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Question
If `(3/2, 5), (7, (-9)/2)` and `(13/2, (-13)/2)` are mid-points of the sides of a triangle, then find the centroid of the triangle
Solution
In ΔABC, Let A(x1, y1), B(x2, y2) and C(x3, y3)
Mid−point of a line = `((x_1 + x_2)/2, (y_1 + y_2)/2)`
Mid−point of AB = `(3/2, 5)`
`((x_1 + x_2)/2, (y_1 + y_2)/2) = (3/2, 5)`
x1 + x2 = 3 → (1)
y1 + y2 = 10 → (2)
Mid−point of BC = `(7, (-9)/2)`
`((x_2 + x_3)/2, (y_2 + y_3)/2) = (7, (-9)/2)`
x2 + x3 = 14 → (3)
y2 + y3 = 10 → (4)
Mid−point of AC = `(13/2, (-13)/2)`
`((x_1 + x_3)/2, (y_1 + y_3)/2) = (13/2, (-13)/2)`
x1 + x3 = 3 → (5)
y1 + y3 = 10 → (6)
Add (1) + (3) + (5) We get
2x1 + 2x2 + 2x3 = 30
2(x1 + x2 + x3) = 30
x1 + x2 + x3 = 15
From (1), x1 + x2 = 3
∴ x3 = 12
From (3), x2 + x3 = 14
∴ x1 = 1
From (5), x1 + x3 = 13
∴ x2 = 2
Add (2), (4) and (6) we get
2y1 + 2y2 + 2y3 = −12
2(y1 + y2 + y3) = −12
∴ y1 + y2 + y3 = −6
From (2), y1 + y2 = 10
∴ y3 = −16
From (4) y2 + y3 = −9
∴ y1 = 3
From (6) y1 + y3 = −13
∴ y2 = 7
The vertices of the A are A(1, 3), B(2, 7) and C(12, −16)
Centroid of ΔABC (G) = `((x_1 + x_2 + x_3)/3, (y_1 + y_2 + y_3)/3)`
= `((1 + 2 + 12)/3, (3 + 7 - 16)/3)`
= `15/3, (-6)/3 (5, -2)`
The Centroid of Δ is (5, −2)
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