Question

A ray of light is sent along the line $x-2y-3=0$ on reaching the line $3x-2y-5=0$ the ray is reflected from it, then the equation of the line containing the reflected ray is

• A. $2x-29y-30=0$
• B. $29x-2y-31=0$
• C. $3x-31y+37=0$
• D. $31x-3y+37=0$

Answer 1

The correct option is B $29x-2y-31=0$

$x-2y-3=0\Rightarrow y=\frac{x}{2}-\frac{3}{2}$

Angle made by line with x-axis $ta{n}^{-1}\left(\frac{1}{2}\right)$

$3x-2y-5=0\Rightarrow y=\frac{3x}{2}-\frac{5}{2}$

Angle made by this line with x-axis $ta{n}^{-1}\frac{3}{2}$

Leth the slope of the unkown line be $m$

$\therefore {\theta }_i=ta{n}^{-1}\left(\frac{3}{2}\right)-ta{n}^{-1}\left(\frac{1}{2}\right)$

Similarly ${\theta }_r=ta{n}^{-1}\left(m\right)-ta{n}^{-1}\left(\frac{3}{2}\right)$

Now, ${\theta }_i={\theta }_r$

$ta{n}^{-1}\left(\frac{3}{2}\right)-ta{n}^{-1}\left(\frac{1}{2}\right)=ta{n}^{-1}\left(m\right)-ta{n}^{-1}\left(\frac{3}{2}\right)$

$ta{n}^{-1}\left(m\right)=2ta{n}^{-1}\frac{3}{2}-ta{n}^{-1}\frac{1}{2}$

$m=tan(2ta{n}^{-1}\frac{3}{2}-ta{n}^{-1}\frac{1}{2})$

Solving,

$m=\frac{29}{2}$

Now, the unknown reflected line is $y=\frac{29}{2}x+c$

Solving the given lines simultaneously we get the point to incidence as $(1,-1)$

Putting this in the equation of reflected line

$-1=\frac{29}{2}+c\Rightarrow c=\frac{-31}{2}$

Thus the equation of reflected lin is $29x-2y-31=0$