JEE Mathematics 2025 Question with Solution

Given: The lines $3x - 4y - \alpha = 0$, $8x - 11y - 33 = 0$ and $2x - 3y + \lambda = 0$ are concurrent. The image of $(1,2)$ in the line $2x - 3y + \lambda = 0$ is $\left(\frac{57}{13}, \frac{-40}{13}\right)$.

Find: $|\alpha\lambda|$.

Use the reflection property. The midpoint of a point and its image in a line lies on that line.

Midpoint of $(1,2)$ and $\left(\frac{57}{13}, \frac{-40}{13}\right)$ is

Since this midpoint lies on the line $2x - 3y + \lambda = 0$,

Now use concurrency of the three lines. For concurrent lines,

Substituting $\lambda = -7$,

Expanding,

Therefore,

So the correct option is B.

Given: The three lines are concurrent, and the image information gives the line $2x - 3y + \lambda = 0$.

Find: $|\alpha\lambda|$.

First solve

and

Multiply the first equation by $8$ and the second by $3$:

Subtracting,

Then from $3x - 4y = \alpha$,

So the intersection point of the first two lines is $\bigl(11\alpha - 132,\, 8\alpha - 99\bigr)$.

Because the third line is concurrent with them, this point also lies on $2x - 3y + \lambda = 0$. Hence,

Now use the image condition. The midpoint of $(1,2)$ and $\left(\frac{57}{13}, \frac{-40}{13}\right)$ is

This lies on $2x - 3y + \lambda = 0$, so

Substitute in $\lambda = 2\alpha - 33$:

Therefore,

Hence the correct option is B.