Question

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

a. It is an scalene triangle
b. It is an right triangle
c. It is an equilateral triangle
d. It is an isosceles triangle

Answer 1

To classify triangle ABC with the given points A(3, -2), B(2, -5), and C(4, -5), we first need to calculate the lengths of the sides of the triangle.

1. **Calculate the distances between the points:**

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

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

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

2. **Summarizing the side lengths:**
- \(AB = \sqrt{10}\)
- \(BC = 2\)
- \(CA = \sqrt{10}\)

3. **Classification of the triangle:**
- Since AB and CA have the same length (\(\sqrt{10}\)), triangle ABC is an **isosceles triangle**.
- Additionally, we can check whether it is a right triangle by applying the Pythagorean theorem:
\[
AC^2 + BC^2 \text{ vs. } AB^2
\]
\[
(\sqrt{10})^2 + 2^2 = 10 + 4 = 14 \text{ vs. } (\sqrt{10})^2 = 10
\]
Since \(14 \neq 10\), triangle ABC is not a right triangle.

Thus, the correct classification for triangle ABC is:

**d. It is an isosceles triangle.**