Question

Choose the best selection for thequadrilateral with vertices at thefollowing points:(4,0), (8,0), (4,-4), (8,-4)Hint: Start by graphing the points.Distance Formula: d= (x2 – x1)2 + (y2 - y1)2A. RectangleB. SquareC. RhombusD. Trapezoid

Answer 1

We are given coordinates of four points and we are asked to find the best selection of the shape.

In order to know what shape this is, we simply graph all the points and see what shape we get.

Using a graphing calculator, the points make up the following shape:

Now that we have the shape, we know that it can either be a Rectangle or a Square.

If the shape is a rectangle, the lengths x and y are not equal. If the shape is a square, lengths x and y are equal.

In order to find x and y, we use the distance formula given to us as:

`\begin{gathered} |d|=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ \text{where,} \\ (x_2,y_2)\to\text{ Second point} \\ (x_1,y_1)\to\text{ First point} \end{gathered}`

Now let us find x and y.

x:

The points that make up the length x are:

(4, 0), and (4, -4)

Therefore, length x can be gotten as:

`\begin{gathered} x_2=4,y_2=0,x_1=4,y_1=-4 \\ |x|=\sqrt[]{(4-4)^2+(0--4)^2} \\ |x|=\sqrt[]{4^2} \\ \\ \therefore|x|=4 \end{gathered}`

Now we find y.

y:

The points that make up the length y are:

(4, 0) and (8, 0)

Therefore, length y can be gotten as:

`\begin{gathered} x_2=4,y_2=0,x_1=4,y_1=-4 \\ |x|=\sqrt[]{(4-4)^2+(0--4)^2} \\ |x|=\sqrt[]{4^2} \\ \\ \therefore|x|=4 \end{gathered}`