sqrt 3 $$
The point which divides the line segment joining the points (1, 1) and (2, 4) in the ratio 3 : 2 is
$$ = \left( {{{3 \times 2 + 2 \times 1} \over {3 + 2}},{{3 \times 4 + 2 \times 1} \over {3 + 2}}} \right)$$
$$ = \left( {{{6 + 2} \over 5},{{12 + 2} \over 5}} \right) = \left( {{8 \over 5},{{14} \over 5}} \right)$$
Since the line 2x + y = k passes through this point,
$$\therefore$$ $$2 \times {8 \over 5} + {{14} \over 5} = k$$ or $${{30} \over 5} = k$$ or, k = 6
Statement-1: The ratio $$PR$$ : $$RQ$$ equals $$2\sqrt 2 :\sqrt 5 $$
Statement-2: In any triangle, bisector of an angle divide the triangle into two similar triangles.
