Let A(-3, 2) and B(-2, 1) be the vertices of | vedclass

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(B) Let the vertex $C$ be $(x_1, y_1)$.The centroid $G$ of triangle $ABC$ with vertices $A(-3, 2)$,$B(-2, 1)$,and $C(x_1, y_1)$ is given by:$G = \left

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$3x + 4y + 3 = 0$

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(B) Let the vertex $C$ be $(x_1, y_1)$.The centroid $G$ of triangle $ABC$ with vertices $A(-3, 2)$,$B(-2, 1)$,and $C(x_1, y_1)$ is given by:$G = \left( \frac{-3 - 2 + x_1}{3}, \frac{2 + 1 + y_1}{3} \right) = \left( \frac{x_1 - 5}{3}, \frac{y_1 + 3}{3} \right)$Since the centroid lies on the line $3x + 4y + 2 = 0$,we substitute the coordinates of $G$ into the equation:$3 \left( \frac{x_1 - 5}{3} \right) + 4 \left( \frac{y_1 + 3}{3} \right) + 2 = 0$Multiply the entire equation by $3$ to clear the denominators:$3(x_1 - 5) + 4(y_1 + 3) + 6 = 0$$3x_1 - 15 + 4y_1 + 12 + 6 = 0$$3x_1 + 4y_1 + 3 = 0$Thus,the vertex $C(x_1, y_1)$ lies on the line $3x + 4y + 3 = 0$.

Let $A(-3, 2)$ and $B(-2, 1)$ be the vertices of a triangle $ABC$. If the centroid of this triangle lies on the line $3x + 4y + 2 = 0$,then the vertex $C$ lies on the line

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