## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

We see that the first two sides haves slopes that are negative reciprocals of one another ($m = -\frac{2}{3}$ and $m = \frac{3}{2}$). Therefore, these sides are perpendicular to one another, and this triangle is a right triangle.
In a right triangle, two of the sides must be perpendicular to one another. This means that the slopes of the two sides are negative reciprocals of one another, or that the product of the slopes of two of the sides should equal $-1$. Let's figure out the slopes of each of the sides. The formula to find the slope of two points on a line is: $m = \dfrac{y_2 - y_1}{x_2 - x_2}$, where $m$ is the slope and $(x_1, y_1)$ and $(x_2, y_2)$ are two points on the line. Let us plug the points $(-2, 5)$ and $(1, 3)$ into the formula: $m = \dfrac{3 - 5}{1 - (-2)}$ Simplify numerator and denominator: $m = \dfrac{-2}{3}$ Rewrite the fraction in more conventional terms: $m = -\dfrac{2}{3}$ Let's plug the points $(1, 3)$ and $(-1, 0)$ into the formula: $m = \dfrac{0 - 3}{-1 - 1}$ Simplify numerator and denominator: $m = \dfrac{-3}{-2}$ Simplify the fraction: $m = \dfrac{3}{2}$ Let us plug the points $(-1, 0)$ and $(-2, 5)$ into the formula: $m = \dfrac{5 - 0}{-2 - (-1)}$ Simplify numerator and denominator: $m = \dfrac{5}{-1}$ Simplify the fraction: $m = -5$ We see that the slopes of the first two sides haves slopes that are negative reciprocals of one another $\left(m = -\frac{2}{3}\text{ and } m = \frac{3}{2}\right)$. Therefore, these sides are perpendicular to one another, and this triangle is a right triangle.