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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(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 $= \frac{1}{2} |a(c + a - (a + b)) + b(a + b - (b + c)) + c(b + c - (c + a))|$Area $= \frac{1}{2} |a(c - b) + b(a - c) + c(b - a)|$Area $= \frac{1}{2} |ac - ab + ba - bc + cb - ca|$Area $= \frac{1}{2} |0| = 0$.Alternatively,using the determinant method:Area $= \frac{1}{2} \left| \begin{matrix} a & b+c & 1 \ b & c+a & 1 \ c & a+b & 1 \end{matrix} ight|$Applying $C_2 	o C_1 + C_2$:Area $= \frac{1}{2} \left| \begin{matrix} a & a+b+c & 1 \ b & a+b+c & 1 \ c & a+b+c & 1 \end{matrix} ight| = \frac{a+b+c}{2} \left| \begin{matrix} a & 1 & 1 \ b & 1 & 1 \ c & 1 & 1 \end{matrix} ight| = 0$ (since two columns are identical).

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

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