Question

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.

Answer 1

Incident ray is along the line
`x-2y-3 = 0 " " (1)`
Mirror is along the line
`3x-2y-5 = 0 " " (2)`
In the figure, incident ray PA strikes the mirror at point A.
Solving the incident ray and mirror line, we get `A-=(1,-1).`
Let the slope of AQ be m.

Also, slope of LM is `(3)/(2)` and slope of PA is `(1)/(2)`.
Now, incident ray and reflected ray make same angle a with the mirror line.
Let the slope of reflected ray be m.
Now, `angle LAP = angleQAM = alpha`
`rArr "tan" alpha = |((3)/(2)-(1)/(2))/(1+(3)/(2) * (1)/(2))| = |(m-(3)/(2))/(1+m(3)/(2))|`
`rArr (4)/(7) = |(2m-3)/(2+3m)|`
`rArr (2m-3)/(2+3m) = +- (4)/(7)`
`m= (1)/(2), (29)/(2)`
But slope of incident ray is `(1)/(2)`
Thus, slope of reflected ray is `(29)/(2)`
Hence, the equation of reflected rays is
`y+1= (29)/(2)(x-1)`
or 29x-2y-31 = 0