Question

The line $3x-2y=24$ meets $x-$axis at $A$ and $y$ axis at $B.$ The perpendicular bisector of $AB$ meets the line through $(0,-1)$ and parallel to $x-$axis at $C.$ Then $C$ is

• A. $\left(\frac{-7}{2},-1\right)$
• B. $\left(\frac{-15}{2},-1\right)$
• C. $\left(\frac{-11}{2},-1\right)$
• D. $\left(\frac{-13}{2},-1\right)$

Answer 1

The correct option is A $\left(\frac{-7}{2},-1\right)$

Slope of the line joining of $AB$ is $\frac{3}{2}$ and it's mid-point is $M(4,-6)$.
The perpendicular bisector of this line will have the slope $-\frac{2}{3}$ and pass through $M(4,-6)$................(using m1.m2= -1 product of slopes of two perpendicular lines)
$\therefore$ The equation of the perpendicular bisector of $AB$ is $2x+3y+10=0\to \left(1\right)$.
The equation of the other line given in the problem is $y=-1\to \left(2\right)$.
Lines $\left(1\right)$ and $\left(2\right)$ intersect at $(-\frac{7}{2},-1)$.