Plot the vertices (–2, –8), (–2, 2), and (8, 2). What is the last vertex of the rectangle? What is the area of the rectangle? square units

There are different types of quadrilaterals, for example: square, rectangle, rhombus, trapezoid and parallelogram. Each type is defined accordingly to its length of sides and angles. For example, in a rectangle, the opposite sides are equal and parallel and their interior angles are equal to 90°.

The area of a rectangle is equal to base x height (A=bh).

Vector Magnitude Formula

It is possible to find the length of the sides from the vector magnitude |v|, since the magnitude of a vector represents the distance between two points.

[tex]|V|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

First, you need to plot the vertices points: A= (–2, –8), B= (–2, 2), and C=(8, 2).See the attached image 01, from this figure you see two sides and you need to find the vertice D.

From properties of rectangle, you have that opposite sides are equal and parallel. Thus, AB=DC and BC=AD.

Considering the properties of rectangle, the vertice D should have the x-coordinate of the point C and the y-coordinate of the point A. Hence, D=( 8,-8).

Now, you should find the length of these sides for calculating the area.

Find AB

[tex]|AB|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ \\ |AB|=\sqrt{(-2-(-2))^2+(2-(-8))^2}\\ \\ |AB|=\sqrt{(-2+2)^2+(2+8)^2}\\ \\ |AB|=\sqrt{0^2+10^2}\\ \\ |AB|=\sqrt{100}\\ \\ |AB|=10[/tex]

If AB=10, then DC also will be equal to 10.

Find BC

[tex]|BC|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ \\ |BC|=\sqrt{(8-(-2))^2+(2-(2))^2}\\ \\ |BC|=\sqrt{(8+2)^2+(2-2)^2}\\ \\ |BC|=\sqrt{10^2+0^2}\\ \\ |BC|=\sqrt{100}\\ \\ |BC|=10[/tex]

If BC=10, then AD also will be equal to 10.

Finally, you should calculate the area of rectangle (A=bh). Where:

b=BC=AD=10 and h=AB=DC= 10. So, the area of rectangle is 10 * 10 =100 square units.