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We have a right triangle, triangle ABC where the legs AB and AC have lengths 6 and 3sqrt3 respectively.

Medians AM and CN meet at point P.

What is the length of CP?

Let $\vec{A} = \binom{0}{0}$
Let $\vec{B} = \binom{6}{0}$
Let $\vec{C} = \binom{0}{3\sqrt{3}}$

$\mathbf{\vec{P} = \ ?}$

$\begin{array}{|rcll|} \hline \vec{P} &=& \frac13 ( \vec{A}+\vec{B}+\vec{C} ) \\ \vec{P} &=& \frac13 \left( \binom{0}{0}+\binom{6}{0}+\binom{0}{3\sqrt{3}} \right) \\ \vec{P} &=& \frac13 \cdot \binom{0+6+0}{0+0+3\sqrt{3} } \\ \vec{P} &=& \frac13 \cdot \binom{6}{3\sqrt{3} } \\ \vec{P} &=& \dbinom{2}{ \sqrt{3} } \\ \hline \end{array}$

CP = ?

$\begin{array}{|rcll|} \hline CP &=& |~\vec{C}-\vec{P}~| \\ CP &=& |~\binom{0}{3\sqrt{3}}-\binom{2}{ \sqrt{3} }~| \\ CP &=& |~\binom{0-2}{3\sqrt{3}-\sqrt{3} } ~| \\ CP &=& |~\binom{-2}{2\sqrt{3} } ~| \\ CP &=& \sqrt{(-2)^2+(2\sqrt{3})^2 } \\ CP &=& \sqrt{4+ 4\cdot3 } \\ CP &=& \sqrt{4+ 12 } \\ CP &=& \sqrt{16} \\ \mathbf{ CP } & \mathbf{=} & \mathbf{4} \\ \hline \end{array}$

the centroid always divide the median in 2:1 ratio 

so,cp:pm=2:1

cm^2=9+27

cm=6

cp=6(2/3)=4