To classify the triangle \( \triangle ABC \) with points A(3,−2), B(2,−5), and C(4,−5), we first calculate the lengths of the sides of the triangle using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
- Calculate AB:
\[ AB = \sqrt{(2 - 3)^2 + (-5 - (-2))^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
- Calculate BC:
\[ BC = \sqrt{(4 - 2)^2 + (-5 - (-5))^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2 \]
- Calculate AC:
\[ AC = \sqrt{(4 - 3)^2 + (-5 - (-2))^2} = \sqrt{(1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
Now we have the lengths of the sides:
- \( AB = \sqrt{10} \)
- \( BC = 2 \)
- \( AC = \sqrt{10} \)
Next, we analyze the side lengths:
- \( AB = AC \) and \( BC \) is different.
This means that \( \triangle ABC \) has two equal sides, hence it is classified as an isosceles triangle.
To confirm if it is a right triangle, we would check if the square of the longest side equals the sum of the squares of the other two sides, using the lengths we calculated.
Let’s consider if \( AB \) or \( AC \) could be the longest side:
- \( BC^2 = 2^2 = 4 \)
- \( AB^2 = AC^2 = (\sqrt{10})^2 = 10 \)
Since \( 10 \) is not equal to \( 4 + 4 \) or any combination thereof, the triangle is not a right triangle.
Thus, the classification of \( \triangle ABC \) is:
It is an isosceles triangle.
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