If triangle HJK has a centroid P then we know that the side HP will be the double of PM so we can made an equation for HM
[tex]\begin{gathered} HM=HP+PM \\ HM=2PM+PM \\ HM=3PM \end{gathered}[/tex]
Now we can replace the value of HM and solve for PM so:
[tex]\begin{gathered} \frac{39}{3}=PM \\ 13=PM \end{gathered}[/tex]
