To classify the quadrilateral formed by the points Q(−8,1), U(3,5), A(7,−1), and Y(−4,−4), we need to check the properties of the shape based on the coordinates of the vertices.
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Calculate the lengths of the sides:
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\( \text{Distance between Q and U} \): \[ \sqrt{(3 - (-8))^2 + (5 - 1)^2} = \sqrt{(3 + 8)^2 + (5 - 1)^2} = \sqrt{11^2 + 4^2} = \sqrt{121 + 16} = \sqrt{137} \]
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\( \text{Distance between U and A} \): \[ \sqrt{(7 - 3)^2 + (-1 - 5)^2} = \sqrt{(4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} \]
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\( \text{Distance between A and Y} \): \[ \sqrt{(-4 - 7)^2 + (-4 - (-1))^2} = \sqrt{(-11)^2 + (-4 + 1)^2} = \sqrt{121 + 9} = \sqrt{130} \]
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\( \text{Distance between Y and Q} \): \[ \sqrt{(-8 - (-4))^2 + (1 - (-4))^2} = \sqrt{(-8 + 4)^2 + (1 + 4)^2} = \sqrt{(-4)^2 + (5)^2} = \sqrt{16 + 25} = \sqrt{41} \]
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Check for parallel sides (to determine if it's a parallelogram):
- Calculate the slopes of the sides:
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Slope of QU: \[ m_{QU} = \frac{5 - 1}{3 - (-8)} = \frac{4}{11} \]
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Slope of UA: \[ m_{UA} = \frac{-1 - 5}{7 - 3} = \frac{-6}{4} = -\frac{3}{2} \]
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Slope of AY: \[ m_{AY} = \frac{-4 - (-1)}{-4 - 7} = \frac{-3}{-11} = \frac{3}{11} \]
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Slope of YQ: \[ m_{YQ} = \frac{1 - (-4)}{-8 - (-4)} = \frac{5}{-4} = -\frac{5}{4} \]
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We see that the slopes do not match in pairs. Therefore, there are no pairs of parallel sides.
- Calculate the slopes of the sides:
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Conclusion: Since the quadrilateral does not have any pairs of parallel sides, it cannot qualify as a parallelogram, rectangle, rhombus, or square.
Thus, the most specific classification for the shape formed by the vertices Q(−8,1), U(3,5), A(7,−1), and Y(−4,−4) is a Quadrilateral.