Question

Find the vertices of a triangle, the mid-points of whose sides are (−1, 2), (−2, 6), and (4, −1).

Answer 1

Let the vertices be A (x₁, y₁), B (x₂, y₂), C (x₃, y₃),

If the mid-points of BC, CA, AB are D (−1, 2), E (−2, 6), F (4, −1) respectively,

Using the midpoint formula:

midpoint of `(x_i, y_i) "and" (x_j, y_j) "is" ((x_i + x_j) / 2, (y_i + y_j) / 2)`

Solving for x-coordinates:

I. From D (−1, 2) being the midpoint of BC:

`(x_2 + x_3) / 2 = −1`

x₂ + x₃ = −2

II. From E (−2, 6) being the midpoint of CA:

`(x_3 + x_1) / 2 = −2`

x₃ + x₁ = −4

III. From F (4,−1) being the midpoint of AB:

`(x_1 + x_2) / 2 = 4`

x₁ + x₂ = 8

Let’s add x-coordinates:

⇒ (x₂ + x₃) + (x₃ + x1) + (x₁ + x₂)

= −2 + (−4) + 8

= 2

⇒ 2(x₁ + x₂ + x₃) = 2

x₁ + x₂ + x₃ = 1

Now finding each xi:

• x₁ = (x₁ + x₂ + x₃) − (x₂ + x₃)
  = 1 − (−2)
  = 3
• x₂ = 1 − (x₃ + x₁)
  = 1 − (−4)
  = 5
• x₃ = 1 − (x₁ + x₂)
  = 1 − 8
  = −7

Solving for y-coordinates:

I. From D (−1, 2) being the midpoint of BC:

`(y_2 + y_3) / 2 = 2`

y₂ + y₃ = 4

II. From E (−2, 6) being the midpoint of CA:

`(y_3 + y_1) / 2 = 6`

y₃ + y₁ = 12

III. From F (4,−1) being the midpoint of AB:

`(y_1 + y_2) / 2 = −1`

y₁ + y₂ = −2

Let’s add y-coordinates:

⇒ (y₂ + y₃) + (y₃ + y1) + (y₁ + y₂)

= 4 + 12 + (−2)

= 14

⇒ 2(y₁ + y₂ + y₃) = 14

y₁ + y₂ + y₃ = 7

Now finding each yi:

• y₁ = (y₁ + y₂ + y₃) − (y₂ + y₃)
  = 7 − 4
  = 3
• y₂ = (y₁ + y₂ + y₃) − (x₃ + x₁)
  = 7 − 12
  = −5
• y₃ = (y₁ + y₂ + y₃) − (x₁ + x₂)
  = 7 − (−2)
  = 9

Therefore, the vertices are: A (x₁, y₁) = (3, 3), B (x₂, y₂) = (5, −5), C (x₃, y₃) = (−7, 9).