Question 15

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:

Solution

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.

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