To classify triangle \( \triangle ABC \), we will first find the lengths of each side using the distance formula. The distance \( d \) 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 side lengths
Length \( AB \):
Given points \( A(3, -2) \) and \( B(2, -5) \):
\[ AB = \sqrt{(2 - 3)^2 + (-5 + 2)^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
Length \( BC \):
Given points \( B(2, -5) \) and \( C(4, -5) \):
\[ BC = \sqrt{(4 - 2)^2 + (-5 + 5)^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2 \]
Length \( CA \):
Given points \( C(4, -5) \) and \( A(3, -2) \):
\[ CA = \sqrt{(3 - 4)^2 + (-2 + 5)^2} = \sqrt{(-1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
Step 2: Determine the side lengths
Now we know the lengths of the sides:
- \( AB = \sqrt{10} \)
- \( BC = 2 \)
- \( CA = \sqrt{10} \)
Step 3: Classify the triangle
We can classify \( \triangle ABC \) based on the lengths of its sides:
- Since \( AB = CA = \sqrt{10} \) and \( BC = 2 \) (where \( \sqrt{10} \approx 3.16 \)), two sides are equal.
Thus, \( \triangle ABC \) is an isosceles triangle.
Step 4: Check if it is also a right triangle
To classify further, we can check if it is also a right triangle using the Pythagorean theorem:
Let \( a = AB \), \( b = CA \), \( c = BC \):
- Assign the longest side, which is either \( CA \) or \( AB \) (\(\sqrt{10}\)), and check if \( c^2 = a^2 + b^2\):
\[ c^2 = (2)^2 = 4 \] \[ a^2 + b^2 = (\sqrt{10})^2 + (\sqrt{10})^2 = 10 + 10 = 20 \]
Since \( c^2 \neq a^2 + b^2 \) (4 is not equal to 20), \( \triangle ABC \) is not a right triangle.
Conclusion
The final classification of \( \triangle ABC \) is that it is an isosceles triangle but not a right triangle.
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