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.
-
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: \[ CA = \sqrt{(3 - 4)^2 + (-2 + 5)^2} = \sqrt{(-1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
-
-
Summarizing the side lengths:
- \(AB = \sqrt{10}\)
- \(BC = 2\)
- \(CA = \sqrt{10}\)
-
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.
1 answer