The points $A(x_{1}, y_{1})$,$B(x_{2}, y_{2})$ and $C(x_{3}, y_{3})$ are the vertices of $\triangle ABC$.
$(i)$ The median from $A$ meets $BC$ at $D$. Find the coordinates of the point $D$.
$(ii)$ Find the coordinates of the point $P$ on $AD$ such that $AP : PD = 2 : 1$.
$(iii)$ Find the coordinates of points $Q$ and $R$ on medians $BE$ and $CF$,respectively,such that $BQ : QE = 2 : 1$ and $CR : RF = 2 : 1$.
$(iv)$ What are the coordinates of the centroid of the triangle $ABC$?

(N/A) Given that the points $A(x_{1}, y_{1})$,$B(x_{2}, y_{2})$ and $C(x_{3}, y_{3})$ are the vertices of $\triangle ABC$.
$(i)$ We know that the median bisects the line segment into two equal parts,i.e.,$D$ is the mid-point of $BC$.
$\therefore$ Coordinate of mid-point of $BC = \left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)$.
$\Rightarrow D = \left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)$.
$(ii)$ Let the coordinates of point $P$ be $(x, y)$.
Given that the point $P(x, y)$ divides the line segment joining $A(x_{1}, y_{1})$ and $D\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)$ in the ratio $2 : 1$.
Using the internal section formula $\left(\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\right)$:
$P = \left[\frac{2\left(\frac{x_{2}+x_{3}}{2}\right) + 1(x_{1})}{2+1}, \frac{2\left(\frac{y_{2}+y_{3}}{2}\right) + 1(y_{1})}{2+1}\right] = \left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$.
$(iii)$ Similarly,for median $BE$,$E$ is the mid-point of $AC$,so $E = \left(\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}\right)$.
Point $Q$ divides $BE$ in ratio $2 : 1$. Using the section formula:
$Q = \left[\frac{2\left(\frac{x_{1}+x_{3}}{2}\right) + 1(x_{2})}{2+1}, \frac{2\left(\frac{y_{1}+y_{3}}{2}\right) + 1(y_{2})}{2+1}\right] = \left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$.
For median $CF$,$F$ is the mid-point of $AB$,so $F = \left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$.
Point $R$ divides $CF$ in ratio $2 : 1$. Using the section formula:
$R = \left[\frac{2\left(\frac{x_{1}+x_{2}}{2}\right) + 1(x_{3})}{2+1}, \frac{2\left(\frac{y_{1}+y_{2}}{2}\right) + 1(y_{3})}{2+1}\right] = \left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$.
$(iv)$ The centroid of a triangle is the point where all medians intersect. Since $P, Q,$ and $R$ are all the same point,the coordinates of the centroid are $\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$.