# A ray of light is sent along the line x-2y-3=0 upon reaching the line 3x-2y-5=0 the ray is reflected from it, find the equation of the line containing the reflected ray.

we assume that the incident ray, the reflected ray and the line representing the surface of reflection are coplanar.

the equation of the line representing the surface is 3x-2y-5=0......(1) or $y=\frac{3}{2}.x-\frac{5}{2}$.

so the slope of the line representing the reflecting surface is 3/2.

thus the slope of the normal is -2/3.

the equation of the incident ray is x-2y-3=0......(2) or $y=\frac{x}{2}-\frac{3}{2}$

so the slope of the incident ray is 1/2.

now solving eq(1) and eq(2), we have:

$x-2y-3=0\phantom{\rule{0ex}{0ex}}3x-2y-5=0\phantom{\rule{0ex}{0ex}}-2x+2=0\Rightarrow x=1and\phantom{\rule{0ex}{0ex}}1-2y-3=0\phantom{\rule{0ex}{0ex}}-2y=2\Rightarrow y=-1$

therefore the coordinates of the point A, where the ray is reflected is (1,-1).

let the slope of the reflected ray be m. then the equation of the reflected ray is $y+1=m(x-1)$

since the incident ray and reflected ray are equally inclined to the normal, i.e.

the angle form the incident ray to the normal = the angle form the normal to the reflected ray

therefore

$\frac{{\displaystyle \frac{1}{2}}-(-{\displaystyle \frac{2}{3}})}{1+{\displaystyle \frac{1}{2}}.(-{\displaystyle \frac{2}{3}})}=\frac{-{\displaystyle \frac{2}{3}}-m}{1+(-{\displaystyle \frac{2}{3}}).m}or\phantom{\rule{0ex}{0ex}}\frac{m+{\displaystyle \frac{2}{3}}}{1-{\displaystyle \frac{2m}{3}}}=-\frac{{\displaystyle \frac{3+4}{6}}}{1-{\displaystyle \frac{1}{3}}}\phantom{\rule{0ex}{0ex}}\frac{3m+2}{3-2m}=-\frac{7}{2}*\frac{1}{2}\phantom{\rule{0ex}{0ex}}\frac{3m+2}{3-2m}=-\frac{7}{4}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow 12m+8+21-14m=0\phantom{\rule{0ex}{0ex}}-2m+29=0\phantom{\rule{0ex}{0ex}}m=\frac{29}{2}$

thus the equation of the reflected ray is

$y+1=\frac{29}{2}.(x-1)\phantom{\rule{0ex}{0ex}}2y+2=29x-29\phantom{\rule{0ex}{0ex}}29x-2y=31$

hope this helps you

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