Find the area of the triangle whose vertices are $(a, b + c), (a, b - c)$ and $(- a, c)$ .

2 a c

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2 b c

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b (a + c)

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c (a - b)

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Solution

The correct option is B $2 a c$
Area of a triangle $(A) = ∣ ∣ ∣ \frac{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})}{2} ∣ ∣ ∣$
Hence, substituting the points $(x_{1}, y_{1}) = (a, b + c)$ , $(x_{2}, y_{2}) = (a, b - c)$ and $(x_{3}, y_{3}) = (- a, c)$
$A = ∣ ∣ ∣ \frac{a (b - c - c) + a (c - b - c) - a (b + c - b + c)}{2} ∣ ∣ ∣$
$= ∣ ∣ ∣ \frac{a (b - 2 c) + a (- b) - a (2 c)}{2} ∣ ∣ ∣$
$= ∣ ∣ ∣ \frac{a b - 2 a c - a b - 2 a c}{2} ∣ ∣ ∣$
$= ∣ ∣ ∣ \frac{- 4 a c}{2} ∣ ∣ ∣$
$= 2 a c$ square units

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