If one vertex of a triangle is (1, 1) and the | vedclass

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(C) Let the vertices of the triangle be $A(1, 1)$,$B(x_1, y_1)$,and $C(x_2, y_2)$.Given that the midpoints of sides $AB$ and $AC$ are $M_1(-1, 2)$ and

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$\left(1, \frac{7}{3}\right)$

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(C) Let the vertices of the triangle be $A(1, 1)$,$B(x_1, y_1)$,and $C(x_2, y_2)$.Given that the midpoints of sides $AB$ and $AC$ are $M_1(-1, 2)$ and $M_2(3, 2)$ respectively.The midpoint formula is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.For side $AB$: $\frac{1+x_1}{2} = -1 \implies 1+x_1 = -2 \implies x_1 = -3$ and $\frac{1+y_1}{2} = 2 \implies 1+y_1 = 4 \implies y_1 = 3$. So,$B = (-3, 3)$.For side $AC$: $\frac{1+x_2}{2} = 3 \implies 1+x_2 = 6 \implies x_2 = 5$ and $\frac{1+y_2}{2} = 2 \implies 1+y_2 = 4 \implies y_2 = 3$. So,$C = (5, 3)$.The centroid $G$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.$G = \left(\frac{1 + (-3) + 5}{3}, \frac{1 + 3 + 3}{3}\right) = \left(\frac{3}{3}, \frac{7}{3}\right) = \left(1, \frac{7}{3}\right)$.

If one vertex of a triangle is $(1, 1)$ and the midpoints of the sides through this vertex are $(-1, 2)$ and $(3, 2)$,then find the centroid of the triangle.

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