Let the lines 3x – 4y – = 0, 8x –11y – 33 = 0 and 2x – 3y + = 0 be concurrent. If the image of the point (1, 2) in the line 2x – 3y + = 0 is , then is equal to [2025]

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Solution

Q.
Let the lines 3x – 4y – $α$ = 0, 8x –11y – 33 = 0 and 2x – 3y + $λ$ = 0 be concurrent. If the image of the point (1, 2) in the line 2x – 3y + $λ$ = 0 is $(\frac{57}{13}, \frac{- 40}{13})$ , then $| α λ |$ is equal to [2025]

1 101

2 113

3 84

4 91

Ans.
(4)

As the three lines are concurrent,

$| \begin{matrix} 3 & - 4 & - α \\ 8 & - 11 & - 33 \\ 2 & - 3 & λ \end{matrix} |$ $= 0$

$\Rightarrow 3 (- 11 λ - 99) + 4 (8 λ + 66) - α (- 24 + 22) = 0$

$\Rightarrow - 33 λ - 297 + 32 λ + 264 + 2 α = 0$

$\Rightarrow - λ - 33 + 2 α = 0$ ...(i)

As image, (1, 2) w.r.t. 2x – 3y + $λ$ = 0 is $(\frac{57}{13}, \frac{- 40}{13})$

$\frac{\frac{57}{13} - 1}{2} = \frac{\frac{- 40}{13} - 2}{- 3} = - 2 (\frac{2 - 6 + λ}{4 + 9})$

$\Rightarrow \frac{22}{13} = \frac{- 2 (- 4 + λ)}{13} \Rightarrow λ = - 7$

Substitute $λ$ = –7 in (i), we get

$7 - 33 + 2 α = 0 \Rightarrow a = 13$

$∴ | α λ | = | 13 \times (- 7) | = 91$

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