Question

Given the points A(3,−2), B(2,−5), and C(4,−5), classify △ABC.

*HINT* Find the distance between each side length

(1 point)
Responses

It is an isosceles triangle.
It is an isosceles triangle.

It is an equilateral triangle.
It is an equilateral triangle.

It is a right triangle.
It is a right triangle.

It is a scalene triangle.

Answer 1

To classify triangle \( \triangle ABC \), we first need to find the lengths of its sides using the distance formula. The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

Let's find the distances for \(AB\), \(BC\), and \(CA\).

1. **Length of \(AB\)**:

\[
A(3, -2) \quad B(2, -5)
\]
\[
AB = \sqrt{(2 - 3)^2 + (-5 - (-2))^2}
\]
\[
AB = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}
\]

2. **Length of \(BC\)**:

\[
B(2, -5) \quad C(4, -5)
\]
\[
BC = \sqrt{(4 - 2)^2 + (-5 - (-5))^2}
\]
\[
BC = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2
\]

3. **Length of \(CA\)**:

\[
C(4, -5) \quad A(3, -2)
\]
\[
CA = \sqrt{(3 - 4)^2 + (-2 - (-5))^2}
\]
\[
CA = \sqrt{(-1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10}
\]

Now we have the lengths of the sides:
- \(AB = \sqrt{10}\)
- \(BC = 2\)
- \(CA = \sqrt{10}\)

Since \(AB\) and \(CA\) are equal and \(BC\) is different, triangle \( \triangle ABC \) is classified as an **isosceles triangle**.

Therefore, the correct response is:

**It is an isosceles triangle.**