Question

Given the points A (3, - 2), B (2, - 5), and C' (4, -5), classify ABC.
(1 point)
• It is a scalene triangle.
• It is an isosceles triangle.
• It is a right triangle.
• It is an equilateral triangle.

Answer 1

To classify 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 and determine their relationships.

1. **Calculate the lengths of the sides** using the distance formula:
\[
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

- Length of AB:
\[
AB = \sqrt{(2 - 3)^2 + (-5 - (-2))^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}
\]

- Length of BC:
\[
BC = \sqrt{(4 - 2)^2 + (-5 - (-5))^2} = \sqrt{(2)^2 + (0)^2} = \sqrt{4} = 2
\]

- Length of AC:
\[
AC = \sqrt{(4 - 3)^2 + (-5 - (-2))^2} = \sqrt{(1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}
\]

2. **Compare the lengths of the sides**:
- \(AB = \sqrt{10}\)
- \(BC = 2\)
- \(AC = \sqrt{10}\)

Here, we observe that:
- \(AB = AC\)
- \(BC\) is different from \(AB\) and \(AC\).

Since two sides (AB and AC) are equal, triangle ABC is classified as an **isosceles triangle**.

3. **Check if it is a right triangle**:
We can check if it forms a right triangle using the Pythagorean theorem. If \(a\) and \(b\) are the lengths of the two shorter sides, and \(c\) is the longest side, then:
\[
a^2 + b^2 = c^2
\]

In our case:
- \(a = AC = \sqrt{10}\)
- \(b = AB = \sqrt{10}\)
- \(c = BC = 2\)

Calculating:
\[
(\sqrt{10})^2 + (\sqrt{10})^2 = 10 + 10 = 20 \quad and \quad (2)^2 = 4
\]

Since \(20 \neq 4\), it is not a right triangle.

**Conclusion**: Triangle ABC is an **isosceles triangle**.