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(C) Let the vertices be $A(-1, 6)$,$B(-3, -9)$,and $C(5, -8)$.The median through $C$ passes through $C(5, -8)$ and the midpoint $M$ of $AB$.The midpoi

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$13x + 14y + 47 = 0$

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(C) Let the vertices be $A(-1, 6)$,$B(-3, -9)$,and $C(5, -8)$.The median through $C$ passes through $C(5, -8)$ and the midpoint $M$ of $AB$.The midpoint $M$ of $AB$ is given by $M = \left( \frac{-1 - 3}{2}, \frac{6 - 9}{2} \right) = \left( -2, -\frac{3}{2} \right)$.The equation of the line passing through $C(5, -8)$ and $M(-2, -\frac{3}{2})$ is given by the two-point form: $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.Substituting the values: $y - (-8) = \frac{-\frac{3}{2} - (-8)}{-2 - 5}(x - 5)$.$y + 8 = \frac{-\frac{3}{2} + 8}{-7}(x - 5) \Rightarrow y + 8 = \frac{\frac{13}{2}}{-7}(x - 5)$.$y + 8 = -\frac{13}{14}(x - 5) \Rightarrow 14(y + 8) = -13(x - 5)$.$14y + 112 = -13x + 65 \Rightarrow 13x + 14y + 47 = 0$.

If the coordinates of the vertices of the triangle $ABC$ are $A(-1, 6)$,$B(-3, -9)$,and $C(5, -8)$,then the equation of the median through $C$ is

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