If Dleft(frac{-1}{2}, frac{5}{2}right),E(7, 3 | vedclass

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(B) Let $A=(x_1, y_1)$,$B=(x_2, y_2)$,and $C=(x_3, y_3)$ be the vertices of $	riangle ABC$.Given that $D\left(-\frac{1}{2}, \frac{5}{2}ight)$,$E(7,

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$11$

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(B) Let $A=(x_1, y_1)$,$B=(x_2, y_2)$,and $C=(x_3, y_3)$ be the vertices of $	riangle ABC$.Given that $D\left(-\frac{1}{2}, \frac{5}{2}ight)$,$E(7, 3)$,and $F\left(\frac{7}{2}, \frac{7}{2}ight)$ are the midpoints of sides $BC$,$CA$,and $AB$ respectively.Using the midpoint formula,we have:For $BC$: $\frac{x_2+x_3}{2} = -\frac{1}{2} \Rightarrow x_2+x_3 = -1$ $(i)$ and $\frac{y_2+y_3}{2} = \frac{5}{2} \Rightarrow y_2+y_3 = 5$ $(ii)$.For $CA$: $\frac{x_3+x_1}{2} = 7 \Rightarrow x_3+x_1 = 14$ $(iii)$ and $\frac{y_3+y_1}{2} = 3 \Rightarrow y_3+y_1 = 6$ $(iv)$.For $AB$: $\frac{x_1+x_2}{2} = \frac{7}{2} \Rightarrow x_1+x_2 = 7$ $(v)$ and $\frac{y_1+y_2}{2} = \frac{7}{2} \Rightarrow y_1+y_2 = 7$ $(vi)$.Adding $(i), (iii), (v)$: $2(x_1+x_2+x_3) = 20 \Rightarrow x_1+x_2+x_3 = 10$ $(vii)$.Subtracting $(i), (iii), (v)$ from $(vii)$ gives $x_1=11, x_2=-4, x_3=3$.Adding $(ii), (iv), (vi)$: $2(y_1+y_2+y_3) = 18 \Rightarrow y_1+y_2+y_3 = 9$ $(viii)$.Subtracting $(ii), (iv), (vi)$ from $(viii)$ gives $y_1=4, y_2=3, y_3=2$.Vertices are $A(11, 4), B(-4, 3), C(3, 2)$.Area $= \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$$= \frac{1}{2} |11(3-2) + (-4)(2-4) + 3(4-3)| = \frac{1}{2} |11(1) + (-4)(-2) + 3(1)| = \frac{1}{2} |11+8+3| = \frac{22}{2} = 11$ sq units.

If $D\left(\frac{-1}{2}, \frac{5}{2}\right)$,$E(7, 3)$,and $F\left(\frac{7}{2}, \frac{7}{2}\right)$ are the midpoints of the sides of $\triangle ABC$,find the area of $\triangle ABC$.

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