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

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$0$

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(D) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Alternatively,using the determinant form: Area $= \frac{1}{2} \left| \begin{array}{ccc} a & b+c & 1 \ b & c+a & 1 \ c & a+b & 1 \end{array} ight|$ Applying the column operation $C_2 	o C_2 + C_1$: Area $= \frac{1}{2} \left| \begin{array}{ccc} a & a+b+c & 1 \ b & a+b+c & 1 \ c & a+b+c & 1 \end{array} ight|$ Taking $(a+b+c)$ common from the second column: Area $= \frac{a+b+c}{2} \left| \begin{array}{ccc} a & 1 & 1 \ b & 1 & 1 \ c & 1 & 1 \end{array} ight|$ Since two columns are identical,the value of the determinant is $0$. Therefore,the area of the triangle is $0$.

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

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