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

Question

Aicharya · Accepted Answer

$\begin{aligned}
& \because \mathrm{PM}=\mathrm{QM} \\ & \text { So, } \mathrm{M}\left(\frac{\frac{57}{13}+1}{2}, \frac{\frac{-40}{13}+2}{2}\right) \\ & =\left(\frac{35}{13}, \frac{-7}{13}\right)
\end{aligned}$
$\because$ M lies on the time
$\begin{gathered}2 x-3 y+\lambda=0 \\ 2\left(\frac{35}{13}\right)-3\left(\frac{-7}{13}\right)+\lambda=0 \\ \lambda=-\frac{70}{13}+\frac{21}{13} \\ =\frac{-91}{13}=-7 \\ \left|\begin{array}{ccc}3 & -4 & -\alpha \\ 8 & -11 & -33 \\ 2 & 3 & \lambda\end{array}\right|=0\end{gathered}$
$\begin{aligned}
& \Rightarrow 3(-11 \lambda-99)+4(8 \lambda+66)-\alpha(-24+22)=0 \\ & \Rightarrow 33 \lambda-297+32 \lambda+264+24 \alpha-22 \alpha=0 \\ & \Rightarrow-\lambda+2 \alpha-33=0 \\ & \therefore \lambda=-7 \\ & -(-7)+2 \alpha-33=0 \\ & 2 \alpha=26 \\ & \alpha=13 \\ & \therefore|\alpha \lambda|=|13 \times(-7)| \\ & =91
\end{aligned}$ .

Let the lines $3 x-4 y-\alpha=0,8 x-11 y-33=0$, and $2 x-3 y+\lambda=0$ be concurrent. If the image of the…

Solution