Question

If a light ray emiting from the point source $A(3,10)$ gets reflected from the straight line $2x+y-6=0$ and then passes through the point $B(7,2)$. Then which of the following is/are true

• A. Equation of the incident ray is $x+y-13=0$
• B. Equation of the incident ray is $3x-y+1=0$
• C. Equation of the reflected ray is $x-2y-3=0$
• D. Equation of the reflected ray is $x+3y-13=0$

Answer 1

Answer: (B), (D)

Equation of the reflected ray is $x+3y-13=0$

Let $A^{\prime },B^{\prime }$ be image's of $A,B$ w.r.t $2x+y-6=0,A^{\prime }\equiv (\alpha ,\beta )$ ,$B^{\prime }\equiv (r,s)$

Now, using image formulae,
$\frac{\alpha -3}{2}=\frac{\beta -10}{1}=\frac{-2(6+10-6)}{5}$$\Rightarrow A^{\prime }=(-5,6)$
$\frac{r-7}{2}=\frac{s-2}{1}=\frac{-2(14+2-6)}{5}$$\Rightarrow B^{\prime }=(-1,-2)$

It can be observed from the diagram points $A^{\prime },P,B$ and $B^{\prime },P,A$ are collinear.
$\therefore$ Equation of incident ray i.e. Line joining Points $A,P,B^{\prime }=3x-y+1=0$
and equation of reflected ray i.e. Line joining Points $B,P,A^{\prime }=x+3y-13=0$