To determine the type of quadrilateral given the vertices [tex]\((-3,0)\)[/tex], [tex]\((-3,-3)\)[/tex], [tex]\((0,-3)\)[/tex], and [tex]\((0,0)\)[/tex], let's go through the steps.
1. Graph the Points:
- Plot the points on a coordinate plane.
- Point [tex]\(A = (-3, 0)\)[/tex]
- Point [tex]\(B = (-3, -3)\)[/tex]
- Point [tex]\(C = (0, -3)\)[/tex]
- Point [tex]\(D = (0, 0)\)[/tex]
2. Calculate the Lengths of All Sides Using the Distance Formula:
- The distance formula is [tex]\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)[/tex].
- Distance [tex]\(AB\)[/tex]:
[tex]\(A (-3, 0)\)[/tex] to [tex]\(B (-3, -3)\)[/tex]
[tex]\[
AB = \sqrt{((-3) - (-3))^2 + ((-3) - 0)^2} = \sqrt{0 + 9} = 3
\][/tex]
- Distance [tex]\(BC\)[/tex]:
[tex]\(B (-3, -3)\)[/tex] to [tex]\(C (0, -3)\)[/tex]
[tex]\[
BC = \sqrt{(0 - (-3))^2 + ((-3) - (-3))^2} = \sqrt{9 + 0} = 3
\][/tex]
- Distance [tex]\(CD\)[/tex]:
[tex]\(C (0, -3)\)[/tex] to [tex]\(D (0, 0)\)[/tex]
[tex]\[
CD = \sqrt{(0 - 0)^2 + (0 - (-3))^2} = \sqrt{0 + 9} = 3
\][/tex]
- Distance [tex]\(DA\)[/tex]:
[tex]\(D (0, 0)\)[/tex] to [tex]\(A (-3, 0)\)[/tex]
[tex]\[
DA = \sqrt{(0 - (-3))^2 + (0 - 0)^2} = \sqrt{9 + 0} = 3
\][/tex]
3. Calculate the Lengths of the Diagonals:
- Diagonal [tex]\(AC\)[/tex]:
[tex]\(A (-3, 0)\)[/tex] to [tex]\(C (0, -3)\)[/tex]
[tex]\[
AC = \sqrt{(0 - (-3))^2 + ((-3) - 0)^2} = \sqrt{9 + 9} = 3\sqrt{2}
\][/tex]
- Diagonal [tex]\(BD\)[/tex]:
[tex]\(B (-3, -3)\)[/tex] to [tex]\(D (0, 0)\)[/tex]
[tex]\[
BD = \sqrt{(0 - (-3))^2 + (0 - (-3))^2} = \sqrt{9 + 9} = 3\sqrt{2}
\][/tex]
4. Analyze the Characteristics:
- All four sides are equal: [tex]\(AB = BC = CD = DA = 3\)[/tex]
- Both diagonals are equal: [tex]\(AC = BD = 3\sqrt{2}\)[/tex]
Given these properties, the quadrilateral has all four sides equal and the diagonals are equal as well, which are the defining characteristics of a square.
Thus, the best selection for the quadrilateral with vertices at the given points is:
Square.
