To classify the quadrilateral formed by the vertices Q(-8, 1), U(3, 5), A(7, -1), and Y(-2, -2), we need to calculate the lengths of the sides and the slopes of the sides.
First, let's find the lengths of the sides using the distance formula, which is:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
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Length of QU: \[ QU = \sqrt{(3 - (-8))^2 + (5 - 1)^2} = \sqrt{(3 + 8)^2 + (5 - 1)^2} = \sqrt{11^2 + 4^2} = \sqrt{121 + 16} = \sqrt{137} \]
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Length of UA: \[ UA = \sqrt{(7 - 3)^2 + (-1 - 5)^2} = \sqrt{(4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} \]
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Length of AY: \[ AY = \sqrt{(-2 - 7)^2 + (-2 - (-1))^2} = \sqrt{(-9)^2 + (-1)^2} = \sqrt{81 + 1} = \sqrt{82} \]
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Length of YQ: \[ YQ = \sqrt{(-8 - (-2))^2 + (1 - (-2))^2} = \sqrt{(-6)^2 + (3)^2} = \sqrt{36 + 9} = \sqrt{45} \]
Now we have the lengths of the sides as follows:
- QU: \(\sqrt{137}\)
- UA: \(\sqrt{52}\)
- AY: \(\sqrt{82}\)
- YQ: \(\sqrt{45}\)
Now, let's calculate the slopes to check for parallelism:
The slope formula is:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \]
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Slope of QU: \[ \text{slope of QU} = \frac{5 - 1}{3 - (-8)} = \frac{4}{11} \]
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Slope of UA: \[ \text{slope of UA} = \frac{-1 - 5}{7 - 3} = \frac{-6}{4} = -\frac{3}{2} \]
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Slope of AY: \[ \text{slope of AY} = \frac{-2 - (-1)}{-2 - 7} = \frac{-1}{-9} = \frac{1}{9} \]
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Slope of YQ: \[ \text{slope of YQ} = \frac{1 - (-2)}{-8 - (-2)} = \frac{3}{-6} = -\frac{1}{2} \]
Now let's analyze the results:
- The lengths of the sides are all different, meaning it cannot be a rhombus or square.
- The slopes are also all different, thus no sides are parallel, meaning it cannot be a parallelogram, rectangle, or square.
Conclusion:
The figure is classified as a quadrilateral since it does not meet the criteria for being a parallelogram, rectangle, rhombus, or square.
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