Question

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

Answer 1

Given the line,

$3x+2y=24⟶\left(1\right)$

for A putting x=0 in $\left(1\right)$,

$2y=24$

$\Rightarrow y=12$

$\therefore$ coordinates of $A=(0,12)$

for B putting y=0 in $\left(1\right)$,

$3y=24$

$\Rightarrow x=8$

$\therefore$ coordinates of $B=(8,0)$

Midpoint of $AB=(\frac{8+0}{2},\frac{0+12}{2})=(4,6)$

Now, equation of line perpendicular to line $\left(1\right)$,

$2x-3y=\lambda$

It will pass through $(4,6)$

so, $2\left(4\right)-3\left(6\right)=\lambda$

$\Rightarrow \lambda =8-18$

$\Rightarrow \lambda =-10$

Equation of the line parallel to X-axis is,

$y=constant$

it will pass through $(0,-1)$

$\Rightarrow -1=constant$

$\Rightarrow y=-1$

To get coordinates of C,putting $y=-1$ in

$2x-3y=-10$

$\Rightarrow 2x+3=-10$

$\Rightarrow x=-\frac{13}{2}$

Hence, we have $C(-\frac{13}{2},-1),A(0,12)andB(8,0)$

$\therefore$ Area of $\triangle ABC

$=|-87-4|$

$=|-91|$

$=91squareunits$