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(A) Let the vertex be $A = (1, 1)$. Let the midpoints of sides $AB$ and $AC$ be $M_1 = (-1, 2)$ and $M_2 = (3, 2)$ respectively.Since $M_1$ is the mid

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

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(A) Let the vertex be $A = (1, 1)$. Let the midpoints of sides $AB$ and $AC$ be $M_1 = (-1, 2)$ and $M_2 = (3, 2)$ respectively.Since $M_1$ is the midpoint of $AB$,we have $\frac{A+B}{2} = M_1 \implies B = 2M_1 - A = 2(-1, 2) - (1, 1) = (-2-1, 4-1) = (-3, 3)$.Since $M_2$ is the midpoint of $AC$,we have $\frac{A+C}{2} = M_2 \implies C = 2M_2 - A = 2(3, 2) - (1, 1) = (6-1, 4-1) = (5, 3)$.The centroid $G$ of the triangle with vertices $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$ is given by $\left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right)$.$G = \left( \frac{1 - 3 + 5}{3}, \frac{1 + 3 + 3}{3} \right) = \left( \frac{3}{3}, \frac{7}{3} \right) = \left( 1, \frac{7}{3} \right)$.

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

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