The line $2 x + 3 y + 12 = 0$ cuts the axes at A and B. The equation to the perpendicular bisector of $\bar{A B}$ is

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$2 x + 3 y + 2 = 0$

$3 x - 2 y + 5 = 0$

$3 x + 2 y + 5 = 0$

$2 x - 3 y + 3 = 0$

answer is A.

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Detailed Solution

Line $2 x + 3 y + 12 = 0$

Let $A (- 6, 0), B (0, - 4)$

Midpoint of $\bar{A B}$ is $P = (- 3, - 2)$

The equation of the line perpendicular to $2 x + 3 y + 12 = 0$

And through $(- 3, - 2)$ is

$\begin{array}{l} \Rightarrow 3 (x + 3) - 2 (y + 2) = 0 \\ \Rightarrow 3 x + 9 - 2 y - 4 = 0 \\ \Rightarrow 3 x - 2 y + 5 = 0 \end{array}$

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