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).

● 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

10th Grade Math

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