To classify the shape formed by the vertices L(-4, 4), A(3, 2), K(5, -5), and E(-2, -3), we can follow these steps:
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Check if it's a Quadrilateral: A quadrilateral has four vertices, which we have.
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Calculate the distances between the points: \[ \text{Distance (d)} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
- Distance LA: \[ d_{LA} = \sqrt{(3 - (-4))^2 + (2 - 4)^2} = \sqrt{(3 + 4)^2 + (2 - 4)^2} = \sqrt{7^2 + (-2)^2} = \sqrt{49 + 4} = \sqrt{53} \]
- Distance AK: \[ d_{AK} = \sqrt{(5 - 3)^2 + (-5 - 2)^2} = \sqrt{(2)^2 + (-7)^2} = \sqrt{4 + 49} = \sqrt{53} \]
- Distance KE: \[ d_{KE} = \sqrt{(-2 - 5)^2 + (-3 - (-5))^2} = \sqrt{(-7)^2 + (2)^2} = \sqrt{49 + 4} = \sqrt{53} \]
- Distance EL: \[ d_{EL} = \sqrt{(-4 - (-2))^2 + (4 - (-3))^2} = \sqrt{(-2)^2 + (7)^2} = \sqrt{4 + 49} = \sqrt{53} \]
Since all sides are equal (\(d_{LA} = d_{AK} = d_{KE} = d_{EL} = \sqrt{53}\)), we can conclude the shape is at least a rhombus.
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Check the diagonals: Calculate the distances of the diagonals to further classify the shape.
- Diagonal LK: \[ d_{LK} = \sqrt{(5 - (-4))^2 + (-5 - 4)^2} = \sqrt{(5 + 4)^2 + (-9)^2} = \sqrt{9^2 + (-9)^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2} \]
- Diagonal AE: \[ d_{AE} = \sqrt{(-2 - 3)^2 + (-3 - 2)^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \]
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Check for right angles: For it to be a rectangle or a square, it must have right angles. We can check if any adjacent sides are perpendicular:
- Vectors LA and AK: \[ \text{Vector LA} = (3 - (-4), 2 - 4) = (7, -2) \] \[ \text{Vector AK} = (5 - 3, -5 - 2) = (2, -7) \]
- The dot product should be zero for perpendicular vectors: \[ LA \cdot AK = 7 \times 2 + (-2) \times (-7) = 14 + 14 = 28 \quad (\text{not perpendicular}) \]
Since all sides are equal, and the diagonals are not the same length, and the angles are not right angles, we can conclude that the quadrilateral is a rhombus.
Conclusion
The most specific classification for the shape formed by the vertices L(-4, 4), A(3, 2), K(5, -5), and E(-2, -3) is a rhombus.
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