Mathematics
The points (2, -1), (-1, 4) and (-2, 2) are mid-points of the sides of a triangle. Find its vertices.
Section Formula
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Answer
Let D = (2, -1), E = (-1, 4) and F = (-2, 2).
Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of △ABC.
Mid-point of AB, i.e. D(2, -1).
⇒ x1 + x2 = 4 ……..(1)
⇒ y1 + y2 = -2 ……..(2)
Mid-point of BC, i.e. E(-1, 4).
⇒ x2 + x3 = -2 ……..(3)
⇒ y2 + y3 = 8 ……..(4)
Mid-point of AC, i.e. F(-2, 2).
⇒ x1 + x3 = -4 ……..(5)
⇒ y1 + y3 = 4 ……..(6)
Adding 1, 3 and 5 we get,
⇒ x1 + x2 + x2 + x3 + x1 + x3 = 4 + (-2) + (-4)
⇒ 2(x1 + x2 + x3) = -2
⇒ x1 + x2 + x3 = -1.
From (1),
⇒ 4 + x3 = -1
⇒ x3 = -5.
Substituting value of x3 in (5) we get,
⇒ x1 + (-5) = -4
⇒ x1 = 1.
Substituting value of x3 in (3) we get,
⇒ x2 + (-5) = -2
⇒ x2 = 3.
Adding (2), (4) and (6) we get,
⇒ y1 + y2 + y2 + y3 + y1 + y3 = -2 + 8 + 4
⇒ 2(y1 + y2 + y3) = 10
⇒ y1 + y2 + y3 = 5
From (2)
⇒ -2 + y3 = 5
⇒ y3 = 7.
Substituting value of y3 in (4) we get,
⇒ y2 + 7 = 8
⇒ y2 = 1.
Substituting value of y3 in (6) we get,
⇒ y1 + 7 = 4
⇒ y1 = -3.
A = (x1, y1) = (1, -3), B = (x2, y2) = (3, 1), C = (x3, y3) = (-5, 7).
Hence, A = (1, -3), B = (3, 1) and C = (-5, 7).
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