To classify triangle ABC with points A (3, -2), B (2, -5), and C (4, -5), we first need to calculate the lengths of the sides AB, BC, and CA.
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Length of AB: \[ AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} = \sqrt{(2 - 3)^2 + (-5 - (-2))^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
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Length of BC: \[ BC = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} = \sqrt{(4 - 2)^2 + (-5 - (-5))^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2 \]
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Length of CA: \[ CA = \sqrt{(x_A - x_C)^2 + (y_A - y_C)^2} = \sqrt{(3 - 4)^2 + (-2 - (-5))^2} = \sqrt{(-1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
Now we have the lengths of the sides:
- AB = \( \sqrt{10} \)
- BC = 2
- CA = \( \sqrt{10} \)
Since sides AB and CA are equal, triangle ABC is classified as an isosceles triangle.
Furthermore, we can check if it is a right triangle by using the Pythagorean theorem:
- If AB = CA, we need to check if \( AB^2 + BC^2 = AC^2 \): \[ AB^2 + BC^2 = (\sqrt{10})^2 + 2^2 = 10 + 4 = 14 \] \[ CA^2 = (\sqrt{10})^2 = 10 \] Since \( 14 \neq 10 \), it is not a right triangle.
Thus, the classification of triangle ABC is C. It is an isosceles triangle.
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