Let $A(4, 2)$,$B(6, 5)$,and $C(1, 4)$ be the vertices of $\Delta ABC$. What do you observe about the point $P$ which divides the median $AD$ in the ratio $2:1$?

(N/A) The coordinates of the vertices are $A(4, 2)$,$B(6, 5)$,and $C(1, 4)$.
First,find the midpoint $D$ of side $BC$ using the midpoint formula: $D = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) = (\frac{6+1}{2}, \frac{5+4}{2}) = (3.5, 4.5)$.
Now,point $P$ divides the median $AD$ in the ratio $2:1$. Using the section formula $P = (\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n})$:
$P = (\frac{2(3.5) + 1(4)}{2+1}, \frac{2(4.5) + 1(2)}{2+1}) = (\frac{7+4}{3}, \frac{9+2}{3}) = (\frac{11}{3}, \frac{11}{3})$.
Next,calculate the centroid $G$ of $\Delta ABC$ using the formula $G = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$:
$G = (\frac{4+6+1}{3}, \frac{2+5+4}{3}) = (\frac{11}{3}, \frac{11}{3})$.
Observation: The point $P$ is the same as the centroid $G$ of the triangle.