Centroid of a Triangle

Centroid of a Triangle

Centroid: The centroid of the triangle formed by the points (x₁, y₁, z₁), (x₂, y₂, z₂), (x₃, y₃, z₃) is $$\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\ \frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3}\ ,\ \frac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3} \right)$$.

Proof: Let A (x₁, y₁, z₁), B (x₂, y₂, z₂) and C (x₃, y₃, z₃)

Let D be the midpoint of BC. Then $$D=\left( \frac{{{x}_{2}}+{{x}_{3}}}{2},\ \frac{{{y}_{2}}+{{y}_{3}}}{2},\ \frac{{{z}_{2}}+{{z}_{3}}}{2} \right)$$.

Since centroid G divides each median in the ratio 2 : 1, it follows that G divide AD in the ratio 2 : 1.

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

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

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

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

The centroid of the triangle $$G=\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3},\frac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3} \right)$$.

Example: Find the centroid of the triangle formed by the point (7, -4, 7), (1, -6, 10), (5, -1, 1).

Solution: Given that A (7, -4, 7), B (1, -6, 10), C (5, -1, 1)

The centroid of the triangle formed by the points (x₁, y₁, z₁), (x₂, y₂, z₂), (x₃, y₃, z₃) is $$\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\ \frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3}\ ,\ \frac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3} \right)$$,

$$G=\left( \frac{7+1+5}{3},\frac{-4-6-1}{3},\frac{7+10+1}{3} \right)$$,

$$G=\left( \frac{13}{3},\frac{-11}{3},\frac{18}{3} \right)$$,

$$G=\left( \frac{13}{3},\frac{-11}{3},6 \right)$$.