Application of Section Formula

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Application of Section Formula:
The coordinates of the centroid of the triangle whose vertices are (x₂, y₁), (x₂, y₂) and (x₃, y₃) are

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Example: If the coordinates of the mid-points of the sides of a triangle are (2, 2), (2, - 3) and (5, 6), Find its centroid.
Solution: Let P (2, 2), Q (2, - 3), R (5, 6) be the mid-points of sides AB, BC and CA respectively of

ABC.
Let A {x₁, y₁), B (x₂, y₂) and C (x₃, y₃) be the vertices of triangle ABC.
Then, P is the mid-point of BC
x₁ + x₂ / 2 = 2, y₁ + y₂ / 2= 2

x₁ + x₂ = 4, and y₁ + y₂ = 4.....(i)
Q is the mid-point of BC
x₂ + x₃ / 2 = 2 and y₂ + y₃ / 2 = - 3

x₂ + x₃ = 4 and y₂ + y₃ = - 6.....(ii)
R is the mid-point of AC
x₁ + x₃ / 2 = 5, y₁ + y₃ / 2= 6

x₁ + x₃ = 10, and y₁ + y₃ = 12.....(iii)
From (i), (ii) and (iii), you get
x₁ + x₂ + x₂ + x₃ + x₃ + x₁ = 4 + 4+ 10 =18
and, y₁ + y₂ + y₂ + y₃ + y₃ + y₁ = 4 - 6 + 12 = 10

x₁ + x₂ + x₃ = 9 and y₁ + y₂ + y₃ = 5
the coordinates of the centroid of

ABC are
[x₁ + x₂ + x₃) / 3, (y₁ + y₂ + y₃) / 3 ]
Or (9/3, 5/3) = (3, 5/3)


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