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Let's determine the type of quadrilateral formed by the vertices [tex]\((2,4)\)[/tex], [tex]\((-4,-2)\)[/tex], [tex]\((-2,4)\)[/tex], and [tex]\((4,-2)\)[/tex].
1. Label the vertices:
- [tex]\(A = (2, 4)\)[/tex]
- [tex]\(B = (-4, -2)\)[/tex]
- [tex]\(C = (-2, 4)\)[/tex]
- [tex]\(D = (4, -2)\)[/tex]
2. Calculate the lengths of the sides:
Using the distance formula [tex]\( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)[/tex]:
- [tex]\(AB\)[/tex]: The distance between [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ AB = \sqrt{((-4 - 2)^2 + (-2 - 4)^2)} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \][/tex]
- [tex]\(BC\)[/tex]: The distance between [tex]\(B\)[/tex] and [tex]\(C\)[/tex]:
[tex]\[ BC = \sqrt{((-2 + 4)^2 + (4 + 2)^2)} = \sqrt{(2)^2 + (6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \][/tex]
- [tex]\(CD\)[/tex]: The distance between [tex]\(C\)[/tex] and [tex]\(D\)[/tex]:
[tex]\[ CD = \sqrt{((4 + 2)^2 + (-2 - 4)^2)} = \sqrt{(6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \][/tex]
- [tex]\(DA\)[/tex]: The distance between [tex]\(D\)[/tex] and [tex]\(A\)[/tex]:
[tex]\[ DA = \sqrt{((2 - 4)^2 + (4 + 2)^2)} = \sqrt{(-2)^2 + (6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \][/tex]
3. Calculate the diagonals:
- [tex]\(AC\)[/tex]: The distance between [tex]\(A\)[/tex] and [tex]\(C\)[/tex]:
[tex]\[ AC = \sqrt{((-2 - 2)^2 + (4 - 4)^2)} = \sqrt{(-4)^2 + 0} = \sqrt{16} = 4 \][/tex]
- [tex]\(BD\)[/tex]: The distance between [tex]\(B\)[/tex] and [tex]\(D\)[/tex]:
[tex]\[ BD = \sqrt{((4 + 4)^2 + (-2 + 2)^2)} = \sqrt{(8)^2 + 0} = \sqrt{64} = 8 \][/tex]
4. Classify the quadrilateral:
- For a quadrilateral to be a square or rectangle, the opposite sides must be equal, and the diagonals should be equal.
- For a quadrilateral to be a parallelogram, the opposite sides must be equal (AB = CD and BC = DA).
- For a trapezoid, only one pair of opposite sides needs to be parallel/equal.
- From the calculations:
- [tex]\(AB = CD = 6\sqrt{2}\)[/tex]
- [tex]\(BC = DA = 2\sqrt{10}\)[/tex]
- [tex]\(AC \neq BD\)[/tex]
Since opposite sides are equal (AB = CD and BC = DA), the given quadrilateral is a parallelogram.
Therefore, the quadrilateral with vertices [tex]\((2,4)\)[/tex], [tex]\((-4,-2)\)[/tex], [tex]\((-2,4)\)[/tex], and [tex]\((4,-2)\)[/tex] is a parallelogram.
Answer: D. parallelogram
1. Label the vertices:
- [tex]\(A = (2, 4)\)[/tex]
- [tex]\(B = (-4, -2)\)[/tex]
- [tex]\(C = (-2, 4)\)[/tex]
- [tex]\(D = (4, -2)\)[/tex]
2. Calculate the lengths of the sides:
Using the distance formula [tex]\( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)[/tex]:
- [tex]\(AB\)[/tex]: The distance between [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ AB = \sqrt{((-4 - 2)^2 + (-2 - 4)^2)} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \][/tex]
- [tex]\(BC\)[/tex]: The distance between [tex]\(B\)[/tex] and [tex]\(C\)[/tex]:
[tex]\[ BC = \sqrt{((-2 + 4)^2 + (4 + 2)^2)} = \sqrt{(2)^2 + (6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \][/tex]
- [tex]\(CD\)[/tex]: The distance between [tex]\(C\)[/tex] and [tex]\(D\)[/tex]:
[tex]\[ CD = \sqrt{((4 + 2)^2 + (-2 - 4)^2)} = \sqrt{(6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \][/tex]
- [tex]\(DA\)[/tex]: The distance between [tex]\(D\)[/tex] and [tex]\(A\)[/tex]:
[tex]\[ DA = \sqrt{((2 - 4)^2 + (4 + 2)^2)} = \sqrt{(-2)^2 + (6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \][/tex]
3. Calculate the diagonals:
- [tex]\(AC\)[/tex]: The distance between [tex]\(A\)[/tex] and [tex]\(C\)[/tex]:
[tex]\[ AC = \sqrt{((-2 - 2)^2 + (4 - 4)^2)} = \sqrt{(-4)^2 + 0} = \sqrt{16} = 4 \][/tex]
- [tex]\(BD\)[/tex]: The distance between [tex]\(B\)[/tex] and [tex]\(D\)[/tex]:
[tex]\[ BD = \sqrt{((4 + 4)^2 + (-2 + 2)^2)} = \sqrt{(8)^2 + 0} = \sqrt{64} = 8 \][/tex]
4. Classify the quadrilateral:
- For a quadrilateral to be a square or rectangle, the opposite sides must be equal, and the diagonals should be equal.
- For a quadrilateral to be a parallelogram, the opposite sides must be equal (AB = CD and BC = DA).
- For a trapezoid, only one pair of opposite sides needs to be parallel/equal.
- From the calculations:
- [tex]\(AB = CD = 6\sqrt{2}\)[/tex]
- [tex]\(BC = DA = 2\sqrt{10}\)[/tex]
- [tex]\(AC \neq BD\)[/tex]
Since opposite sides are equal (AB = CD and BC = DA), the given quadrilateral is a parallelogram.
Therefore, the quadrilateral with vertices [tex]\((2,4)\)[/tex], [tex]\((-4,-2)\)[/tex], [tex]\((-2,4)\)[/tex], and [tex]\((4,-2)\)[/tex] is a parallelogram.
Answer: D. parallelogram
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