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

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

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Solution:

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