Question

Find the possible pairs of coordinates of the fourth vertex D of the parallelogram, if three of its vertices are A(5, 6), B(1, –2) and C(3, –2).

Answer 1

ABCD is a parallelogram.
∴ AD = BC and CD = AB      ...(Opposite sides of the parallelogram is congruent.)

∴ AD = BC
∴ `sqrt((a - 5)^2 + (b - 6)^2) = sqrt((3 - 1)^2 + [- 2 - (-2)]^2)`  ...(Distance Formula)

Squaring bothe the sides,
∴ (a - 5)² + (b - 6)² = (3 - 1)² + (- 2 + 2)² 
∴ a² - 10a + 25 + b² - 12b + 36 = 4 + 0
∴ a² + b² - 10a - 12b  + 57 = 0     ...(I)

∴ CD = AB 
`∴ sqrt((a - 3)^2 + [b - (- 2)]^2) = sqrt((5 - 1)^2 + [6 - (-2)]^2)`  ...(Distance Formula)

Squaring bothe the sides,
∴ (a - 3)² + (b + 2)² = (5 - 1)²  + (6 + 2)² 
∴  a² - 6a + 9 + b² + 4b + 4 = 16 + 64
∴  a² - 6a + b² + 4b = 80 - 9 - 4
∴  a² + b² - 6a + 4b - 67 = 0    ...(II)

Point D lies on the line passing through the point A. So, the ordinate of the point D will also be same as that of point A which is 6. So, b = 6.

Putting the value of b in (I), we get,
a² + 6² - 10a - 12 × 6 + 57 = 0
a² + 36 - 10a - 72 + 57 = 0
a² - 10a - 21 = 0
a² - 7a - 3a + 21 = 0
a(a - 7) - 3(a - 7) = 0
(a - 7)(a - 3) = 0
a = 3, 7

Thus, the possible values of point D are (3, 6) and (7, 6).

Answer 2

Let the points A(5, 6), B(1, -2) and C(3, -2) be the three vertices of a parallelogram.

The fourth vertex can be point D or point D₁ or point D₂ as shown in the figure.

Let D(x₁, y₁), D₁(x₂, y₂) and D₂(x₃, y₃).

Consider the parallelogram ACBD.

The diagonals of a parallelogram bisect each other. 

∴ midpoint of DC = midpoint of AB

`∴ ((x_1 + 3)/2, (y_1 - 2)/2) = ((5 + 1)/2, (6 - 2)/2)`

`∴ ((x_1 + 3)/2, (y_1 - 2)/2) = (6/2, 4/2)`

`∴ (x_1 + 3)/2 = 6/2 and (y_1 - 2)/2 = 4/2`

∴ x₁ + 3 = 6 and y₁ - 2 = 4
∴ x₁ = 3 and y₁ = 6

Co-ordinates of point D(x₁, y₁) are (3, 6).

Consider the parallelogram ABD₁C.
The diagonals of a parallelogram bisect each other.
∴ midpoint of AD₁ = midpoint of BC

`∴ ((x_2 + 5)/2, (y_2 + 6)/2) = ((3 + 1)/2, (-2 - 2)/2)`

`∴ ((x_2 + 5)/2, (y_2 + 6)/2) = (4/2, (-4)/2)`

`∴ (x_2 + 5)/2 = 4/2 and (y_2 + 6)/2 = (-4)/2`

∴ x₂ + 5 = 4 and y₂ + 6 = -4
∴ x₂ = - 1 and y₂ = - 10

∴ Co-ordinates of D (x₂, y₂) are (-1,-10).

Consider the parallelogram ABCD₂.
The diagonals of a parallelogram bisect each other.
∴ midpoint of BD₂ = midpoint of AC

`∴ ((x_3 + 1)/2, (y_3 - 2)/2) = ((5 + 3)/2, (6 - 2)/2)`

`∴ ((x_3 + 1)/2, (y_3 - 2)/2) = (8/2, 4/2)`

`∴ (x_3 + 1)/2 = 8/2 and (y_3 - 2)/2 = 4/2`

∴ x₃ + 1 = 8 and y₃ - 2 = 4
∴ x₃ = 7 and y₃ = 6

∴ co-ordinates of point D₂(x₃ , y₃) are (7, 6).

∴ The possible pairs of co-ordinates of the fourth vertex D of the parallelogram are (3, 6), (-1, -10) and (7, 6).