Calculate coordinates of a centroid of $ABC$

Question

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?

Answer 1

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: $$\color{blue}{AS=2\cdot SD}$$

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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

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

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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

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

Answer 2

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!

Answer 3

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.

Answer 4

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