How to Find the Medians of a Triangle Using Vector Calculation

You can do this in two ways: By using what you know about the median theorem, or by doing all the dirty work yourselves. Either way, you need to find the midpoints $K$, $L$ and $M$ and the vectors

They give you

$$\begin{array}{llll}\hfill \overrightarrow{OK}& =\overrightarrow{OA}+\frac{1}{2}\overrightarrow{AC}& \hfill & \\ \hfill & =\left(-2,-4\right)+\frac{1}{2}\left(3,10\right)& \hfill & \\ \hfill & =\left(-\frac{1}{2},1\right)& \hfill & \\ \hfill & \Downarrow & \hfill & \\ \hfill K& =\left(-\frac{1}{2},1\right),& \hfill & \\ \hfill \overrightarrow{OL}& =\overrightarrow{OB}+\frac{1}{2}\overrightarrow{BC}& \hfill & \\ \hfill & =\left(3,-5\right)+\frac{1}{2}\left(-2,11\right)& \hfill & \\ \hfill & =\left(2,\frac{1}{2}\right)& \hfill & \\ \hfill & \Downarrow & \hfill & \\ \hfill L& =\left(2,\frac{1}{2}\right),& \hfill & \\ \hfill \overrightarrow{OM}& =\overrightarrow{OA}+\frac{1}{2}\overrightarrow{AB}& \hfill & \\ \hfill & =\left(-2,-4\right)+\frac{1}{2}\left(5,-1\right)& \hfill & \\ \hfill & =\left(\frac{1}{2},\frac{-9}{2}\right)& \hfill & \\ \hfill & \Downarrow & \hfill & \\ \hfill M& =\left(\frac{1}{2},-\frac{9}{2}\right).& \hfill & \end{array}$$

This in turn gives you

Because the intersection $P$ has two coordinates, $x$ and $y$, you need two equations to find the set of coordinates. You can do this by creating two expressions, using two different sums of vectors that both become $P$. Then you can solve the system of equations you end up with to find the two coordinates of $P$. Solve the system of equations: That $t=s=\frac{2}{3}$ only confirms what the median theorem told us: The medians intersect $\frac{2}{3}$ of the way from each of the corners to the midpoints of their opposite sides. Finally, you can put the values you got for $t$ or $s$ back into one of the expressions you created for $P$ to find its coordinates:

$$\begin{array}{llll}\hfill P=\left(x,y\right)& =\left(1,6\right)-\frac{2}{3}\left(2,21\right)& \hfill & \\ \hfill & =\left(1,6\right)-\left(\frac{4}{3},14\right)& \hfill & \\ \hfill & =\left(\frac{-1}{3},-8\right).& \hfill & \end{array}$$