ABC is a triangle and the coordinates of A, B and C are (a, b-2c), (a, b+4c) and (-2a,3c) respectively where a, b and c are positive numbers.
The area of the triangle ABC is:
The length of AB = $$\left(b+4c\right)-\left(b-2c\right)=6c$$ (X-coordinates of A&B are same).
The altitude of triangle ABC, CD = a-(-2a)=3a.
Area of triangle ABC = $$\ \frac{\ AB\ \times\ CD}{2}$$ = $$\ \frac{6c\ \times\ 3a\ }{2}=9ac$$
Option (D) is correct.