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QJEEM24Ma98cf: JEE Main 2024 Mathematics - Class 11 JEEM24Ma98cf Solution: Let a ray of light passing through the point reflects on the…

Let a ray of light passing through the point (3,10)(3,10) reflects on the line 2x+y=62 x+y=6 and the reflected ray passes through the point (7,2)(7,2) . If the equation of the incident ray is ax+by+1=0a x+b y+1=0 , then a2+b2+3aba^2+b^2+3 a b is equal to _________.

Step-by-Step Verified Solution

Equation of incident ray : ax+by+1=0a x+b y+1=0

Using mirror image,

m−72=n−21=−2(14+2−6)5\frac{m-7}{2}=\frac{n-2}{1}=\frac{-2(14+2-6)}{5}

m−72=−4amp;n−2=−4m=−8+7amp;n=−2m=−1amp;\begin{array}{l|l}\frac{m-7}{2}=-4 & n-2=-4 \\m=-8+7 & n=-2 \\m=-1 &\end{array}

JEE Main 2024 (Online) 8th April Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 16 English Explanation

∗{ }^* Note: It can be observed from diagram A,P,BA, P, B ' are collinear.

Equation of Incident Ray,

Using two-point form,

amp;(y−10)=10+23+1(x−3)amp;(y−10)=124(x−3)amp;y−10=+3(x−3)amp;y−10=+3x−9amp;3x−y+1=0\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,

amp;a=3amp;b=−1\begin{aligned}& a=3 \\& b=-1\end{aligned}

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