Question

What is the area of a rectangle with vertices at ​ (6, −3) ​, ​ (3, −6) ​ , (−1, −2) , and (2, 1) ? Enter your answer in the box.

Answer 1

We are given rectangle with vertices (6, −3) ​, ​ (3, −6) ​ , (−1, −2) , and (2, 1) .

`\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad √(\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2)`

`\mathrm{The\:distance\:between\:}\left(6,\:-3\right)\mathrm{\:and\:}\left(3,\:-6\right)\mathrm{\:is\:}`

`=√(\left(3-6\right)^2+\left(-6-\left(-3\right)\right)^2)\`

Length of the rectangle is
`=3√(2)`.

Width of the rectangle is

`=√(\left(-1-3\right)^2+\left(-2-\left(-6\right)\right)^2)`

`=4√(2)`

Area of the rectangle = length × width.

Plugging values of length and width in above formula, we get

Area =
`3√(2)* 4√(2)\:`= 12(2) =24 square units.

Therefore, area of the rectangle is 24 square units.