Question

If points A(–5, y), В(2, –2), C(8, 4) and D(x, 5) taken in order, form a parallelogram ABCD, then find the values of x and y. Hence, find the lengths of the sides of the parallelogram.

Answer 1

We know that diagonals of a parallelogram bisect each other.

So, the coordinates of the midpoint of AC = coordinates of the midpoint of BD.

`(∵ "x- coordinate of O" = (x_1 + x_2)/2 = "y-coordinate of O" = (y_1 + y_2)/2)`

⇒ `((-5 + 8)/2, (y + 4)/2) = ((2 + x)/2, (-2 + 5)/2)`

⇒ `(3/2, (y + 4)/2) = ((x + 2)/2, 3/2)`

⇒ `3/2 = (x + 2)/2`

⇒ 2(x + 2) = 3 × 2   ...(By cross multiplying)

⇒ 2x + 4 = 6

⇒ 2x = 6 – 4

⇒ `x = 2/2`

⇒ x = 1

And `(y + 4)/2 = 3/2`

⇒ 2(y + 4) = 3 × 2   ...(By cross multiplying)

⇒ 2y + 8 = 6

⇒ 2y = 6 – 8

⇒ `y = (-2)/2`

⇒ y = –1

Therefore, the value of x = 1 and y = –1.

To find the lengths of the sides of the parallelogram, we use the distance formula.

We know that in the parallelogram ABCD

AB = CD and BC = AD

So, the lengths of sides

AB = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

= `sqrt([2 - (-5)]^2 + [(-2) - (-1)^2]`

= `sqrt((2 + 5)^2 + (-2 + 1)^2`

= `sqrt((7)^2 + (-1)^2`

= `sqrt(49 + 1)`

= `sqrt(50)`

= `5sqrt(2)`

And BC = `sqrt((8 - 2)^2 + [4 - (-2)]^2`

= `sqrt((6)^2 + (4 + 2)^2`

= `sqrt(36 + (6)^2`

= `sqrt(36 + 36)`

= `sqrt(72)`

= `6sqrt(2)`

Therefore, the lengths of sides of the parallelogram are AB = CD = `5sqrt(2)` and BC = AD = `6sqrt(2)`.