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 $2 \mathrm{x}-3 \mathrm{y}+\lambda=0$ is $\left(\frac{57}{13}, \frac{-40}{13}\right)$, then $|\alpha \lambda|$ is equal to :

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Straight Lines

Let the lines

3 x - 4 y - α = 0, 8 x - 11 y - 33 = 0

, and

2 x - 3 y + λ = 0

be concurrent. If the image of the point ( 1,2 ) in the line

2 x - 3 y + λ = 0

(\frac{57}{13}, \frac{- 40}{13})

, then

∣ α λ ∣

is equal to :

Answer: (B)

The image of

(1, 2)

2 x - 3 y + λ = 0

(\frac{57}{13}, \frac{- 40}{13})

, so the midpoint of these two points lies on the line:

M = (\frac{1 + \frac{57}{13}}{2}, \frac{2 - \frac{40}{13}}{2}) = (\frac{35}{13}, - \frac{7}{13}) .

Substituting into

2 x - 3 y + λ = 0

2 (\frac{35}{13}) - 3 (- \frac{7}{13}) + λ = 0 \Rightarrow \frac{70}{13} + \frac{21}{13} + λ = 0 \Rightarrow λ = - 7.

For concurrency, all three lines meet at one point. Find the intersection of

8 x - 11 y - 33 = 0

and

2 x - 3 y - 7 = 0

(using

λ = - 7

):
from the second,

x = \frac{7 + 3 y}{2}

; substituting into the first,

8 \cdot \frac{7 + 3 y}{2} - 11 y - 33 = 0 \Rightarrow 4 (7 + 3 y) - 11 y - 33 = 0 \Rightarrow 28 + y - 33 = 0 \Rightarrow y = 5, x = \frac{7 + 15}{2} = 11.

This point

(11, 5)

must also lie on

3 x - 4 y - α = 0

3 (11) - 4 (5) - α = 0 \Rightarrow 33 - 20 - α = 0 \Rightarrow α = 13.

∣ α λ ∣ = ∣13 \times (- 7) ∣ = 91.

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option