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.

Answer 1

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.