JEE Main 2025 (Online) 24th January Morning ShiftQuestion 1
Let the lines $3x-4y-\alpha=0$, $8x-11y-33=0$, and $2x-3y+\lambda=0$ be concurrent. If the image of the point $(1,2)$ in the line $2x-3y+\lambda=0$ is $\left(\frac{57}{13},\frac{-40}{13}\right)$, then $|\alpha\lambda|$ is equal to
Points and Straight Lines
Solution
The midpoint $\left(\frac{35}{13},-\frac{7}{13}\right)$ lies on $2x-3y+\lambda=0$, so $\lambda=-7$. Then $8x-11y=33$ and $2x-3y=7$ meet at $(11,5)$, giving $\alpha=33-20=13$ and $|\alpha\lambda|=91$.