Question

Let the lines \[ 3x - 4y - \alpha = 0, \quad 8x - 11y - 33 = 0, \quad 2x - 3y + \lambda = 0 \] be concurrent. If the image of the point \( (1, 2) \) in the line \[ 2x - 3y + \lambda = 0 \text{ is } \left( \frac{57}{13}, \frac{-40}{13} \right), \text{ then } |\alpha \lambda| \text{ is equal to:} \]

• A. 84
• B. 91
• C. 113
• D. 101

Hint:

When solving for concurrent lines, use the determinant condition. For the image of a point with respect to a line, use the formula involving the coefficients of the line and the coordinates of the point. Always check the sign and the value of \( \lambda \) and \( \alpha \).

Answer 1

The Correct Option is B

Step 1: Let us solve \( 3x - 4y = \alpha \) and \( 8x - 11y = 33 \). 
Multiply the first equation by 8 and the second by 3. \[ 24x - 32y = 8\alpha \quad \text{and} \quad 24x - 33y = 99 \] Subtracting, \[ y = 8\alpha - 99 \] Then \[ 3x = 4(8\alpha - 99) + \alpha = 33\alpha - 396 \] or \[ x = 11\alpha - 132 \] So the intersection is \( (11\alpha - 132, 8\alpha - 99) \). 
Step 2: Use the concurrency condition. For the three lines to be concurrent, the determinant formed by their coefficients must be zero: \[ \begin{vmatrix} 3 & -4 & -\alpha \\ 8 & -11 & -33 \\ 2 & -3 & \lambda \end{vmatrix} = 0 \] Expanding the determinant: \[ 3(-11\lambda - 99) - (-4)(8\lambda + 66) - \alpha(-24 + 22) = 0 \] \[ -33\lambda - 297 + 32\lambda + 264 + 2\alpha = 0 \] \[ \lambda - 33 + 2\alpha = 0 \] So \( \lambda = 33 - 2\alpha \). 
Step 3: Use the reflection property. The image of the point \( (1, 2) \) in the line \( 2x - 3y + \lambda = 0 \) is given as \[ \left( \frac{57}{13}, \frac{-40}{13} \right) \] The midpoint of the point and its image is \[ \left( \frac{1 + 57/13}{2}, \frac{2 - 40/13}{2} \right) = \left( \frac{70/13}{2}, \frac{-14/13}{2} \right) = \left( \frac{35}{13}, \frac{-7}{13} \right) \] This midpoint lies on the line \( 2x - 3y + \lambda = 0 \), so \[ 2\left( \frac{35}{13} \right) - 3\left( \frac{-7}{13} \right) + \lambda = 0 \] \[ \frac{70}{13} + \frac{21}{13} + \lambda = 0 \] \[ \lambda = -\frac{91}{13} = -7 \] Then the slope of the line joining the point and its image is \[ \frac{2 - (-40/13)}{1 - 57/13} = \frac{26 + 40}{-44} = \frac{66}{-44} = -\frac{3}{2} \] The slope of the line \( 2x - 3y + \lambda = 0 \) is \( \frac{2}{3} \). Since \[ -\frac{3}{2} \cdot \frac{2}{3} = -1 \] the two lines are perpendicular. 
Step 4: Solve for \(\alpha\) and calculate the desired value. We have \( \lambda = 33 - 2\alpha \) and \( \lambda = -7 \), so \[ -7 = 33 - 2\alpha \implies 2\alpha = 40 \implies \alpha = 20 \] Then \[ |\alpha \lambda| = |20 \cdot (-7)| = |-140| = 140 \] Using concurrency as \( (11\alpha-132, 8\alpha -99) \) on \( 2x - 3y + \lambda = 0 \), then \[ 2(11\alpha-132) - 3(8\alpha-99) + \lambda = 0 \] \[ 22\alpha - 264 - 24\alpha + 297 + \lambda = 0 \] \[ -2\alpha + 33 + \lambda = 0 \] So \( \lambda = 2\alpha - 33 \). Since the midpoint between \( (1, 2) \) and \( \left(\frac{57}{13}, \frac{-40}{13} \right) \) lies on \( 2x - 3y + \lambda = 0 \), \[ 2\left( \frac{35}{13} \right) - 3\left( \frac{-7}{13} \right) + \lambda = 0 \] \[ \frac{70}{13} + \frac{21}{13} + \lambda = 0 \] So \( \lambda = -\frac{91}{13} = -7 \). \[ -7 = 2\alpha - 33 \implies 2\alpha = 26 \implies \alpha = 13 \] Then \[ |\alpha\lambda| = |13 \cdot (-7)| = 91 \] Final Answer: The correct answer is (2) 91.