Question

A light beam emanating from the point A$(3,10)$ reflects from the line $2x+y-6=0$ and then passes through the point B$(5,6)$. The equation of the incident and reflected beams respectively are:

• A. $4x-3y+18=0andy=6$
• B. $x-2y+8=0andx=5$
• C. $x+2y-8=0andy=6$
• D. None of these

Answer 1

The correct option is A $4x-3y+18=0andy=6$

Let The images of point $A(3,10)$ and $B(5,6)$ about the line $2x+y-6=0$ are $A^{\prime }(\alpha ,\beta )$ and $B^{\prime }(\gamma ,\delta )$ respectively.

Then, $\frac{\alpha -3}{2}=\frac{\beta -10}{1}=\frac{-2(6+10-6)}{2^2+1^2}=4$

$\therefore \alpha =-5,\beta =6$

Hence $A^{\prime }(-5,6)$

Also, $\frac{\gamma -5}{2}=\frac{\delta -6}{1}=\frac{-2(10+6-6)}{2^2+1^2}=-4$

$\therefore \gamma =-3,\delta =2$

Hence $B^{\prime }(-3,2)$

From Above figure you can see, the equation of incident ray $A{B}^{\prime }$ is,

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

$\Rightarrow 4x-3y+18=0$

Similarly the equation of reflected ray $A^{\prime }B$ is,

$\Rightarrow y-6=\frac{6-6}{5+5}(x+5)$

$\Rightarrow y=6$

Hence the incident ray and reflected ray are $\Rightarrow 4x-3y+18=0$ and $\Rightarrow y=6$ respectively. Correct option is $A$.