# Centroid of a Triangle

The Centroid of a triangle is the point of intersection of the medians of a triangle.

To find the centroid of a triangle

Let A (x$$_{1}$$, y$$_{1}$$), B (x$$_{2}$$, y$$_{2}$$) and C (x$$_{3}$$, y$$_{3}$$) are  the three vertices of the ∆ABC .

Let D be the midpoint of side BC.

Since, the coordinates of B (x$$_{2}$$, y$$_{2}$$) and C (x$$_{3}$$, y$$_{3}$$), the coordinate of the point D are ($$\frac{x_{2} + x_{3}}{2}$$, $$\frac{y_{2} + y_{3}}{2}$$).

Let G(x, y) be the centroid of the triangle ABC.

Then, from the geometry, G is on the median AD and it divides AD in the ratio 2 : 1, that is AG : GD = 2 : 1.

Therefore, x = $$\left \{\frac{2\cdot \frac{(x_{2} + x_{3})}{2} + 1 \cdot x_{1}}{2 + 1}\right \}$$ = $$\frac{x_{1} + x _{2} + x_{3}}{3}$$

y = $$\left \{\frac{2\cdot \frac{(y_{2} + y_{3})}{2} + 1 \cdot y_{1}}{2 + 1}\right \}$$ = $$\frac{y_{1} + y _{2} + y_{3}}{3}$$

Therefore, the coordinate of the G are ($$\frac{x_{1} + x _{2} + x_{3}}{3}$$, $$\frac{y_{1} + y _{2} + y_{3}}{3}$$)

Hence, the centroid of a triangle whose vertices are (x$$_{1}$$, y$$_{1}$$), (x$$_{2}$$, y$$_{2}$$) and (x$$_{3}$$, y$$_{3}$$) has the coordinates ($$\frac{x_{1} + x _{2} + x_{3}}{3}$$, $$\frac{y_{1} + y _{2} + y_{3}}{3}$$).

Note: The centroid of a triangle divides each median in the ratio 2 : 1 (vertex to base).

Solved examples to find the centroid of a triangle:

1. Find the co-ordinates of the point of intersection of the medians of trangle ABC; given A = (-2, 3), B = (6, 7) and C = (4, 1).

Solution:

Here, (x$$_{1}$$  = -2, y$$_{1}$$ = 3), (x$$_{2}$$  = 6, y$$_{2}$$ = 7) and  (x$$_{3}$$  = 4, y$$_{3}$$ = 1),

Let G (x, y) be the centroid of the triangle ABC. Then,

x = $$\frac{x_{1} + x _{2} + x_{3}}{3}$$ = $$\frac{(-2) + 6 + 4}{3}$$ = $$\frac{8}{3}$$

y = $$\frac{y_{1} + y _{2} + y_{3}}{3}$$ = $$\frac{3 + 7 + 1}{3}$$ = $$\frac{11}{3}$$

Therefore, the coordinates of the centroid G of the triangle ABC are ($$\frac{8}{3}$$, $$\frac{11}{3}$$)

Thus, the coordinates of the point of intersection of the medians of triangle are ($$\frac{8}{3}$$, $$\frac{11}{3}$$).

2. The three vertices of the triangle ABC are (1, -4), (-2, 2) and (4, 5) respectively. Find the centroid and the length of the median through the vertex A.

Solution:

Here, (x$$_{1}$$  = 1, y$$_{1}$$ = -4), (x$$_{2}$$  = -2, y$$_{2}$$ = 2) and  (x$$_{3}$$  = 4, y$$_{3}$$ = 5),

Let G (x, y) be the centroid of the triangle ABC. Then,

x = $$\frac{x_{1} + x _{2} + x_{3}}{3}$$ = $$\frac{1 + (-2) + 4}{3}$$ = $$\frac{3}{3}$$ = 1

y = $$\frac{y_{1} + y _{2} + y_{3}}{3}$$ = $$\frac{(-4) + 2 + 5}{3}$$ = $$\frac{3}{3}$$ = 1

Therefore, the coordinates of the centroid G of the triangle ABC are (1, 1).

D is the middle point of the side BC of the triangle ABC.

Therefore, the coordinates of D are ($$\frac{(-2) + 4}{2}$$, $$\frac{2 + 5}{2}$$) = (1, $$\frac{7}{2}$$)

Therefore, the length of the median AD = $$\sqrt{(1 - 1)^{2} + (-4 - \frac{7}{2})^{2}}$$ = $$\frac{15}{2}$$ units.

3. Two vertices of a triangle are (1, 4) and (3, 1). If the centroid of the triangle is the origin, find the third vertex.

Solution:

Let the coordinates of the third vertex are (h, k).

Therefore, the coordinates of the centroid of the triangle ($$\frac{1 + 3 + h}{3}$$, $$\frac{4 + 1 + k}{3}$$)

According to the problem we know that the centroid of the given triangle is (0, 0)

Therefore,

$$\frac{1 + 3 + h}{3}$$ = 0 and $$\frac{4 + 1 + k}{3}$$ = 0

⟹ h = -4 and k = -5

Therefore, the third vertex of the given triangle are (-4, -5).

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Distance and Section Formulae