Given line 3x+2y=24
So, A≡(0,12),B≡(8,0)
Mid-point of line segment AB is (4,6)
The equation of the perpendicular bisector of AB is
y−6=−8−00−12(x−4)⇒3y−18=2x−8⇒2x−3y+10=0
This line meets the y=−1 at C≡(−132,−1)
Thus, the area of the △ABC is
12∣∣∣x1x2x3x1y1y2y3y1∣∣∣
=12∣∣
∣∣08−1320120−112∣∣
∣∣=12|−96−8−78|=91 sq. units