To classify triangle ABC with the vertices A (3, -2), B (2, -5), and C (4, -5), we first need to determine the lengths of its sides.
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Find the lengths of the sides (AB, BC, and CA):
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AB: \[ AB = \sqrt{(2 - 3)^2 + (-5 - (-2))^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
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BC: \[ BC = \sqrt{(4 - 2)^2 + (-5 - (-5))^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2 \]
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CA: \[ CA = \sqrt{(4 - 3)^2 + (-5 - (-2))^2} = \sqrt{(1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
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Summary of side lengths:
- \( AB = \sqrt{10} \)
- \( BC = 2 \)
- \( CA = \sqrt{10} \)
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Classify the triangle:
- Since \( AB = CA \), triangle ABC is Isosceles (two sides are equal).
- It cannot be equilateral since not all sides are equal.
- It cannot be scalene since at least two sides are equal.
- To check if it's a right triangle, we can use the Pythagorean theorem, but we note that \(AB^2 + BC^2 = (\sqrt{10})^2 + 2^2 = 10 + 4 = 14\) does not equal \(CA^2 = (\sqrt{10})^2 = 10\). Thus it’s not a right triangle.
Thus, the classification of triangle ABC is D - it is an isosceles triangle.