ABC is a triangle. G is the centroid. D is the midpoint of BC. If A = (2, 3) and G = (7, 5), then the point D is

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(2)

$Here (x_{1}, y_{1}) = (2, 3), (p, q) = (7, 5)$

$Centroid, G (p, q) = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$

$\Rightarrow$ $\frac{x_{1} + x_{2} + x_{3}}{3} = 7, \frac{y_{1} + y_{2} + y_{3}}{3} = 5$

$\Rightarrow$ $2 + x_{2} + x_{3} = 21, 3 + y_{2} + y_{3} = 15$

$\Rightarrow$ $x_{2} + x_{3} = 19, y_{2} + y_{3} = 12$

$∴$ $D \equiv (\frac{19}{2}, \frac{12}{2}) = (\frac{19}{2}, 6)$

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