Asked by PJ
                Given the points A (3,-2), B (2,-5), and C (4,-5), classify  △ABC .(1 point)
Responses
It is an equilateral triangle.
It is an equilateral triangle.
It is an scalene triangle.
It is an scalene triangle.
It is an right triangle.
It is an right triangle.
It is an isosceles triangle.
            
        Responses
It is an equilateral triangle.
It is an equilateral triangle.
It is an scalene triangle.
It is an scalene triangle.
It is an right triangle.
It is an right triangle.
It is an isosceles triangle.
Answers
                                                    There are no human answers yet.
                                            
                
                    Answered by
            GPT-4o mini
            
    To classify triangle \( \triangle ABC \) formed by the points \( A(3,-2) \), \( B(2,-5) \), and \( C(4,-5) \), we need to calculate the lengths of the sides of the triangle using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
1. **Calculate \( AB \)**:
\[
AB = \sqrt{(2 - 3)^2 + (-5 - (-2))^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}
\]
2. **Calculate \( BC \)**:
\[
BC = \sqrt{(4 - 2)^2 + (-5 - (-5))^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2
\]
3. **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:
- \( AB = \sqrt{10} \)
- \( BC = 2 \)
- \( AC = \sqrt{10} \)
### Classification:
- Since \( AB = AC \), two sides are equal, which means that \( \triangle ABC \) is an isosceles triangle.
Therefore, the correct classification of \( \triangle ABC \) is:
**It is an isosceles triangle.**
    
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
1. **Calculate \( AB \)**:
\[
AB = \sqrt{(2 - 3)^2 + (-5 - (-2))^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}
\]
2. **Calculate \( BC \)**:
\[
BC = \sqrt{(4 - 2)^2 + (-5 - (-5))^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2
\]
3. **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:
- \( AB = \sqrt{10} \)
- \( BC = 2 \)
- \( AC = \sqrt{10} \)
### Classification:
- Since \( AB = AC \), two sides are equal, which means that \( \triangle ABC \) is an isosceles triangle.
Therefore, the correct classification of \( \triangle ABC \) is:
**It is an isosceles triangle.**
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