The combined equation of the straight lines passing through the point (4 , 3) and each line making intercepts on the coordinate axes whose sum is -1 is

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$(3 x - 2 y + 6) (x - 2 y + 2) = 0$

$(3 x - 2 y + 6) (x - 2 y - 2) = 0$

$(3 x - 2 y - 6) (x - 2 y + 2) = 0$

$(3 x - 2 y - 6) (x - 2 y - 2) = 0$

answer is A.

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Detailed Solution

$\begin{array}{l} L e t E q u a t i o n o f l i n e \frac{x}{a} + \frac{y}{b} = 1 \\ I f p a s s e s t h r o u g h (4, 3) \\ \Rightarrow \frac{4}{a} + \frac{3}{b} = 1 \\ a n d g i v e n a + b = - 1 \Rightarrow b = - 1 - a \\ \Rightarrow \frac{4}{a} + \frac{3}{- 1 - a} = 1 \\ \Rightarrow \frac{4 (1 + a) - 3 a}{a (1 + a)} = 1 \\ \Rightarrow 4 + 4 a - 3 a = a + a^{2} \\ \Rightarrow a^{2} = 4 \Rightarrow a = \pm 2 \end{array}$

$\begin{array}{l} I f a = 2 t h e n b = - 1 - 2 = - 3 \\ E q u a t i o n i s \frac{x}{2} - \frac{y}{3} = 1 \\ \Rightarrow 3 x - 2 y - 6 = 0 \\ I f a = - 2 t h e n b = - 1 + 2 \\ E q u a t i o n i s \frac{x}{- 2} + \frac{y}{1} = 1 \\ \Rightarrow x - 2 y + 2 = 0 \\ C o m b a i n e d e q u a t i o n i s (3 x - 2 y - 6) (x - 2 y + 2) = 0 \end{array}$