Let the area of the triangle with vertices A( | vedclass

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(C) 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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$-64$

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(C) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 4$. Substituting the vertices $A(1, \alpha)$,$B(\alpha, 0)$,and $C(0, \alpha)$: $\frac{1}{2} |1(0 - \alpha) + \alpha(\alpha - \alpha) + 0(\alpha - 0)| = 4$ $\frac{1}{2} |-\alpha| = 4 \Rightarrow |\alpha| = 8 \Rightarrow \alpha = \pm 8$. If $\alpha = 8$,the points are $(8, -8)$,$(-8, 8)$,and $(64, \beta)$. Since these points are collinear,the slope between any two points must be equal. Slope of the line passing through $(8, -8)$ and $(-8, 8)$ is $m = \frac{8 - (-8)}{-8 - 8} = \frac{16}{-16} = -1$. The equation of the line is $y - 8 = -1(x - (-8)) \Rightarrow y - 8 = -x - 8 \Rightarrow y = -x$. For the point $(64, \beta)$ to lie on this line,$\beta = -64$. If $\alpha = -8$,the points are $(-8, 8)$,$(8, -8)$,and $(64, \beta)$,which results in the same line $y = -x$,so $\beta = -64$.

Let the area of the triangle with vertices $A(1, \alpha)$,$B(\alpha, 0)$,and $C(0, \alpha)$ be $4 \text{ sq. units}$. If the points $(\alpha, -\alpha)$,$(-\alpha, \alpha)$,and $(\alpha^2, \beta)$ are collinear,then $\beta$ is equal to:

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