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The vertices of a $\triangle ABC$ are A$(-1, 3)$, B$(2, -3)$ and C$(4, 5)$. Find the coordinates of a point P on median AD such that AP: PD = $2:3$.
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Solution:
Coordinates of D are $(\frac{4+2}{2}, \frac{5-3}{2})$ i.e. $(3, 1)$ (1 Mark)
Coordinates of P are $(\frac{2\times3+3\times(-1)}{2+3}, \frac{2\times1 + 3\times3}{2+3})$ i.e. $(\frac{3}{5}, \frac{11}{5})$ (1 Mark)
Coordinates of D are $(\frac{4+2}{2}, \frac{5-3}{2})$ i.e. $(3, 1)$ (1 Mark)
Coordinates of P are $(\frac{2\times3+3\times(-1)}{2+3}, \frac{2\times1 + 3\times3}{2+3})$ i.e. $(\frac{3}{5}, \frac{11}{5})$ (1 Mark)