$\begin{array}{l} C o n s i d e r t h e f a m i l y o f l i n e s 5 x + 3 y - 2 + λ_{1} (3 x - y - 4) = 0 a n d x - y + 1 + λ_{2} (2 x - y - 2) = 0. E q u a t i o n o f a \\ s t r a i g h t l i n e t h a t b e l o n g t o b o t h f a m i l i e s i s \end{array}$

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$62 x - 25 y + 86 = 0$

$5 x - 2 y - 7 = 0$

$25 x - 62 y + 86 = 0$

$25 x - 62 y = 86$

answer is D.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n 5 x + 3 y - 2 λ_{1} (3 x - y - 4) = 0 \\ L_{1} = 5 x + 3 y - 2 = 0 \\ L_{2} = 3 x - y - 4 = 0 \\ S o l v i n g L_{1} a n d L_{2} \\ x y 1 \\ \begin{matrix} 3 & - 2 & 5 & 3 \\ - 1 & - 4 & 3 & - 1 \end{matrix} \\ \Rightarrow \frac{x}{- 12 - 2} = \frac{y}{- 6 + 20} = \frac{1}{- 5 - 9} \\ \Rightarrow \frac{x}{- 14} = \frac{y}{14} = \frac{1}{- 14} \\ A (x, y) = (1, - 1) \\ a n d x - y + 1 + λ_{2} (2 x - y - 2) = 0 \\ L_{3} = x - y + 1 = 0 \\ L_{4} = 2 x - y - 2 = 0 \\ s o l v i n g L_{3} a n d L_{4} \\ x y 1 \\ \begin{matrix} - 1 & 1 & 1 & - 1 \\ - 1 & - 2 & 2 & - 1 \end{matrix} \end{array}$

$\begin{array}{l} \Rightarrow \frac{x}{2 + 1} = \frac{y}{2 + 2} = \frac{1}{- 1 + 2} \\ \Rightarrow \frac{x}{3} = \frac{y}{4} = \frac{1}{1} \\ B (x, y) = (3, 4) \\ E q u a t i o n o f A B i s \\ y + 1 = \frac{4 + 1}{3 - 1} (x - 1) \\ \Rightarrow y + 1 = \frac{5}{2} (x - 1) \\ \Rightarrow 2 y + 2 = 5 x - 5 \\ \Rightarrow 5 x - 2 y - 7 = 0 \end{array}$