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JEE Main 202524 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Straight Lines

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Straight Lines

Let the lines 3 x-4 y-α=0,8 x-11 y-33=0 , and 2 x-3 y+λ=0 be...

Let the lines 3x−4y−α=0,8x−11y−33=03 x-4 y-\alpha=0,8 x-11 y-33=0, and 2x−3y+λ=02 x-3 y+\lambda=0 be concurrent. If the image of the point (1,2)(1,2) in the line 2x−3y+λ=02 x-3 y+\lambda=0 is \left(\frac{57}{13}, \frac{-40}{13}\right),then, then|\alpha \lambda| is equal to

Options

  1. A8484
  2. B113113
  3. C9191
  4. D101101

Correct answer

C. 9191

Step-by-step solution

∵PM=QM So, M(5713+12,−4013+22)=(3513,−713)\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}\becauseMliesonthelineM lies on the line\begin{gathered}2 x-3 y+\lambda=0 \\ 2\left(\frac{35}{13}\right)-3\left(\frac{-7}{13}\right)+\lambda=0 \\ \lambda=-\frac{70}{13}+\frac{21}{13} \\ =\frac{-91}{13}=-7 \\ \left|3−4−α8−11−3323λ\begin{array}{ccc}3 & -4 & -\alpha \\ 8 & -11 & -33 \\ 2 & 3 & \lambda\end{array}\right|=0\end{gathered} ⇒3(−11λ−99)+4(8λ+66)−α(−24+22)=0⇒33λ−297+32λ+264+24α−22α=0⇒−λ+2α−33=0∴λ=−7−(−7)+2α−33=02α=26α=13∴∣αλ∣=∣13×(−7)∣=91\begin{aligned} & \Rightarrow 3(-11 \lambda-99)+4(8 \lambda+66)-\alpha(-24+22)=0 \\ & \Rightarrow 33 \lambda-297+32 \lambda+264+24 \alpha-22 \alpha=0 \\ & \Rightarrow-\lambda+2 \alpha-33=0 \\ & \therefore \lambda=-7 \\ & -(-7)+2 \alpha-33=0 \\ & 2 \alpha=26 \\ & \alpha=13 \\ & \therefore|\alpha \lambda|=|13 \times(-7)| \\ & =91 \end{aligned}$
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