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Select the correct answer from each drop-down menu.

Given: [tex]W(-1,1), X(3,4), Y(6,0)[/tex], and [tex]Z(2,-3)[/tex] are the vertices of quadrilateral [tex]WXYZ[/tex].

Prove: [tex]WXYZ[/tex] is a square.

Using the distance formula, I found that [tex]\square[/tex]

Answer :

To determine if the quadrilateral WXYZ is a square, we need to check two things: first, that all the sides are equal in length, and second, that the diagonals are equal in length. Here's how we can do that:

1. List the points:
We have the vertices:
[tex]\( W(-1, 1) \)[/tex],
[tex]\( X(3, 4) \)[/tex],
[tex]\( Y(6, 0) \)[/tex],
[tex]\( Z(2, -3) \)[/tex].

2. Calculate the side lengths:
Use the distance formula to find the length of each side:

The distance formula between two points [tex]\( (x_1, y_1) \)[/tex] and [tex]\( (x_2, y_2) \)[/tex] is:
[tex]\[
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\][/tex]

- Length of [tex]\( WX \)[/tex]:
[tex]\[
\text{WX} = \sqrt{(3 - (-1))^2 + (4 - 1)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5
\][/tex]

- Length of [tex]\( XY \)[/tex]:
[tex]\[
\text{XY} = \sqrt{(6 - 3)^2 + (0 - 4)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5
\][/tex]

- Length of [tex]\( YZ \)[/tex]:
[tex]\[
\text{YZ} = \sqrt{(2 - 6)^2 + (-3 - 0)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = 5
\][/tex]

- Length of [tex]\( ZW \)[/tex]:
[tex]\[
\text{ZW} = \sqrt{(2 - (-1))^2 + (-3 - 1)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5
\][/tex]

All sides are equal in length, which is 5.

3. Calculate the lengths of the diagonals:

- Length of diagonal [tex]\( WY \)[/tex]:
[tex]\[
\text{WY} = \sqrt{(6 - (-1))^2 + (0 - 1)^2} = \sqrt{7^2 + (-1)^2} = \sqrt{49 + 1} = 7.07
\][/tex]

- Length of diagonal [tex]\( XZ \)[/tex]:
[tex]\[
\text{XZ} = \sqrt{(2 - 3)^2 + (-3 - 4)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = 7.07
\][/tex]

The diagonals are equal in length, which is approximately 7.07.

Since all four sides of the quadrilateral WXYZ are equal, and both diagonals are equal, we can conclude that WXYZ is indeed a square.

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