JEE MAIN - Mathematics (2024 - 8th April Evening Shift - No. 30)
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
Explanation
Equation of incident ray : $$a x+b y+1=0$$
Using mirror image,
$$\frac{m-7}{2}=\frac{n-2}{1}=\frac{-2(14+2-6)}{5}$$
$$\begin{array}{l|l} \frac{m-7}{2}=-4 & n-2=-4 \\ m=-8+7 & n=-2 \\ m=-1 & \end{array}$$_8th_April_Evening_Shift_en_30_1.png)
$${ }^*$$ Note: It can be observed from diagram $$A, P, B$$' are collinear.
Equation of Incident Ray,
Using two-point form,
$$\begin{aligned} & (y-10)=\frac{10+2}{3+1}(x-3) \\ & (y-10)=\frac{12}{4}(x-3) \\ & y-10=+3(x-3) \\ & y-10=+3 x-9 \\ & 3 x-y+1=0 \end{aligned}$$
On comparing,
$$\begin{aligned} & a=3 \\ & b=-1 \end{aligned}$$
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