ABC is a triangle with A(3,2) B(2,-3) and C(-4,1).AD is the median in which G is the centroid such that AG:GD=2:1. Find the coordinates of G and D.
Given $A(3,2)$, $B(2,-3)$ and $C(-4,1)$.
1) Point $D$ is the midpoint of $BC$:
\[
D = \left(\frac{2+(-4)}{2}, \; \frac{-3+1}{2}\right) = (-1,-1).
\]
2) The centroid $G$ divides $AD$ in the ratio $AG:GD = 2:1$.
So,
\[
G = A + \frac{2}{3}(D-A).
\]
Substituting $A(3,2)$ and $D(-1,-1)$:
\[
G = (3,2) + \tfrac{2}{3}\big((-1,-1) - (3,2)\big)
= \left(\tfrac{1}{3}, \; 0\right).
\]
\[
\boxed{D(-1,-1), \quad G\left(\tfrac{1}{3},0\right)}
\]