To classify the given quadrilateral with vertices U(0, 2), N(4, 7), I(6, 3), and T(3, -2), we need to calculate the lengths of its sides and the slopes of its sides to determine the nature of the quadrilateral.
- Find the lengths of the sides:
- Length \( UN \): \[ UN = \sqrt{(4 - 0)^2 + (7 - 2)^2} = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \]
- Length \( NI \): \[ NI = \sqrt{(6 - 4)^2 + (3 - 7)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]
- Length \( IT \): \[ IT = \sqrt{(3 - 6)^2 + (-2 - 3)^2} = \sqrt{(-3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34} \]
- Length \( TU \): \[ TU = \sqrt{(0 - 3)^2 + (2 - (-2))^2} = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
Now we have the lengths:
- \( UN = \sqrt{41} \)
- \( NI = 2\sqrt{5} \)
- \( IT = \sqrt{34} \)
- \( TU = 5 \)
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Check if opposite sides are equal (for parallelogram) and calculate the slopes:
- Slope \( UN \): \[ \text{slope of } UN = \frac{7 - 2}{4 - 0} = \frac{5}{4} \]
- Slope \( IT \): \[ \text{slope of } IT = \frac{-2 - 3}{3 - 6} = \frac{-5}{-3} = \frac{5}{3} \]
- Slope \( NI \): \[ \text{slope of } NI = \frac{3 - 7}{6 - 4} = \frac{-4}{2} = -2 \]
- Slope \( TU \): \[ \text{slope of } TU = \frac{2 - (-2)}{0 - 3} = \frac{4}{-3} = -\frac{4}{3} \]
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Check for parallelism:
- \( UN \) and \( IT \) are not parallel.
- \( NI \) and \( TU \) are also not parallel.
Since opposite sides are not equal and the slopes of opposite sides are not equal (which means no pairs of sides are parallel), the shape is not a parallelogram, rectangle, rhombus, or square.
Conclusion:
The most specific classification for this figure with the given vertices is a Quadrilateral.