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

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_________

Solution


For $B^{\prime} \quad \frac{x-7}{2}=\frac{y-2}{1}=-2\left(\frac{14+2-6}{5}\right)$ $\begin{aligned} & \frac{x-7}{2}=\frac{y-2}{1}=-4 \\ & x=-1 \quad y=-2 \quad B^{\prime}(-1,-2)\end{aligned}$ incident ray $\mathrm{AB}^{\prime}$ $\begin{aligned} & M_{\mathrm{AB}^{\prime}}=3 \\ & y+2=3(x+1) \\ & 3 x-y+1=0 \\ & a=3 b=-1 \\ & a^2+b^2+3 a b=9+1-9=1\end{aligned}$

Asked in: JEE Main 2024 (08 Apr Shift 2)

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