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(B) The equation $y = |x|$ represents two rays: $y = x$ (for $x \ge 0$) and $y = -x$ (for $x < 0$).$1$. Intersection of $y = x$ and $x + 2y = 2$:Subst

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$\frac{4}{3} 	ext{ units}^2$

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(B) The equation $y = |x|$ represents two rays: $y = x$ (for $x \ge 0$) and $y = -x$ (for $x $1$. Intersection of $y = x$ and $x + 2y = 2$:Substituting $y = x$ into the line equation: $x + 2(x) = 2 \implies 3x = 2 \implies x = \frac{2}{3}$.Thus,$y = \frac{2}{3}$. The intersection point is $A(\frac{2}{3}, \frac{2}{3})$.$2$. Intersection of $y = -x$ and $x + 2y = 2$:Substituting $y = -x$ into the line equation: $x + 2(-x) = 2 \implies -x = 2 \implies x = -2$.Thus,$y = 2$. The intersection point is $B(-2, 2)$.$3$. The triangle is formed by vertices $O(0, 0)$,$A(\frac{2}{3}, \frac{2}{3})$,and $B(-2, 2)$.Using the area formula for a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ as $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$:Area $= \frac{1}{2} |0(\frac{2}{3} - 2) + \frac{2}{3}(2 - 0) + (-2)(0 - \frac{2}{3})|$Area $= \frac{1}{2} |0 + \frac{4}{3} + \frac{4}{3}| = \frac{1}{2} |\frac{8}{3}| = \frac{4}{3} 	ext{ units}^2$.

Find the area of the triangle formed by the two rays of the equation $y = |x|$ and the line $x + 2y = 2$.

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