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QJEEM25Ma7c2d: JEE Main 2025 Mathematics - Class 11 JEEM25Ma7c2d Solution: Let the lines , and be concurrent. If the image of the point in …

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 (5713,−4013)\left(\frac{57}{13}, \frac{-40}{13}\right) , then ∣αλ∣|\alpha \lambda| is equal to

Multiple Choice Options

  • A 91
  • B 113
  • C 101
  • D 84

Step-by-Step Verified Solution

JEE Main 2025 (Online) 24th January Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 2 English Explanation

∵PM=QM\because \mathrm{PM}=\mathrm{QM}

So, M(5713+12,−4013+22)M\left(\frac{\frac{57}{13}+1}{2}, \frac{\frac{-40}{13}+2}{2}\right)

=(3513,−713)=\left(\frac{35}{13}, \frac{-7}{13}\right)

∵M\because \mathrm{M} lies on the time

amp;2x−3y+λ=0amp;2(3513)−3(−713)+λ=0amp;λ=−7013+2113amp;=−9113=−7\begin{aligned}& 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\end{aligned}

amp;∣3amp;−4amp;−α8amp;−11amp;−332amp;3amp;λ∣=0amp;⇒3(−11λ−99)+4(8λ+66)−α(−24+22)=0amp;⇒33λ−297+32λ+264+24α−22α=0amp;⇒−λ+2α−33=0.... (1)amp;∴λ=−7amp;−(−7)+2α−33=0amp;2α=26amp;α=13amp;∴∣αλ∣=∣13×(−7)∣amp;=91\begin{aligned}& \left|\begin{array}{ccc}3 & -4 & -\alpha \\8 & -11 & -33 \\2 & 3 & \lambda\end{array}\right|=0 \\& \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 \quad\text{.... (1)}\\& \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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