Question

A light beam emanating from the point $A(3,10)$ reflects from the straight line $2x+y-6=0$ and then passes through the point $B(4,3)$. The equation of the reflected beam is $x+3y-\lambda =0$, then the value of $\lambda$ is?

• A. $11$
• B. $12$
• C. $13$
• D. $14$

Answer 1

Answer: (C)

$13$

let the slope of incident ray be $m_1$ and reflected ray be $m_2$ eq. of

two lines are :

$\begin{array}{c}y-10=m_1(x-3)\\ y-2=m_2(x-7)\end{array}$

Point of intersection on $2x+y-6=0$ is $(a,6-2a)$

Normal to $2x+y-6=0$ has slope $=\frac{1}{2}$

$\because$ Incident and reflected bean make equal angles with normal.

$\Rightarrow \frac{m_1-1/2}{1+m_1{x}_{\frac{1}{2}}}=\frac{1/2\sigma -m_2}{1+m_2{x}_{1/2}}$

$\Rightarrow 3(m_1+m_2)+4m_1{m}_2-4=0$

Also, incident ray passes through (3,10) and $(a,6-2)$ a)

$\Rightarrow m_1=\frac{-4-2a}{a-3}$

Similarly, $m_2=\frac{4-2a}{a-7}$

Substitutiung $m_1,m_2$ in above equ;

$3(\frac{-4-2a}{a-3}+\frac{4-2a}{a-7})+4\left(\frac{-4-2a}{a-3}\right)\left(\frac{4-2a}{a-7}\right)-4>0$

we get, $a=1$

$\Rightarrow m_1=3,m_2=\frac{1}{3}$

$\therefore$ Reflected ray is

$(y-2)=\frac{-1}{3}(x-7)$

$\Rightarrow x+3y-13=0$

$\Rightarrow \lambda =13$

option $C$ is correct