JEE Mains · Maths · STD 11 - 9. straight line
Let the lines \(3 x-4 y-\alpha=0,8 x-11 y-33=0\), and \(2 x-3 y+\lambda=0\) be concurrent. If the image of the point
\((1,2)\) in the line \(2 x-3 y+\lambda=0\) is \(\left(\frac{57}{13}, \frac{-40}{13}\right)\), then \(|\alpha \lambda|\) is equal to
- A \(84\)
- B \(113\)
- C \(91\)
- D \(101\)
Answer & Solution
Correct Answer
(C) \(91\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \because \mathrm{PM}=\mathrm{QM} \\ & \text { So, } \mathrm{M}\left(\frac{\frac{57}{13}+1}{2}, \frac{\frac{-40}{13}+2}{2}\right) \\ & =\left(\frac{35}{13}, \frac{-7}{13}\right) \end{aligned}\) \(\because\) M lies on the line…
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