To classify triangle \( \triangle ABC \) with the points \( A(3, -2) \), \( B(2, -5) \), and \( C(4, -5) \), we first need to find the lengths of the sides of the triangle.
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Calculate the lengths of the sides:
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Length \( AB \): \[ AB = \sqrt{(2 - 3)^2 + (-5 + 2)^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
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Length \( BC \): \[ BC = \sqrt{(4 - 2)^2 + (-5 + 5)^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2 \]
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Length \( AC \): \[ AC = \sqrt{(4 - 3)^2 + (-5 + 2)^2} = \sqrt{(1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
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Compare the lengths of the sides:
- \( AB = \sqrt{10} \)
- \( BC = 2 \)
- \( AC = \sqrt{10} \)
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Classify the triangle:
- Since \( AB = AC \) (both are \( \sqrt{10} \)), \( \triangle ABC \) is an isosceles triangle because it has at least two equal sides.
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Check if it's a right triangle: To check if it's a right triangle, we can use the Pythagorean theorem. We need to check if: \[ AB^2 + BC^2 = AC^2 \]
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Calculate \( AB^2 + BC^2 \): \[ AB^2 = 10, \quad BC^2 = 4 \] \[ AB^2 + BC^2 = 10 + 4 = 14 \]
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Calculate \( AC^2 \): \[ AC^2 = 10 \]
Since \( AB^2 + BC^2 \neq AC^2 \), it is not a right triangle.
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Therefore, the classification of triangle \( \triangle ABC \) is isosceles triangle.
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