To classify triangle \( \triangle ABC \) using the given points \( A(3, -2) \), \( B(2, -5) \), and \( C(4, -5) \), we need to calculate the lengths of the sides of the triangle.
The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Let's calculate the lengths of the sides \( AB \), \( BC \), and \( CA \):
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Length of \( AB \): \[ AB = \sqrt{(2 - 3)^2 + (-5 - (-2))^2} = \sqrt{(-1)^2 + (-5 + 2)^2} = \sqrt{1 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
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Length of \( BC \): \[ BC = \sqrt{(4 - 2)^2 + (-5 - (-5))^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2 \]
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Length of \( CA \): \[ CA = \sqrt{(4 - 3)^2 + (-5 - (-2))^2} = \sqrt{(1)^2 + (-5 + 2)^2} = \sqrt{1 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
Now we have the lengths of the sides:
- \( AB = \sqrt{10} \)
- \( BC = 2 \)
- \( CA = \sqrt{10} \)
Since \( AB \) and \( CA \) are equal, this means \( \triangle ABC \) is an isosceles triangle.
Classification of Triangle:
Based on the calculated lengths of the sides:
- It is an isosceles triangle (since two sides are equal).
Let's also determine if it is a right triangle. For a triangle to be a right triangle, the following condition using the Pythagorean theorem must hold:
\[ \text{(length of the longest side)}^2 = \text{(length of one side)}^2 + \text{(length of the other side)}^2 \]
In this case, \( BC \) (2) is the shortest side. Checking: \[ BC^2 + AB^2 = 2^2 + (\sqrt{10})^2 = 4 + 10 = 14 \] \[ CA^2 = (\sqrt{10})^2 = 10 \]
Since \( 14 \neq 10 \), it is not a right triangle.
Thus, the classification of triangle \( \triangle ABC \) is: It is an isosceles triangle.
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