To classify triangle ABC formed by the points A(3, -2), B(2, -5), and C'(4, -5), we need to calculate the lengths of the sides and determine their relationships.
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Calculate the lengths of the sides using the distance formula: \[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
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Length of AB: \[ AB = \sqrt{(2 - 3)^2 + (-5 - (-2))^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
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Length of BC: \[ BC = \sqrt{(4 - 2)^2 + (-5 - (-5))^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2 \]
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Length of 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:
- \(AB = \sqrt{10}\)
- \(BC = 2\)
- \(AC = \sqrt{10}\)
Here, we observe that:
- \(AB = AC\)
- \(BC\) is different from \(AB\) and \(AC\).
Since two sides (AB and AC) are equal, triangle ABC is classified as an isosceles triangle.
- Check if it is a right triangle: We can check if it forms a right triangle using the Pythagorean theorem. If \(a\) and \(b\) are the lengths of the two shorter sides, and \(c\) is the longest side, then: \[ a^2 + b^2 = c^2 \]
In our case:
- \(a = AC = \sqrt{10}\)
- \(b = AB = \sqrt{10}\)
- \(c = BC = 2\)
Calculating: \[ (\sqrt{10})^2 + (\sqrt{10})^2 = 10 + 10 = 20 \quad and \quad (2)^2 = 4 \]
Since \(20 \neq 4\), it is not a right triangle.
Conclusion: Triangle ABC is an isosceles triangle.
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