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The vertices of triangle XYZ are $X(-2,6)$, $Y(4,10)$, $Z(14,6)$. Find the coordinates of the centroid.

The centroid is the point of concurrency of all of the medians. The answer key says that the answer is $(\frac{16}{3},\frac{22}{3})$. This certainly appears to be the answer when you plot the medians on graph paper, but I don't see how you can show that this is definitively the answer.

My attempt was to take one of the medians and use the Centroid Theorem. So, I use $(4,10)$ and $(6,6)$ and find that the distance between these points is $\sqrt{20}$ or $2\sqrt{5}$. So, $\frac{2}{3}$ of the distance from the vertex $(4,10)$ to the midpoint $(6,6)$ is $\frac{4}{3}\sqrt{5}$. My question is, how can I be sure that point is $(\frac{16}{3},\frac{22}{3})$? It certainly appears that the medians intersect at that point, but I think the answer key may be making assumptions. What am I missing?

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    Hint: the line joining $A$ and the midpoint of the $BC$ segment, i.e. $\frac{B+C}{2}$, goes through $\frac{A+B+C}{3} = \frac{2}{3}\cdot\frac{B+C}{2}+\frac{1}{3}\cdot A$. The same holds for the other medians, hence $G=\frac{A+B+C}{3}$ is the centroid of $ABC$. – 2017-01-20
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    Answer to your first sentence: you do not need to compute medians. The coordinates are (mean of $x$ coord. of the vertices, mean of $y$ coord. of the vertices) – 2017-01-20
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    Ah, @JeanMarie this would have been helpful if it were in the book. It does make sense that the mean of the coordinates would be the centroid. However, I'm certain that we were supposed to find the centroid using the fact that the centroid is located 2/3 the distance from vertex to midpoint. Do you see a way to show it using this theorem? – 2017-01-20
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    No better way than as indicated by @Jack d'Aurizio – 2017-01-20
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    You could write the equation of the lines on which the median lies and test for the centroid being on those lines. – 2017-01-21

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