The line 3x + 2y = 24 meets the y-axis at A a | vedclass

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(B) The coordinates of $A$ are found by setting $x=0$ in $3x + 2y = 24$,giving $A(0, 12)$.Setting $y=0$ gives $B(8, 0)$.The midpoint of $AB$ is $M\lef

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$91$

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(B) The coordinates of $A$ are found by setting $x=0$ in $3x + 2y = 24$,giving $A(0, 12)$.Setting $y=0$ gives $B(8, 0)$.The midpoint of $AB$ is $M\left(\frac{0+8}{2}, \frac{12+0}{2}ight) = (4, 6)$.The slope of $AB$ is $m = \frac{0-12}{8-0} = -\frac{3}{2}$.The slope of the perpendicular bisector is $m' = \frac{2}{3}$.The equation of the perpendicular bisector is $y - 6 = \frac{2}{3}(x - 4)$,which simplifies to $2x - 3y + 10 = 0$.The line through $(0, -1)$ parallel to the $x$-axis is $y = -1$.Substituting $y = -1$ into the bisector equation: $2x - 3(-1) + 10 = 0 \implies 2x + 13 = 0 \implies x = -\frac{13}{2}$.Thus,$C$ is $\left(-\frac{13}{2}, -1ight)$.The area of $	riangle ABC$ with vertices $(0, 12), (8, 0), \left(-\frac{13}{2}, -1ight)$ is:$	ext{Area} = \frac{1}{2} |0(0 - (-1)) + 8(-1 - 12) + (-\frac{13}{2})(12 - 0)|$$= \frac{1}{2} |0 + 8(-13) - \frac{13}{2}(12)|$$= \frac{1}{2} |-104 - 78| = \frac{1}{2} |-182| = 91 \, sq. \, units$.

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

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