If left( frac{3}{2}, 0 right),left( frac{3}{2 | vedclass

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(A) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.The midpoints of the sides are given as $M_1 = \left( \frac{3}{2

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$\left( \frac{2}{3}, 4 \right)$

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(A) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.The midpoints of the sides are given as $M_1 = \left( \frac{3}{2}, 0 \right)$,$M_2 = \left( \frac{3}{2}, 6 \right)$,and $M_3 = (-1, 6)$.The centroid of a triangle formed by the midpoints of the sides of a triangle is the same as the centroid of the original triangle.Let the centroid be $G(x, y)$.The centroid $G$ of the triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is $\left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right)$.The centroid of the triangle formed by the midpoints $M_1(x'_1, y'_1)$,$M_2(x'_2, y'_2)$,and $M_3(x'_3, y'_3)$ is $\left( \frac{x'_1+x'_2+x'_3}{3}, \frac{y'_1+y'_2+y'_3}{3} \right)$.Substituting the given midpoint values:$x = \frac{\frac{3}{2} + \frac{3}{2} - 1}{3} = \frac{3 - 1}{3} = \frac{2}{3}$$y = \frac{0 + 6 + 6}{3} = \frac{12}{3} = 4$Thus,the centroid is $\left( \frac{2}{3}, 4 \right)$.

If $\left( \frac{3}{2}, 0 \right)$,$\left( \frac{3}{2}, 6 \right)$,and $(-1, 6)$ are the midpoints of the sides of a triangle,find the centroid of the triangle.

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