Let A(a, 0),B(b, 2b+1),and C(0, b),where b ne | 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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$\frac{-2b^2}{b+1}$

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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 $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 1$. Substituting the given points $A(a, 0)$,$B(b, 2b+1)$,and $C(0, b)$: $\frac{1}{2} |a(2b+1 - b) + b(b - 0) + 0(0 - (2b+1))| = 1$ $\frac{1}{2} |a(b+1) + b^2| = 1$ $|a(b+1) + b^2| = 2$ This implies $a(b+1) + b^2 = 2$ or $a(b+1) + b^2 = -2$. Case $1$: $a(b+1) = 2 - b^2 \Rightarrow a_1 = \frac{2 - b^2}{b+1}$. Case $2$: $a(b+1) = -2 - b^2 \Rightarrow a_2 = \frac{-2 - b^2}{b+1}$. The sum of all possible values of $a$ is $a_1 + a_2 = \frac{2 - b^2 - 2 - b^2}{b+1} = \frac{-2b^2}{b+1}$.

Let $A(a, 0)$,$B(b, 2b+1)$,and $C(0, b)$,where $b \neq 0$ and $|b| \neq 1$,be points such that the area of triangle $ABC$ is $1 \, \text{sq. unit}$. Then,the sum of all possible values of $a$ is:

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