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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
  x
2 + 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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