Question

show that a triangle with vertices at the points $(-1,-5),(0,2)$, and $(3,-2)$ is a right triangle.
two segments are perpendicular if they lie in the same plane and meet at a $90^{circ}$ angle.
(type an integer or a simplified fraction.)
how do the slopes of perpendicular segments compare?
the product of the slopes of two perpendicular segments is $-1$
(type an integer or a simplified fraction.)
find the slope of the segment through $(-1,-5)$ and $(0,2)$.
the slope of the segment through $(-1,-5)$ and $(0,2)$ is $7$
(type an integer or a simplified fraction.)
find the slope of the segment through $(3,-2)$, and $(0,2)$.
the slope of the segment through $(3,-2)$, and $(0,2)$ is $-\frac{4}{3}$
(type an integer or a simplified fraction.)
find the slope of the segment through $(-1,-5)$ and $(3,-2)$.
the slope of the segment through $(-1,-5)$ and $(3,-2)$ is $\square$
(type an integer or a simplified fraction.)

Explanation:

Step1: Use the slope formula

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). For the points \((-1,-5)\) and \((3,-2)\), let \((x_1,y_1)=(-1,-5)\) and \((x_2,y_2)=(3,-2)\).

$$m=\frac{-2-(-5)}{3 - (-1)}$$

Step2: Simplify the expression

Simplify the numerator \(-2-(-5)=-2 + 5=3\) and the denominator \(3-(-1)=3 + 1=4\).

$$m=\frac{3}{4}$$

Answer:

\(\frac{3}{4}\)