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Question
Find the co-ordinates of centroid of a triangle if points D(–7, 6), E(8, 5) and F(2, –2) are the mid-points of the sides of that triangle.
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Solution
Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of MBC.
Given that D(–7, 6), E(8, 5) and F(2, –2) are the midpoints of sides AB, BC and AC respectively.
Let G(x, y) be the centroid.
D is now the midpoint of AB.
∴ By midpoint formula.
–7 = `(x_1 + x_2)/2`
∴ x1 + x2 = –7 × 2
∴ x1 + x2 = –14 ......(i)
Similarly x2 + x3 = 16 ......(ii)
And x1 + x3 = 4 ......(iii)
Adding equations (i), (ii), and (iii), we get
x1 + x2 + x2 + x3 + x1 + x3 = –14 + 16 + 4
∴ 3x1 + 2x2 + 2x3 = 6
∴ 2(x1 + x2 + x3) = 6
∴ x1 + x2 + x3 = `6/2`
∴ x1 + x2 + x3 = 3 ......(iv)
D is the midpoint of AB.
∴ By the midpoint formula,
6 = `("y"_1 + "y"_2)/2`
∴ y1 + y2 = 12 ......(v)
Similarly, y2 + y3 = 10 ......(vi)
y1 + y3 = –4 ......(vii)
Adding equations (v), (vi), and (vii), we get
y1 + y2 + y2 + y3 + y1 + y4 = 12 + 10 – 4
∴ 2y1 + 2y2 + 2y3 = 18
∴ 2(y1 + y2 + y3) = 18
∴ y1 + y2 + y3 = `18/2`
∴ y1 + y2 + y3 = 9 ......(viii)
G(x, y) is the centroid of ΔABC.
∴ By centroid formula,
a = `(x_1 + x_2 + x_3)/3`
∴ x = `3/3` ......[From (iv)]
∴ x = 1
b = `("y"_1 + "y"_2 + "y"_3)/3`
∴ y = `9/3` ......[From (viii)]
∴ y = 3
