In the following acute triangle \(ABC\),
\(D\), \(E\) and \(F\)
are the midpoints of \(AB\), \(BC\) and \(CA\).
So, \(CD\), \(AE\) and \(BF\) are the medians of this triangle.
The medians intersect at the point \(G\).
This point \(G\) is the centroid of \(△ABC\).
Note: A median is a segment that is drawn from
a vertex of a triangle to the mid-point of the opposite side of that vertex.
Each median of a triangle divides the triangle into
two smaller triangles that have equal areas.
A triangle has three medians because it has three sides.
The three medians divide the triangle into 6 smaller
triangles of equal area.
If coordinates of the vertices \(A\), \(B\) and \(C\) are
\( (x_1, y_1) \), \( (x_2, y_2) \) and \( (x_3, y_3) \) respectively,
then the coordinates of the centroid \(G\) can be calculated
with the following formula:
\[ G(x, y) = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) \]
For a triangle with vertices \(A(-30, 3)\), \(B(0, 3)\) and \(C(10, 30)\),
the coordinates of the centroid \(G\) may be calculated as follows:
Here, \(x_1=-30,\; x_2=0,\; x_3=10,\; y_1=3,\; y_2=3,\; y_3=30 \)
Coordinates of the centroid \(G\): \begin{align} G(x, y) &= \left( \frac{-30 + 0 + 10}{3}, \frac{3 + 3 + 30}{3}\right)\\\\ &= \left( \frac{-20}{3}, \frac{36}{3}\right) \\\\ &= ( -6.67, 12 ) \\\\ \end{align} So the coordinates of the centroid of the above triangle is \(( -6.67, 12 )\)
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