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

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] The mid-point of $(1,2)$ and its image is $\left(\frac{35}{13},-\frac7{13}\right)$, which lies on $2x-3y+\lambda=0$: $\frac{70+21}{13}+\lambda=0\Rightarrow\lambda=-7$. Then $8x-11y=33$ and $2x-3y=7$ meet at $(11,5)$. Substituting in $3x-4y-\alpha=0$: $\alpha=33-20=13$. So $|\alpha\lambda|=91$. Answer (3).

Q. 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

Explanation

Reviewed by Dr. Abubacker

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