Question

Let the line $3x+2y=24$ meets $y-$ 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 $△ABC$ is (in sq. units)

Answer 1

Given line $3x+2y=24$
Now, the points
$A=(0,12),B=(8,0)$
Midpoint of line segment $AB$
$=(4,6)$
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
$C=(-\frac{13}{2},-1)$

Thus the area of the $△ABC$
$=\frac{1}{2}\left|\begin{array}{cccc}{x}_1& x_2& x_3& x_1\\ y_1& y_2& y_3& y_1\end{array}\right|=\frac{1}{2}\left|\begin{array}{cccc}0& 8& -\frac{13}{2}& 0\\ 12& 0& -1& 12\end{array}\right|=\frac{1}{2}|-96-8-78|=91\text{sq. units}$