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).

● **Distance and Section Formulae**

**Distance Formula****Distance Properties in some Geometrical Figures****Conditions of Collinearity of Three Points****Problems on Distance Formula****Distance of a Point from the Origin****Distance Formula in Geometry****Section Formula****Midpoint Formula****Centroid of a Triangle****Worksheet on Distance Formula****Worksheet on Collinearity of Three Points****Worksheet on Finding the Centroid of a Triangle****Worksheet on Section Formula**

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