To classify triangle \( \triangle ABC \) formed by points \( A(3, -2) \), \( B(2, -5) \), and \( C(4, -5) \), we will calculate the lengths of its sides using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Calculating the lengths of the sides:
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Length \( AB \): \[ AB = \sqrt{(2 - 3)^2 + (-5 - (-2))^2} = \sqrt{(-1)^2 + (-5 + 2)^2} = \sqrt{1 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
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Length \( BC \): \[ BC = \sqrt{(4 - 2)^2 + (-5 - (-5))^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4 + 0} = \sqrt{4} = 2 \]
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Length \( CA \): \[ CA = \sqrt{(4 - 3)^2 + (-5 - (-2))^2} = \sqrt{(1)^2 + (-5 + 2)^2} = \sqrt{1 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
Summary of side lengths:
- \( AB = \sqrt{10} \)
- \( BC = 2 \)
- \( CA = \sqrt{10} \)
Classification:
Now we see if any sides are equal:
- \( AB = CA \) (both are \( \sqrt{10} \))
- \( BC \) is different.
Since \( AB \) and \( CA \) are equal, triangle \( \triangle ABC \) is isoceles.
Next, we check if it is a right triangle using the Pythagorean theorem:
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\( AB^2 + BC^2 \) vs \( CA^2 \): \[ (\sqrt{10})^2 + 2^2 = 10 + 4 = 14 \quad \text{and} \quad (\sqrt{10})^2 = 10 \]
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\( AB^2 + CA^2 \) vs \( BC^2 \): \[ (\sqrt{10})^2 + (\sqrt{10})^2 = 10 + 10 = 20 \quad \text{and} \quad 2^2 = 4 \]
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\( BC^2 + CA^2 \) vs \( AB^2 \): \[ 2^2 + (\sqrt{10})^2 = 4 + 10 = 14 \quad \text{and} \quad (\sqrt{10})^2 = 10 \]
None of these satisfy the Pythagorean theorem, so it is not a right triangle.
Thus, the classification of triangle \( \triangle ABC \) is:
It is an isosceles triangle.