# Centroid of Triangle & Centroid of Tetrahedron

Centroid of Triangle: The coordinates of the centroid of the triangle with vertices A (x1, y1, z1) B (x2, y2, z2) and C (x3, y3, z3)

$$\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\,\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3},\,\frac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3} \right)$$

Proof: Let D be the mid-point of AC then coordinates of D are $$\left( \frac{{{x}_{2}}+{{x}_{3}}}{2},\,\frac{{{y}_{2}}+{{y}_{3}}}{2},\,\frac{{{z}_{2}}+{{z}_{3}}}{2} \right)$$

Let G be the centroid of ΔABCThen, G divides AD in the ratio 2:1

So, coordinate of D are $$\left( \frac{1.{{x}_{1}}+2\left( \frac{{{x}_{2}}+{{x}_{3}}}{2} \right)}{1+2},\,\frac{1.{{y}_{1}}+2\left( \frac{{{y}_{2}}+{{y}_{3}}}{2} \right)}{1+2},\,\frac{1.{{z}_{1}}+2\left( \frac{{{z}_{2}}+{{z}_{3}}}{2} \right)}{1+2} \right)$$

i.e, $$\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\,\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3},\,\frac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3} \right)$$

Centroid of Tetrahedron: Let ABCD be a tetrahedron such that the coordinates of its vertices are A (x1, y1, z1) B (x2, y2, z2) C (x3, y3, z3) and D (x4, y4, z4) the coordinates of the centroid of faces ABC, DAB, DBC and DCA are respectively$${{G}_{1}}=\left[ \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\,\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3},\,\frac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3} \right]$$,

$${{G}_{2}}=\left[ \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{4}}}{3},\,\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{4}}}{3},\,\frac{{{z}_{1}}+{{z}_{2}}+{{z}_{4}}}{3} \right]$$,

$${{G}_{3}}=\left[ \frac{{{x}_{2}}+{{x}_{3}}+{{x}_{4}}}{3},\,\frac{{{y}_{2}}+{{y}_{3}}+{{y}_{4}}}{3},\,\frac{{{z}_{2}}+{{z}_{3}}+{{z}_{4}}}{3} \right]$$,

$${{G}_{4}}=\left[ \frac{{{x}_{4}}+{{x}_{3}}+{{x}_{1}}}{3},\,\frac{{{y}_{4}}+{{y}_{3}}+{{y}_{1}}}{3},\,\frac{{{z}_{4}}+{{z}_{3}}+{{z}_{1}}}{3} \right]$$.

Now, coordinates of point G dividing DG, in the ratio 3:1 are

$$\left[ \frac{1.{{x}_{4}}+3\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3} \right)}{1+3},\,\frac{1.{{y}_{4}}+3\left( \frac{{{y}_{1}}+{{y}_{2}}+{{y}_{4}}}{3} \right)}{1+3},\,\frac{1.{{z}_{4}}+3\left( \frac{{{z}_{2}}+{{z}_{3}}+{{z}_{4}}}{3} \right)}{1+3} \right]$$,

= $$\left[ \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}}{4},\,\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}}{4},\,\frac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}+{{z}_{4}}}{4} \right]$$,

Similarly the point dividing CG1, CG2, AG3 and BG4 in the ratio 3:1 has the same coordinates.

Hence, the point $$G\left[ \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}}{4},\,\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}}{4},\,\frac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}+{{z}_{4}}}{4} \right]$$.