To classify triangle ABC formed by the points A(3, 2), B(2, -5), and C(4, -5), we can use the distance formula 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:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Step 1: Calculate the lengths of the sides.
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Length of side AB: \[ AB = \sqrt{(2 - 3)^2 + (-5 - 2)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \]
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Length of side BC: \[ BC = \sqrt{(4 - 2)^2 + (-5 - (-5))^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2 \]
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Length of side AC: \[ AC = \sqrt{(4 - 3)^2 + (-5 - 2)^2} = \sqrt{(1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \]
Step 2: Classify the triangle.
Now we have:
- \( AB = 5\sqrt{2} \)
- \( BC = 2 \)
- \( AC = 5\sqrt{2} \)
Here, \( AB \) and \( AC \) are equal, which indicates that triangle ABC is an isosceles triangle (two sides are equal).
Next, we can check if the triangle is acute, right, or obtuse by examining the squares of the lengths:
- \( AB^2 = (5\sqrt{2})^2 = 50 \)
- \( AC^2 = (5\sqrt{2})^2 = 50 \)
- \( BC^2 = 2^2 = 4 \)
Now we compare:
- \( AB^2 + BC^2 = 50 + 4 = 54 \)
- \( AC^2 = 50 \)
Since \( AB^2 + BC^2 > AC^2 \), triangle ABC is an acute triangle as well.
Conclusion: Triangle ABC is an isosceles acute triangle.