Let $A(4, 2)$,$B(6, 5)$,and $C(1, 4)$ be the vertices of $\Delta ABC$. Find the coordinates of points $Q$ and $R$ on medians $BE$ and $CF$ respectively such that $BQ : QE = 2 : 1$ and $CR : RF = 2 : 1$.

(N/A) The median $BE$ of the triangle divides the side $AC$ into two equal parts. Therefore,$E$ is the midpoint of side $AC$.
Coordinates of $E = \left(\frac{4+1}{2}, \frac{2+4}{2}\right) = \left(\frac{5}{2}, 3\right)$.
Point $Q$ divides the side $BE$ in a ratio $2 : 1$. Using the section formula:
Coordinates of $Q = \left(\frac{2 \times \frac{5}{2} + 1 \times 6}{2+1}, \frac{2 \times 3 + 1 \times 5}{2+1}\right) = \left(\frac{5+6}{3}, \frac{6+5}{3}\right) = \left(\frac{11}{3}, \frac{11}{3}\right)$.
The median $CF$ of the triangle divides the side $AB$ into two equal parts. Therefore,$F$ is the midpoint of side $AB$.
Coordinates of $F = \left(\frac{4+6}{2}, \frac{2+5}{2}\right) = \left(5, \frac{7}{2}\right)$.
Point $R$ divides the side $CF$ in a ratio $2 : 1$. Using the section formula:
Coordinates of $R = \left(\frac{2 \times 5 + 1 \times 1}{2+1}, \frac{2 \times \frac{7}{2} + 1 \times 4}{2+1}\right) = \left(\frac{10+1}{3}, \frac{7+4}{3}\right) = \left(\frac{11}{3}, \frac{11}{3}\right)$.