To classify the quadrilateral formed by the vertices K(3, 2), I(9, 4), N(6, 0), and G(1, -2), we need to analyze the sides and angles of the shape.
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Calculate the length of each side:
- Distance KI: \[ d(KI) = \sqrt{(9-3)^2 + (4-2)^2} = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \]
- Distance IN: \[ d(IN) = \sqrt{(6-9)^2 + (0-4)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
- Distance NG: \[ d(NG) = \sqrt{(1-6)^2 + (-2-0)^2} = \sqrt{(-5)^2 + (-2)^2} = \sqrt{25 + 4} = \sqrt{29} \]
- Distance GK: \[ d(GK) = \sqrt{(3-1)^2 + (2-(-2))^2} = \sqrt{(2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]
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Check if opposite sides are equal (necessary for parallelogram):
- Compare sides KI and NG: \[ d(KI) = 2\sqrt{10} \quad \text{and} \quad d(NG) = \sqrt{29} \quad (not\ equal) \]
- Compare sides IN and GK: \[ d(IN) = 5 \quad \text{and} \quad d(GK) = 2\sqrt{5} \quad (not\ equal) \]
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Determine if it is a trapezoid:
- Check if slopes of opposite sides are equal for parallelogram: \[ \text{slope of } KI = \frac{4-2}{9-3} = \frac{2}{6} = \frac{1}{3} \] \[ \text{slope of } NG = \frac{-2-0}{1-6} = \frac{-2}{-5} = \frac{2}{5} \] \[ \text{slope of } IN = \frac{0-4}{6-9} = \frac{-4}{-3} = \frac{4}{3} \] \[ \text{slope of } GK = \frac{2-(-2)}{3-1} = \frac{4}{2} = 2 \]
Since neither pairs of opposite sides are equal nor parallel, this quadrilateral does not classify as a parallelogram.
Conclusion: Based on these calculations, the figure cannot be classified as a quadrilateral, parallelogram, rectangle, rhombus, or square. Thus, the most specific classification of the given quadrilateral is just a Quadrilateral.