A triangle has a vertex at (1, 2) and the mid | vedclass

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(C) Let the vertices of the triangle be $A(1, 2)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.Given that the midpoints of sides $AB$ and $AC$ are $E(-1, 1)$ and $

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$\left( \frac{1}{3}, 2 \right)$

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(C) Let the vertices of the triangle be $A(1, 2)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.Given that the midpoints of sides $AB$ and $AC$ are $E(-1, 1)$ and $F(2, 3)$ respectively.Using the midpoint formula for $AB$:$\frac{x_2 + 1}{2} = -1 \implies x_2 + 1 = -2 \implies x_2 = -3$$\frac{y_2 + 2}{2} = 1 \implies y_2 + 2 = 2 \implies y_2 = 0$So,$B = (-3, 0)$.Using the midpoint formula for $AC$:$\frac{x_3 + 1}{2} = 2 \implies x_3 + 1 = 4 \implies x_3 = 3$$\frac{y_3 + 2}{2} = 3 \implies y_3 + 2 = 6 \implies y_3 = 4$So,$C = (3, 4)$.The centroid $G$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$.$G = \left( \frac{1 - 3 + 3}{3}, \frac{2 + 0 + 4}{3} \right) = \left( \frac{1}{3}, \frac{6}{3} \right) = \left( \frac{1}{3}, 2 \right)$.

$A$ triangle has a vertex at $(1, 2)$ and the midpoints of the two sides through it are $(-1, 1)$ and $(2, 3)$. Then the centroid of this triangle is

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