Question
Given the points A(3,−2) , B(2,−5) , and C(4,−5) , classify △ABC .(1 point) Responses It is a right triangle. It is a right triangle. It is a scalene triangle. It is a scalene triangle. It is an isosceles triangle. It is an isosceles triangle. It is an equilateral triangle.
Answers
GPT-4o mini
To classify triangle \( \triangle ABC \) formed by the points \( A(3, -2) \), \( B(2, -5) \), and \( C(4, -5) \), we first need to calculate the lengths of each side using the distance formula:
The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
1. **Length of side AB**:
\[
AB = \sqrt{(2 - 3)^2 + (-5 - (-2))^2} = \sqrt{(-1)^2 + (-5 + 2)^2} = \sqrt{1 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}
\]
2. **Length of side BC**:
\[
BC = \sqrt{(4 - 2)^2 + (-5 - (-5))^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2
\]
3. **Length of side AC**:
\[
AC = \sqrt{(4 - 3)^2 + (-5 - (-2))^2} = \sqrt{(1)^2 + (-5 + 2)^2} = \sqrt{1 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}
\]
Now we have the lengths of the sides:
- \( AB = \sqrt{10} \)
- \( BC = 2 \)
- \( AC = \sqrt{10} \)
Now let's classify the triangle based on its sides:
- Since \( AB = AC \), we have two equal sides. Therefore, triangle \( \triangle ABC \) is an **isosceles triangle**.
Next, we also need to check if it's a right triangle by using the Pythagorean theorem. If one side is the longest, say \( c \), then a triangle with sides \( a \), \( b \), and \( c \) is a right triangle if \( a^2 + b^2 = c^2 \).
Using the side lengths we found:
- \( BC \) (the shortest side) = 2
- \( AB \) = \( \sqrt{10} \)
- \( AC \) = \( \sqrt{10} \)
Checking if it satisfies the Pythagorean theorem:
Let \( a = BC = 2\) and \( b = AC = \sqrt{10} \) and the largest side \( c = AC = \sqrt{10} \).
Calculating:
\[
2^2 + \sqrt{10}^2 = 4 + 10 = 14
\]
\[
(\sqrt{10})^2 = 10
\]
Since \( 14 \neq 10 \), this confirms it is **not** a right triangle.
Thus, the classification of triangle \( \triangle ABC \) is:
**It is an isosceles triangle.**
The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
1. **Length of side AB**:
\[
AB = \sqrt{(2 - 3)^2 + (-5 - (-2))^2} = \sqrt{(-1)^2 + (-5 + 2)^2} = \sqrt{1 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}
\]
2. **Length of side BC**:
\[
BC = \sqrt{(4 - 2)^2 + (-5 - (-5))^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2
\]
3. **Length of side AC**:
\[
AC = \sqrt{(4 - 3)^2 + (-5 - (-2))^2} = \sqrt{(1)^2 + (-5 + 2)^2} = \sqrt{1 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}
\]
Now we have the lengths of the sides:
- \( AB = \sqrt{10} \)
- \( BC = 2 \)
- \( AC = \sqrt{10} \)
Now let's classify the triangle based on its sides:
- Since \( AB = AC \), we have two equal sides. Therefore, triangle \( \triangle ABC \) is an **isosceles triangle**.
Next, we also need to check if it's a right triangle by using the Pythagorean theorem. If one side is the longest, say \( c \), then a triangle with sides \( a \), \( b \), and \( c \) is a right triangle if \( a^2 + b^2 = c^2 \).
Using the side lengths we found:
- \( BC \) (the shortest side) = 2
- \( AB \) = \( \sqrt{10} \)
- \( AC \) = \( \sqrt{10} \)
Checking if it satisfies the Pythagorean theorem:
Let \( a = BC = 2\) and \( b = AC = \sqrt{10} \) and the largest side \( c = AC = \sqrt{10} \).
Calculating:
\[
2^2 + \sqrt{10}^2 = 4 + 10 = 14
\]
\[
(\sqrt{10})^2 = 10
\]
Since \( 14 \neq 10 \), this confirms it is **not** a right triangle.
Thus, the classification of triangle \( \triangle ABC \) is:
**It is an isosceles triangle.**