JEE Mains · Maths · STD 11 - 9. straight line
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 \(a x+\) by \(+1=0\), then \(a^2+b^2+3 a b\) is equal to ...........
- A \(50\)
- B \(10\)
- C \(1\)
- D \(5\)
Answer & Solution
Correct Answer
(C) \(1\)
Step-by-step Solution
Detailed explanation
\(\text { For } B^{\prime} \quad \)\( \frac{x-7}{2}=\frac{y-2}{1}=-2\left(\frac{14+2-6}{5}\right) \) \( \frac{x-7}{2}=\frac{y-2}{1}=-4 \) \( x=-1 \quad y=-2 \quad B^{\prime}(-1,-2)\) incident ray \(\mathrm{AB}^{\prime}\) \( \mathrm{M}_{\mathrm{AB}^{\prime}}=3 \)…
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