Let the lines 3 x-4 y-=0,8 x-11 y-33=0, and 2 x-3 y+=0 be concurrent.

Question

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

Accepted Answer

$$\because \mathrm{PM}=\mathrm{QM}$$

So, $M\left(\frac{\frac{57}{13}+1}{2}, \frac{\frac{-40}{13}+2}{2}\right)$

$$=\left(\frac{35}{13}, \frac{-7}{13}\right)$$

$\because \mathrm{M}$ lies on the time

$$\begin{aligned}
& 2 x-3 y+\lambda=0 \
& 2\left(\frac{35}{13}\right)-3\left(\frac{-7}{13}\right)+\lambda=0 \
& \lambda=-\frac{70}{13}+\frac{21}{13} \
& =\frac{-91}{13}=-7
\end{aligned}$$

$$\begin{aligned}
& \left|\[\begin{array}{ccc} 3 & -4 & -\alpha \ 8 & -11 & -33 \ 2 & 3 & \lambda \end{array}\end{aligned}$$\right|=0 \
\& \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 \quad\text{.... (1)}\
\& \therefore \lambda=-7 \
\& -(-7)+2 \alpha-33=0 \
\& 2 \alpha=26 \
\& \alpha=13 \
\& \therefore|\alpha \lambda|=|13 \times(-7)| \
\& =91
\]

Solutions · 1