In triangle ABC, the coordinates of A are (1, | vedclass

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(D) Let $G$ be the centroid of $	riangle ABC$. The centroid is the intersection of the medians. Given the equations of the medians $x+y=5$ and $x=4$.

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$\left(\frac{11}{2}, \frac{1}{2}\right)$

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(D) Let $G$ be the centroid of $	riangle ABC$. The centroid is the intersection of the medians. Given the equations of the medians $x+y=5$ and $x=4$. Substituting $x=4$ into $x+y=5$, we get $4+y=5$, so $y=1$. Thus, the centroid $G$ is $(4, 1)$.Let $A = (1, 2)$ and $D = (\alpha, \beta)$ be the midpoint of $BC$. The centroid $G$ divides the median $AD$ in the ratio $2:1$. Using the section formula, $G = \left(\frac{2\alpha + 1}{3}, \frac{2\beta + 2}{3}ight) = (4, 1)$.Equating the coordinates: $\frac{2\alpha + 1}{3} = 4 \implies 2\alpha + 1 = 12 \implies 2\alpha = 11 \implies \alpha = \frac{11}{2}$.$\frac{2\beta + 2}{3} = 1 \implies 2\beta + 2 = 3 \implies 2\beta = 1 \implies \beta = \frac{1}{2}$.Therefore, the midpoint $D$ of $BC$ is $\left(\frac{11}{2}, \frac{1}{2}ight)$.

In $\triangle ABC$, the coordinates of $A$ are $(1, 2)$ and the equations of the medians through $B$ and $C$ are $x+y=5$ and $x=4$ respectively. Then the midpoint of $BC$ is

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