Calculate coordinates of a centroid of $ABC$ We have a triangle $ABC$, where on the cartesian coordinate system: 
$A$ lies on $[-3, -2]$,
$B$ lies on $[1, 1]$,
$C$ lies on $[0, -6]$.
How do we calculate coordinates for the centroid of this triangle?
 A: It's quite easy to find a formula online. What you might not find that easily is the proof of this formula. So far I only know one with analytic geometry and a simpler one with vectors, so I'll present this one.

Let $\triangle ABC$ be a triangle with centroid $S$. Denote by $D, E, F$ the midpoints of $BC, AC, AB$ respectively. Let furthermore, $O(0, 0)$ be the origin of the coordinate system.

I will only use an additional and well-known fact which I prove here: $$\color{blue}{AS=2\cdot SD}$$

Vectors
Observe now that
\begin{align*}
\vec{OS}&=\vec{OA}+\vec{AS}\\
&=\vec{OA}+\color{blue}{\frac23\vec{AD}}\\
&=\vec{OA}+\frac{2}{3}\cdot \big(\vec{AC}+\frac12\vec{CB}\big)\\
&=\vec{OA}+\frac23\cdot \bigg(\vec{OC}-\vec{OA}\;+\;\frac12\big(\vec{OB}-\vec{OC}\big)\bigg)\\
&=\vec{OA}+\frac23\cdot \bigg(\frac 12\vec{OC}+\frac12\vec{OB}-\vec{OA}\bigg)\\
&=\frac13\cdot \bigg(\vec{OA}+\vec{OB}+\vec{OC}\bigg)\\
&=\frac13\cdot \bigg[\begin{pmatrix}x_a\\y_a\end{pmatrix}+\begin{pmatrix}x_b\\y_b\end{pmatrix}+\begin{pmatrix}x_c\\y_c\end{pmatrix}\bigg]\\
&=\frac13\cdot\begin{pmatrix}x_a + x_b + x_c\\y_a+y_b+y_c\end{pmatrix}
\end{align*}

The coordinates are thus

$$S\bigg(\frac{x_a+x_b+x_c}3, \frac{y_a+y_b+y_c}3\bigg)$$


Analityc Geometry
Define $A(x_a,y_a)$ and the points $B,C$ similarly. We, hence, have $$D\bigg(\frac{x_b+x_c}{2}, \frac{y_b+y_c}{2}\bigg)$$ from the midpoint formula. Furthermore, using $AS=2\cdot SD$, we obtain $$S\bigg(\frac13 x_a+\frac23 \cdot\frac{x_b+x_c}{2}, y_a+\frac23 \cdot\frac{y_b+y_c}{2}\bigg)\implies S\bigg(\frac{x_a+x_b+x_c}3, \frac{y_a+y_b+y_c}3\bigg)$$
Taking your values I obtained

$$S=\bigg(-\frac23,-\frac73\bigg)$$

A: Here's another answer, starting with a little Euclidean geometry before you deal with the coordinates.

The centroid is the intersection of the medians. It is $2/3$ of the
  way along each median from its vertex to the opposite side.

The midpoint $M$ on edge $AB$ has coordinates the average of the coordinates of $A$ and $B$. Then to travel $2/3$ of the way from the third vertex $C$ to $M$ you find the weighted average combining $1/3$ of each  coordinate of $C$ with $2/3$ of the corresponding coordinate of $M$.
You will find lots of ways to understand your question here: https://www.cut-the-knot.org/triangle/medians.shtml
A: The co-ordinates are,
$C :C (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$
Substitute the values and you may find it! 
A: Hint:
To at least understand the formula given in an answer, try to make a diagram of the triangle and think about it. This might help you.
