To prove that quadrilateral [tex]\(WXYZ\)[/tex] is a square, we need to show both that all four sides are of equal length and that the diagonals are also equal in length.
### Step-by-Step Solution
1. Identify 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 lengths of the sides using the distance formula:
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]
3. Side lengths:
- [tex]\(d_{WX}\)[/tex]:
[tex]\[
\sqrt{(3 - (-1))^2 + (4 - 1)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5.0
\][/tex]
- [tex]\(d_{XY}\)[/tex]:
[tex]\[
\sqrt{(6 - 3)^2 + (0 - 4)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.0
\][/tex]
- [tex]\(d_{YZ}\)[/tex]:
[tex]\[
\sqrt{(2 - 6)^2 + (-3 - 0)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5.0
\][/tex]
- [tex]\(d_{ZW}\)[/tex]:
[tex]\[
\sqrt{(2 - (-1))^2 + (-3 - 1)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.0
\][/tex]
4. Calculate the lengths of the diagonals:
- [tex]\(d_{WY}\)[/tex]:
[tex]\[
\sqrt{(6 - (-1))^2 + (0 - 1)^2} = \sqrt{7^2 + (-1)^2} = \sqrt{49 + 1} = \sqrt{50} = 7.0710678118654755
\][/tex]
- [tex]\(d_{XZ}\)[/tex]:
[tex]\[
\sqrt{(2 - 3)^2 + (-3 - 4)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 7.0710678118654755
\][/tex]
5. Verify the conditions of a square:
- All sides are equal: [tex]\(d_{WX} = d_{XY} = d_{YZ} = d_{ZW} = 5.0\)[/tex]
- Both diagonals are equal: [tex]\(d_{WY} = d_{XZ} = 7.0710678118654755\)[/tex]
Since all four sides are equal and both diagonals are equal, quadrilateral [tex]\(WXYZ\)[/tex] meets the criteria to be classified as a square.
Thus, quadrilateral [tex]\(WXYZ\)[/tex] is indeed a square.
