Question

If the line $2x+3y+12=0$ cuts the axes at $A$ and $B$, then the equation of the perpendicular bisector of $AB$ is

• A. $3x-2y+5=0$
• B. $3x-2y+7=0$
• C. $3x-2y+9=0$
• D. $3x-2y+8=0$

Answer 1

The correct option is A $3x-2y+5=0$

$2x+3y+12=0$
At $x-$axis, $y=0$,

So, $2x+12=0$
$\Rightarrow x=-6,A=(-6,0)$
And at $y-$axis, $x=0,$

So, $3y+12=0$
$\Rightarrow y=-4,B(0,-4)$
So, mid point of $AB=(\frac{-6}{2},\frac{-4}{2})\equiv (-3,-2)$

Slope of $AB=-\frac{4}{6}\equiv -\frac{2}{3}$

$\therefore$ Equation of the perpendicular bisector of $AB$ is
$y+2=\frac{3}{2}(x+3)$
$\Rightarrow 2y+4=3x+9$
$\Rightarrow 2y-3x-5=0$

Hence, option A is the correct answer.