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If $4 a^{2} + 9 b^{2} - c^{2} + 12 a b = 0$ , then the lines $a x + b y + c = 0$ pass through the fixed point

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$(1, - 2) (- 1, 2)$

$(1, 2) (- 1, - 2)$

$(2, 3) (- 2, - 3)$

$(2, - 3) (- 2, 3)$

answer is B.

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Detailed Solution

$\begin{array}{l} 4 a^{2} + 9 b^{2} - c^{2} + 12 a b = 0 \\ \Rightarrow {(2 a + 3 b)}^{2} - c^{2} = 0 \\ \Rightarrow 2 a + 3 b = \pm c \\ \Rightarrow a (2) + b (3) + c = 0, a (- 2) + b (- 3) + c = 0 \end{array}$

The point $(2, 3) (- 2, - 3)$ lies on $a x + b y + c = 0$ .

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