Let A(1, 1),B(-4, 3),and C(-2, -5) be the ver | vedclass

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(C) The area of $\Delta ABC$ is given by $\Delta_{2} = \frac{1}{2} |1(3 - (-5)) + (-4)(-5 - 1) + (-2)(1 - 3)| = \frac{1}{2} |8 + 24 + 4| = 18$. Given

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$\frac{1}{2}$

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(C) The area of $\Delta ABC$ is given by $\Delta_{2} = \frac{1}{2} |1(3 - (-5)) + (-4)(-5 - 1) + (-2)(1 - 3)| = \frac{1}{2} |8 + 24 + 4| = 18$. Given $\frac{\Delta_{1}}{\Delta_{2}} = \frac{4}{7}$,we have $\Delta_{1} = \frac{4}{7} 	imes 18 = \frac{72}{7}$. Since $P$ lies on $BC$,the ratio of areas $\frac{	ext{Area}(\Delta APB)}{	ext{Area}(\Delta ABC)} = \frac{BP}{BC} = \frac{4}{7}$. Using the section formula,$P$ divides $BC$ in the ratio $4:3$. $P = \left( \frac{4(-2) + 3(-4)}{4+3}, \frac{4(-5) + 3(3)}{4+3} ight) = \left( \frac{-20}{7}, \frac{-11}{7} ight)$. The line $AP$ passes through $A(1, 1)$ and $P\left( \frac{-20}{7}, \frac{-11}{7} ight)$. The slope $m_{AP} = \frac{1 - (-11/7)}{1 - (-20/7)} = \frac{18/7}{27/7} = \frac{2}{3}$. The equation of $AP$ is $y - 1 = \frac{2}{3}(x - 1) \Rightarrow 2x - 3y + 1 = 0$. The line $AC$ passes through $A(1, 1)$ and $C(-2, -5)$. The slope $m_{AC} = \frac{1 - (-5)}{1 - (-2)} = \frac{6}{3} = 2$. The equation of $AC$ is $y - 1 = 2(x - 1) \Rightarrow 2x - y - 1 = 0$. The $x$-axis is $y = 0$. The intersection points are: $AP \cap x	ext{-axis}: 2x + 1 = 0 \Rightarrow x = -1/2$. Point $Q(-1/2, 0)$. $AC \cap x	ext{-axis}: 2x - 1 = 0 \Rightarrow x = 1/2$. Point $R(1/2, 0)$. $AP \cap AC$: Point $A(1, 1)$. The area of $\Delta AQR$ with vertices $A(1, 1)$,$Q(-1/2, 0)$,and $R(1/2, 0)$ is $\frac{1}{2} |1(0 - 0) + (-1/2)(0 - 1) + (1/2)(1 - 0)| = \frac{1}{2} |1/2 + 1/2| = \frac{1}{2}$.

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