Use the coordinates to compute the perimeter of a paralleleogram whose vertices are located at the ordered pairs (0, 0), (1, 2), (2, 1) and (1, -1).

1 answer

To find the perimeter of the parallelogram with vertices at (0, 0), (1, 2), (2, 1), and (1, -1), we first need to calculate the lengths of its sides.

We can pair the vertices into two adjacent pairs:

  • \( A(0, 0) \) and \( B(1, 2) \)
  • \( C(2, 1) \) and \( D(1, -1) \)
  1. Calculate the distance \( AB \): \[ AB = \sqrt{(1 - 0)^2 + (2 - 0)^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \]

  2. Calculate the distance \( BC \): \[ BC = \sqrt{(2 - 1)^2 + (1 - 2)^2} = \sqrt{(1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

  3. Calculate the distance \( CD \): \[ CD = \sqrt{(1 - 2)^2 + (-1 - 1)^2} = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \]

  4. Calculate the distance \( DA \): \[ DA = \sqrt{(0 - 1)^2 + (0 - (-1))^2} = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

Now we have the side lengths:

  • \( AB = \sqrt{5} \)
  • \( BC = \sqrt{2} \)
  • \( CD = \sqrt{5} \) (same as \( AB \))
  • \( DA = \sqrt{2} \) (same as \( BC \))

The perimeter \( P \) of the parallelogram is given by the sum of the lengths of all sides: \[ P = AB + BC + CD + DA = \sqrt{5} + \sqrt{2} + \sqrt{5} + \sqrt{2} \] \[ P = 2\sqrt{5} + 2\sqrt{2} \]

Thus, the perimeter of the parallelogram is: \[ \boxed{2\sqrt{5} + 2\sqrt{2}} \]