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 coordinates 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
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