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Question
The points A(−3, 6), B(0, 7) and C(1, 9) are the mid-points of the sides DE, EF and FD of a triangle DEF. Show that the quadrilateral ABCD is a parallelogram.
Solution
Let D be (x1 y1), E(x2, y2) and F(x3, y3)
Mid−point of a line = `((x_1 + x_2)/2, (y_1 + y_2)/2)`
Mid−point of DE = `((x_1 + x_2)/2, (y_1 + y_2)/2)`
(−3, 6) = `((x_1 + x_2)/2, (y_1 + y_2)/2)`
x1 + x2 = −6 → (1)
y1 + y2 = 12 → (2)
Mid−point of EF = `((x_2 + x_3)/2, (y_2 + y_3)/2)`
(0, 7) = `((x_2 + x_3)/2, (y_2 + y_3)/2)`
∴ x2 + x3 = 0 → (3)
y2 + y3 = 14 → (4)
Mid−point of FD = `((x_1 + x_3)/2, (y_1 + y_3)/2)`
(1, 9) = `((x_1 + x_3)/2, (y_1 + y_3)/2)`
x1 + x3 = 2 → (5)
y1 + y3 = 18 → (6)
Add (1) + (3) + (5)
2x1 + 2x2 + 2x3 = −6 + 0 + 2
2(x1 + x2 + x3) = −4
x1 + x2 + x3 = −2
From (1) x1 + x2 = −6
x3 = −2 + 6 = 4
From (3) x2 + x3 = 0
x1 = −2 + 0
x1 = −2
From (5) x1 + x3 = 2
x2 + 2 = −2
x2 = −2 − 2 = −4
∴ x1 = −2, x2 = −4, x3 = 4
Add (2) + (4) + 6
2y1 + 2y2 + 2y3 = 12 + 14 + 18
2(y1 + y2 + y3) = 44
y1 + y2 + y3 = `44/2` = 22
From (2) y1 + y2 = 12
12 + y3 = 22
⇒ y3 = 22 – 12
y3 = 10
From (4) y2 + y3 = 14
y1 + 14 = 22
⇒ y1 = 22 – 14
y1 = 8
From (6)
y1 + y3 = 18
y2 + 18 = 22
⇒ = 22 – 18
y2 = 4
∴ y1 = 8, y2 = 4, y3 = 10
The vertices D is (−2, 8)
Mid−point of AC = `((-3 + 1)/2, (6 + 9)/2)`
= `(-1, 15/2)`
Mid−point of BD = `((0 - 2)/2, (7 + 8)/2)`
= `((-2)/2, 15/2)`
= `(-1, 15/2)`
Mid-point of diagonal AC = Mid-point of diagonal BD
The quadrilateral ABCD is a parallelogram.
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