Question 63
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 = $$\ \dfrac{\ AB\ \times\ CD}{2}$$ = $$\ \dfrac{6c\ \times\ 3a\ }{2}=9ac$$
Option (D) is correct.
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