Find the area of $△ A B C$ in which A =(2, -1), B=(3, 4) and C=(-3, -2)

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Solution

The correct option is D 12
Area of a triangle with vertices $(x_{1}, y_{1})$ ; $(x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is $∣ ∣ \frac{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})}{2} ∣ ∣$
Hence, substituting the points $(x_{1}, y_{1}) = (2, - 1)$ ; $(x_{2}, y_{2}) = (3, 4)$ and $(x_{3}, y_{3}) = (- 3, - 2)$ in the area formula, we get
Area of triangle ABC $= ∣ ∣ \frac{(2) (4 + 2) + (3) (- 2 + 1) - 3 (- 1 - 4)}{2} ∣ ∣ = ∣ ∣ \frac{12 - 3 + 15}{2} ∣ ∣ = 12 s q u n i t s$

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