The area of the triangle formed by the points | vedclass

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(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_

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$0$

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(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Substituting the given points $(a, b + c)$,$(b, c + a)$,and $(c, a + b)$:Area $= \frac{1}{2} |a(c + a - a - b) + b(a + b - b - c) + c(b + c - c - a)|$$= \frac{1}{2} |a(c - b) + b(a - c) + c(b - a)|$$= \frac{1}{2} |ac - ab + ba - bc + cb - ca|$$= \frac{1}{2} |0| = 0$.Since the area is $0$,the points are collinear.

The area of the triangle formed by the points $(a, b + c)$,$(b, c + a)$,and $(c, a + b)$ is equal to:

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