Straight Lines – Concurrency of lines – Image of a point – JEE Main 24 January 2025 Shift 1

Solution:

Let the line $L$ be $2x-3y+\lambda=0$.
The image of point $A(1, 2)$ in $L$ is $B(\frac{57}{13}, \frac{-40}{13})$.
The midpoint $M$ of segment $AB$ must lie on the line $L$.


Coordinates of midpoint $M$:
$$ x_M = \frac{1 + \frac{57}{13}}{2} = \frac{\frac{70}{13}}{2} = \frac{35}{13} $$
$$ y_M = \frac{2 + \frac{-40}{13}}{2} = \frac{\frac{-14}{13}}{2} = \frac{-7}{13} $$
Substitute $M(\frac{35}{13}, \frac{-7}{13})$ into $2x-3y+\lambda=0$:
$$ 2\left(\frac{35}{13}\right) – 3\left(\frac{-7}{13}\right) + \lambda = 0 $$
$$ \frac{70}{13} + \frac{21}{13} + \lambda = 0 $$
$$ \frac{91}{13} + \lambda = 0 \Rightarrow 7 + \lambda = 0 \Rightarrow \lambda = -7 $$

Now, the three lines are concurrent:
1) $3x-4y-\alpha=0$
2) $8x-11y-33=0$
3) $2x-3y-7=0$ (Substituting $\lambda = -7$)

The determinant of the coefficients must be zero:
$$ \begin{vmatrix} 3 & -4 & -\alpha \\ 8 & -11 & -33 \\ 2 & -3 & -7 \end{vmatrix} = 0 $$
Expanding along the first row:
$$ 3[(-11)(-7) – (-33)(-3)] – (-4)[(8)(-7) – (-33)(2)] + (-\alpha)[(8)(-3) – (-11)(2)] = 0 $$
$$ 3[77 – 99] + 4[-56 + 66] – \alpha[-24 + 22] = 0 $$
$$ 3[-22] + 4[10] – \alpha[-2] = 0 $$
$$ -66 + 40 + 2\alpha = 0 $$
$$ -26 + 2\alpha = 0 \Rightarrow 2\alpha = 26 \Rightarrow \alpha = 13 $$

We need to find $|\alpha\lambda|$:
$$ |\alpha\lambda| = |13 \times (-7)| = |-91| = 91 $$

Ans. (2)