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

Answer 1

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

Given : $3x-2y=24\Rightarrow A\equiv (8,0)$ and $B\equiv (0,-12)$
$\Rightarrow$mid point of $AB\equiv (4,-6)$ and slope of $AB=\frac{3}{2}$
$\Rightarrow$slope of perndicular $=-\frac{2}{3}$
$\Rightarrow$equation of perpendicular bisector is
$(y+6)=\frac{-2}{3}(x-4)\Rightarrow 2x+3y+10=0$
and the line parallel to $x-$axis passing through $(0,-1)$ is $y=-1$ solving with $2x+3y+10=0$
$\Rightarrow C\equiv (\frac{-7}{2},-1)$