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To determine the type of parallelogram [tex]\(ABCD\)[/tex] with vertices at [tex]\(A(0,0)\)[/tex], [tex]\(B(3,6)\)[/tex], [tex]\(C(5,5)\)[/tex], and [tex]\(D(2,-1)\)[/tex], we need to follow a detailed step-by-step process.
1. Calculate the lengths of diagonals [tex]\(\overline{AC}\)[/tex] and [tex]\(\overline{BD}\)[/tex]:
- 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]
- For diagonal [tex]\(\overline{AC}\)[/tex]:
[tex]\[ A(0,0) \quad \text{and} \quad C(5,5) \][/tex]
[tex]\[ AC = \sqrt{(5-0)^2 + (5-0)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \approx 7.071 \][/tex]
- For diagonal [tex]\(\overline{BD}\)[/tex]:
[tex]\[ B(3,6) \quad \text{and} \quad D(2,-1) \][/tex]
[tex]\[ BD = \sqrt{(2-3)^2 + (-1-6)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \approx 7.071 \][/tex]
Since [tex]\(AC = BD\)[/tex], we see that the diagonals are equal in length.
2. Check if the diagonals [tex]\(\overline{AC}\)[/tex] and [tex]\(\overline{BD}\)[/tex] are perpendicular:
- To determine if two lines are perpendicular, the product of their slopes should be [tex]\(-1\)[/tex].
- The slope formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
- For diagonal [tex]\(\overline{AC}\)[/tex]:
[tex]\[ A(0,0) \quad \text{and} \quad C(5,5) \][/tex]
[tex]\[ \text{slope of } \overline{AC} = \frac{5 - 0}{5 - 0} = \frac{5}{5} = 1 \][/tex]
- For diagonal [tex]\(\overline{BD}\)[/tex]:
[tex]\[ B(3,6) \quad \text{and} \quad D(2,-1) \][/tex]
[tex]\[ \text{slope of } \overline{BD} = \frac{-1 - 6}{2 - 3} = \frac{-7}{-1} = 7 \][/tex]
- Check the product of the slopes:
[tex]\[ \text{slope of } \overline{AC} \times \text{slope of } \overline{BD} = 1 \times 7 = 7 \neq -1 \][/tex]
Therefore, [tex]\(\overline{AC}\)[/tex] is not perpendicular to [tex]\(\overline{BD}\)[/tex].
3. Determine the type of the parallelogram:
- Since [tex]\(AC = BD\)[/tex] and they are not perpendicular, the only possible conclusion is that quadrilateral [tex]\(ABCD\)[/tex] is a rhombus.
Therefore, the correct conclusion is:
[tex]\[ AC = BD; \text{ therefore, } ABCD \text{ is a rhombus.} \][/tex]
1. Calculate the lengths of diagonals [tex]\(\overline{AC}\)[/tex] and [tex]\(\overline{BD}\)[/tex]:
- 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]
- For diagonal [tex]\(\overline{AC}\)[/tex]:
[tex]\[ A(0,0) \quad \text{and} \quad C(5,5) \][/tex]
[tex]\[ AC = \sqrt{(5-0)^2 + (5-0)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \approx 7.071 \][/tex]
- For diagonal [tex]\(\overline{BD}\)[/tex]:
[tex]\[ B(3,6) \quad \text{and} \quad D(2,-1) \][/tex]
[tex]\[ BD = \sqrt{(2-3)^2 + (-1-6)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \approx 7.071 \][/tex]
Since [tex]\(AC = BD\)[/tex], we see that the diagonals are equal in length.
2. Check if the diagonals [tex]\(\overline{AC}\)[/tex] and [tex]\(\overline{BD}\)[/tex] are perpendicular:
- To determine if two lines are perpendicular, the product of their slopes should be [tex]\(-1\)[/tex].
- The slope formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
- For diagonal [tex]\(\overline{AC}\)[/tex]:
[tex]\[ A(0,0) \quad \text{and} \quad C(5,5) \][/tex]
[tex]\[ \text{slope of } \overline{AC} = \frac{5 - 0}{5 - 0} = \frac{5}{5} = 1 \][/tex]
- For diagonal [tex]\(\overline{BD}\)[/tex]:
[tex]\[ B(3,6) \quad \text{and} \quad D(2,-1) \][/tex]
[tex]\[ \text{slope of } \overline{BD} = \frac{-1 - 6}{2 - 3} = \frac{-7}{-1} = 7 \][/tex]
- Check the product of the slopes:
[tex]\[ \text{slope of } \overline{AC} \times \text{slope of } \overline{BD} = 1 \times 7 = 7 \neq -1 \][/tex]
Therefore, [tex]\(\overline{AC}\)[/tex] is not perpendicular to [tex]\(\overline{BD}\)[/tex].
3. Determine the type of the parallelogram:
- Since [tex]\(AC = BD\)[/tex] and they are not perpendicular, the only possible conclusion is that quadrilateral [tex]\(ABCD\)[/tex] is a rhombus.
Therefore, the correct conclusion is:
[tex]\[ AC = BD; \text{ therefore, } ABCD \text{ is a rhombus.} \][/tex]
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