Let the lines 3x - 4y - alpha = 0,8x - 11y - | vedclass

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(D) Let $P = (1, 2)$ and $Q = \left(\frac{57}{13}, \frac{-40}{13}\right)$. The line $2x - 3y + \lambda = 0$ is the perpendicular bisector of the segme

Vedclass · Accepted Answer

$91$

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(D) Let $P = (1, 2)$ and $Q = \left(\frac{57}{13}, \frac{-40}{13}ight)$. The line $2x - 3y + \lambda = 0$ is the perpendicular bisector of the segment $PQ$. The midpoint $M$ of $PQ$ is given by: $M = \left(\frac{1 + \frac{57}{13}}{2}, \frac{2 - \frac{40}{13}}{2}ight) = \left(\frac{\frac{70}{13}}{2}, \frac{\frac{-14}{13}}{2}ight) = \left(\frac{35}{13}, \frac{-7}{13}ight)$. Since $M$ lies on the line $2x - 3y + \lambda = 0$: $2\left(\frac{35}{13}ight) - 3\left(\frac{-7}{13}ight) + \lambda = 0$ $\frac{70}{13} + \frac{21}{13} + \lambda = 0$ $\Rightarrow \frac{91}{13} + \lambda = 0$ $\Rightarrow 7 + \lambda = 0$ $\Rightarrow \lambda = -7$. Since the three lines are concurrent,the determinant of their coefficients must be zero: $\begin{vmatrix} 3 & -4 & -\alpha \ 8 & -11 & -33 \ 2 & -3 & \lambda \end{vmatrix} = 0$ Substituting $\lambda = -7$: $\begin{vmatrix} 3 & -4 & -\alpha \ 8 & -11 & -33 \ 2 & -3 & -7 \end{vmatrix} = 0$ $3(77 - 99) + 4(-56 + 66) - \alpha(-24 + 22) = 0$ $3(-22) + 4(10) - \alpha(-2) = 0$ $-66 + 40 + 2\alpha = 0$ $2\alpha = 26 \Rightarrow \alpha = 13$. Thus,$|\alpha \lambda| = |13 	imes (-7)| = |-91| = 91$.

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:

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