Question

A light beam eminating 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 :

• A. $3x-y+1=0$
• B. $x+3y-13=0$
• C. $3x+y-15=0$
• D. $x-3y+5=0$

Answer 1

Answer: (B)

Image of $A(3,10)$ in $2x+y-6=0$
$\frac{x-3}{2}=\frac{y-10}{1}=-2\left(\frac{6+10-6}{2^2+1^2}\right)$
$\frac{x-3}{2}=\frac{y-10}{1}=-4$
$A^{\prime }=(-5,6)$
Equation of $A^{\prime }B$ is $y-3=\left(\frac{6-3}{-5-4}\right)(x-4)$
$y-3=-\frac{1}{3}(x-4)$
$3y-9=-x+4$
$\Rightarrow x+3y-13=0$