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 ax + by \(+1=0\), then \(\mathrm{a}^{2}+\mathrm{b}^{2}+3 \mathrm{ab}\) is equal to_.

Answer 1

Correct answer is : 1 

For B'

\( \begin{aligned} \frac{x-7}{2}=\frac{y-2}{1}=-2\left(\frac{14+2-6}{5}\right) \end{aligned} \)

\( \begin{aligned} \frac{x-7}{2}=\frac{y-2}{1}=-4 \\ \end{aligned} \)

\( \begin{aligned} x=-1 \quad y=-2 \quad B^{\prime}(-1,-2) \end{aligned} \)

incident ray \(\mathrm{AB}^{\prime}\)

\(\mathrm{M}_{\mathrm{AB}^{\prime}}=3\)

\(\mathrm{y}+2=3(\mathrm{x}+1)\)

\(3 \mathrm{x}-\mathrm{y}+1=0\)

\(\mathrm{a}=3 \mathrm{~b}=-1\)

\(\mathrm{a}^{2}+\mathrm{b}^{2}+3 \mathrm{ab}=9+1-9=1\)