Question

Let the line $3x+2y=24$ meet $y-$axis at $A$ and $x-$axis at $B.$ If the perpendicular bisector of $AB$ meets the line $y=-1$ at $C,$ then the area of the $△ABC$ (in sq. units) is

Answer 1

Given line $3x+2y=24$
So, $A\equiv (0,12),B\equiv (8,0)$
Mid-point of line segment $AB$ is $(4,6)$
The equation of the perpendicular bisector of $AB$ is
$y-6=-\frac{8-0}{0-12}(x-4)\Rightarrow 3y-18=2x-8\Rightarrow 2x-3y+10=0$
This line meets the $y=-1$ at $C\equiv (-\frac{13}{2},-1)$

Thus, the area of the $△ABC$ is
$\frac{1}{2}\left|\begin{array}{cccc}{x}_1& x_2& x_3& x_1\\ y_1& y_2& y_3& y_1\end{array}\right|$

$=\frac{1}{2}\left|\begin{array}{cccc}0& 8& -\frac{13}{2}& 0\\ 12& 0& -1& 12\end{array}\right|=\frac{1}{2}|-96-8-78|=91$ sq. units