Question

A light beam, emanating from the point $(3,10)$ reflects from the straight line $2x+y-6=0$ then passes through the point $(7,2)$. Find the equation of the incident and reflected rays.

Answer 1

Given ray start from point $A(3,10)$ and after reflection it end at $B(7,2)$

Here the ray from point A incident on line $2x+y-6=0$ at point say $I(a,b)$ and after reflection end at $B$

If we join A and B triangle $\Delta AIB$ is formed

Since angle of incidence is equal to angle of reflection

So normal of line $2x+y-6=0$ is act as median of $\Delta AIB$

So Noraml cuts line AB at mid-point of line segment AB

midpoint $C(5,6)$

Equation of normal IC perpendicular to line $2x+y-6=0$ is

$x-2y-\lambda =0$

Above equation or median passing through C

$5-12-\lambda =0$

$\lambda =-7$

Hence equation of median or normal $IC:x-2y+7=0$

Since Point I lie on $2x+y-6=0$

Hence $2a+b-6=0------\left(1\right)$

Since Point I also lie on $x-2y+7=0$

Hence $a-2b+7=0------\left(2\right)$

On solving eq (1) and (2)

$2(2b-7)+b-6=0$

$4b-14+b-6=0$

$5b=20$

$b=4$

from eq (2)

$a=2b-7=1$

Incident point $I(1,4)$

Equation of incident ray from $A(3,10)$ and $I(1,4)$

$y-10=\frac{10-4}{3-1}(x-3)$

$y-10=\frac{6}{2}(x-3)$

$y-10=3x-9$

$3x-y+1=0$

Equation of reflected ray from $B(7,2)$ and $I(1,4)$

$y-2=\frac{2-4}{7-1}(x-7)$

$y-2=\frac{-2}{6}(x-7)$

$3y-6=-x+7$

$x+3y-13=0$