Key Concepts and Formulas

• Image of a Point in a Line: If $P(x_1, y_1)$ has an image $Q(x_2, y_2)$ in the line $ax + by + c = 0$, then the midpoint of $PQ$ lies on the line.
• Concurrency of Three Lines: Three lines $A_1x + B_1y + C_1 = 0$, $A_2x + B_2y + C_2 = 0$, and $A_3x + B_3y + C_3 = 0$ are concurrent if and only if $\begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix} = 0$

Step-by-Step Solution

Step 1: Find the midpoint of the point and its image.

We are given the point $P(1, 2)$ and its image $Q\left(\frac{57}{13}, \frac{-40}{13}\right)$. We need to find the midpoint $M$ of the line segment $PQ$. The midpoint formula is given by $M\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$. Therefore, the coordinates of $M$ are: $x_M = \frac{1 + \frac{57}{13}}{2} = \frac{\frac{13 + 57}{13}}{2} = \frac{\frac{70}{13}}{2} = \frac{35}{13}$ $y_M = \frac{2 + \frac{-40}{13}}{2} = \frac{\frac{26 - 40}{13}}{2} = \frac{\frac{-14}{13}}{2} = \frac{-7}{13}$ So, the midpoint is $M\left(\frac{35}{13}, \frac{-7}{13}\right)$.

Step 2: Find the value of $\lambda$ using the midpoint.

Since the midpoint $M$ lies on the line $2x - 3y + \lambda = 0$, we can substitute the coordinates of $M$ into the equation of the line to solve for $\lambda$. $2\left(\frac{35}{13}\right) - 3\left(\frac{-7}{13}\right) + \lambda = 0$ $\frac{70}{13} + \frac{21}{13} + \lambda = 0$ $\frac{91}{13} + \lambda = 0$ $7 + \lambda = 0$ Thus, $\lambda = -7$.

Step 3: Set up the determinant for the concurrency condition.

We are given three concurrent lines: $3x - 4y - \alpha = 0$ $8x - 11y - 33 = 0$ $2x - 3y + \lambda = 0$ For these lines to be concurrent, the determinant of their coefficients must be zero. $\begin{vmatrix} 3 & -4 & -\alpha \\ 8 & -11 & -33 \\ 2 & -3 & \lambda \end{vmatrix} = 0$

Step 4: Expand the determinant.

Expanding the determinant along the first row: $3((-11)(\lambda) - (-33)(-3)) - (-4)((8)(\lambda) - (-33)(2)) + (-\alpha)((8)(-3) - (-11)(2)) = 0$ $3(-11\lambda - 99) + 4(8\lambda + 66) - \alpha(-24 + 22) = 0$ $-33\lambda - 297 + 32\lambda + 264 - \alpha(-2) = 0$ $-\lambda - 33 + 2\alpha = 0$

Step 5: Solve for $\alpha$.

Substitute $\lambda = -7$ into the equation: $-(-7) - 33 + 2\alpha = 0$ $7 - 33 + 2\alpha = 0$ $-26 + 2\alpha = 0$ $2\alpha = 26$ $\alpha = 13$

Step 6: Calculate $|\alpha \lambda|$.

We have $\alpha = 13$ and $\lambda = -7$. $|\alpha \lambda| = |13 \times (-7)| = |-91| = 91$

Common Mistakes & Tips

• Be careful with signs when calculating the midpoint and expanding the determinant.
• Double-check all arithmetic operations to avoid errors.
• Remember the condition for concurrency of three lines.

Summary

We first found the value of $\lambda$ using the image property of a point in a line. Then, we used the concurrency condition of three lines to find the value of $\alpha$. Finally, we calculated $|\alpha \lambda|$, which resulted in 91.

The final answer is \boxed{91}, which corresponds to option (A).