Let $A(4, 2)$,$B(6, 5)$,and $C(1, 4)$ be the vertices of $\Delta ABC$. Find the coordinates of the point $P$ on the median $AD$ such that $AP : PD = 2 : 1$.

(N/A) $1$. First,find the coordinates of the midpoint $D$ of the side $BC$. Since $D$ is the midpoint of $BC$ with $B(6, 5)$ and $C(1, 4)$,the coordinates of $D$ are $\left(\frac{6+1}{2}, \frac{5+4}{2}\right) = \left(\frac{7}{2}, \frac{9}{2}\right)$.
$2$. Now,we need to find the coordinates of point $P$ that divides the line segment $AD$ in the ratio $m:n = 2:1$,where $A(4, 2)$ and $D(\frac{7}{2}, \frac{9}{2})$.
$3$. Using the section formula,the coordinates of $P$ are given by $\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$.
$4$. Substituting the values: $P = \left(\frac{2 \times \frac{7}{2} + 1 \times 4}{2+1}, \frac{2 \times \frac{9}{2} + 1 \times 2}{2+1}\right) = \left(\frac{7+4}{3}, \frac{9+2}{3}\right) = \left(\frac{11}{3}, \frac{11}{3}\right)$.