Question

show that the points A(2,-2), B(14,10) C(11,13) and D(-1,1) are the vertices of a rectangle

Answer 1

Calculate the distance between these points by distance formula,
It's proved... Check the photo...

Answer 2

Answer:

The opposite sides and diagonals are equal, therefore ABCD is a rectangle.

Step-by-step explanation:

The given vertices are A(2,-2), B(14,10) C(11,13) and D(-1,1).

We need to show that ABCD is a rectangle.

Opposite sides of a rectangle are equal and parallel. Length of the diagonals are equal.

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula we get

$AB=\sqrt{\left(14-2\right)^2+\left(10-\left(-2\right)\right)^2}=12\sqrt{2}$

$BC=\sqrt{\left(11-14\right)^2+\left(13-10\right)^2}=3\sqrt{2}$

$CD=\sqrt{\left(-1-11\right)^2+\left(1-13\right)^2}=12\sqrt{2}$

$AD=\sqrt{\left(-1-2\right)^2+\left(1-\left(-2\right)\right)^2}=3\sqrt{2}$

Since AB=CD and BC=AD, therefore opposite sides are equal.

$AC=\sqrt{\left(11-2\right)^2+\left(13-\left(-2\right)\right)^2}=3\sqrt{34}$

$BD=\sqrt{\left(-1-14\right)^2+\left(1-10\right)^2}=3\sqrt{34}$

AC=BD, it means diagonals are equal.

Since, the opposite sides and diagonals are equal, therefore ABCD is a rectangle.