Straight lines

Intersection with the $y-axis$ (x=0).

$2y=24$

$y=12$

$A(0,12)$

Intersection with the $x-axis$ (y=0).

$3x=24$

$x=8$

$B(8,0)$

Coordinates of the perpendicular bisector of line $AB$ :

$x=\dfrac{0+8}{2}=4$

$y=\dfrac{0+12}{2}=6$

$D(4,6)$

Slope of line $AB$ :

$m=\dfrac{12-0}{0-8}=-\dfrac{12}{8}=-\dfrac{3}{2}$

The slope of the perpendicular line is the negative reciprocal of the slope of $AB$ , so the slope is $\dfrac{2}{3}$ .

The equation of the perpendicular bisector is $y=\dfrac{2}{3}x+\dfrac{10}{3}$ .

The intersection of $y=\dfrac{2}{3}x+\dfrac{10}{3}$ and $y=-1$ is

$-1=\dfrac{2}{3}x+\dfrac{10}{3}$

$x=-6.5$

$C(-6.5,-1)$

The area of $\triangle ABC$ is:

$A=\dfrac{1}{2} \begin{vmatrix} 0 & -6.5 & 8 & 0\\ 12 & -1 & 0 & 12 \end{vmatrix} =\dfrac{1}{2}[0+0+96-(-78-8+0)]=\dfrac{1}{2}(182)=\boxed{91}$

Plugging in x=0 in the equation we get y=12. Therefore A = (0,12). Similarly, plugging in y=0, we get x = 8. B = (8,0)

The slope of AB = -3/2. Therefore the slope of perpendicular bisector = 2/3

The perpendicular bisector passes through the midpoint of AB , which is (4,6)

Therefore, the equation of perpendicular bisector 0is :

y-6=(2/3) *(x-4)

3y-18 = 2x-8

3y-2x=10

To find coordinates of Point C, we substitute Y=-1

-3-2x=10

-2x=13

X=-6.5

C=(-6.5,-1)

Using formula for Euclidean Distance, we get

AC^2 = 6.5^2 + 13^2

= 211.25

BC^2 = 14.5^2+1

= 211.25

AB^2=64+144

= 208

We then use Heron's formula for area of a triangle knowing the length of the sides and obtain the area as sqrt(8281) = 91