JEE MAIN - Mathematics (2025 - 24th January Morning Shift - No. 18)

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
91
113
101
84

Explanation

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

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

So, $M\left(\frac{\frac{57}{13}+1}{2}, \frac{\frac{-40}{13}+2}{2}\right)$

$$=\left(\frac{35}{13}, \frac{-7}{13}\right)$$

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

$$\begin{aligned} & 2 x-3 y+\lambda=0 \\ & 2\left(\frac{35}{13}\right)-3\left(\frac{-7}{13}\right)+\lambda=0 \\ & \lambda=-\frac{70}{13}+\frac{21}{13} \\ & =\frac{-91}{13}=-7 \end{aligned}$$

$$\begin{aligned} & \left|\begin{array}{ccc} 3 & -4 & -\alpha \\ 8 & -11 & -33 \\ 2 & 3 & \lambda \end{array}\right|=0 \\ & \Rightarrow 3(-11 \lambda-99)+4(8 \lambda+66)-\alpha(-24+22)=0 \\ & \Rightarrow 33 \lambda-297+32 \lambda+264+24 \alpha-22 \alpha=0 \\ & \Rightarrow-\lambda+2 \alpha-33=0 \quad\text{.... (1)}\\ & \therefore \lambda=-7 \\ & -(-7)+2 \alpha-33=0 \\ & 2 \alpha=26 \\ & \alpha=13 \\ & \therefore|\alpha \lambda|=|13 \times(-7)| \\ & =91 \end{aligned}$$

Comments (0)

Advertisement