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The line $3x + 2y = 24$ meets $y$-axis at $A$ and $x$-axis at $B$. The perpendicular bisector of $AB$ meets the line through $(0, - 1)$ parallel to $x$-axis at $C$. The area of the triangle $ABC$ is ............... $\mathrm{sq. \, units}$
$182$
$91$
$48$
None of these
Solution
(b) The coordinates of $A$ and $B$ are $(0,\,12)$and $(8,0)$ respectively. The equation of the perpendicular bisector of $AB$ is $y – 6 = \frac{2}{3}(x – 4)$ or $2x – 3y + 10 = 0$ …..$(i)$
Equation of a line passing through $(0, -1)$ and parallel to $x$-axis is $y = – 1$. This meets $(i)$ at $C$, Therefore the coordinates of $C$ are $\left( { – \frac{{13}}{2}, – 1} \right)$.
Hence the area of the triangle $ABC$ is
$\Delta = \frac{1}{2}\left| {\begin{array}{*{20}{c}}0&{12}&1\\8&0&1\\{ – \frac{{13}}{2}}&{ – 1}&1\end{array}\,} \right| = 91$ sq. units.
