Let a ray of light passing through the point (3,10) reflects on the line 2 x+y=6 and the r…

Question

Let a ray of light passing through the point $(3,10)$ reflects on the line $2 x+y=6$ and the reflected ray passes through the point $(7,2)$. If the equation of the incident ray is $a x+b y+1=0$, then $a^2+b^2+3 a b$ is equal to \_\_...

Answer: 1

Accepted Answer

Equation of incident ray : $ax+by+1=0$

Using mirror image,

$\frac{m-7}{2}=\frac{n-2}{1}=\frac{-2(14+2-6)}{5}$

$\begin{gathered} \frac{m-7}{2}=-4 n-2=-4 \ m=-8+7 n=-2 \ m=-1 \end{gathered}$

${}^{\ast }$ Note: It can be observed from diagram $A,P,B$' are collinear.

Equation of Incident Ray,

Using two-point form,

$\begin{gathered} (y-10)=\frac{10+2}{3+1}(x-3) \ (y-10)=\frac{12}{4}(x-3) \ y-10=+3(x-3) \ y-10=+3x-9 \ 3x-y+1=0 \end{gathered}$

On comparing,

$\begin{gathered} a=3 \ b=-1 \end{gathered}$

Solutions · 1