Question

A ray of light is incident along a line which meets another line, $7x-y+1=0$, at the point $(0,1)$. The ray is then reflected from his point along the line, $y+2x=1$. Then the equation of the line of incidence of the ray of light is :

• A. $41x-38y+38=0$
• B. $41x+25y-25=0$
• C. $41x+38y-38=0$
• D. $41x-25y+25=0$

Answer 1

Answer: (A)

Incidene line
$L_1+\lambda L_2=0$
$(7x-y+1)+\lambda (v+2x-1)=0$

Let a point $(1,-1)$ on $y+2x=1$
And image of $(1,-1)$ lie on incidence line in
$7x-v+1=0$
$\frac{x-1}{1}=\frac{y+1}{-1}=\frac{-2(7+1+1)}{50}=x=\frac{-38}{25},y=\frac{-16}{25}$
$\left(7\left(\frac{-38}{25}\right)+\frac{16}{25}+1\right)+\lambda \left(\frac{-16}{25}-\frac{76}{25}-1\right)$
$\lambda =\frac{-225}{117}$
$(7x-y+1)\frac{-225}{117}(y+2x-1)=0$
$369x-342y+342=0$
$41x-38y+38=0$