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 $\square$. (type an integer or a simplified fraction.)

Explanation:

Step1: Recall the slope formula

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\), where \((x_1,y_1)\) and \((x_2,y_2)\) are two points on the line.

Step2: Identify the points

For the segment through \((3,-2)\) and \((0,2)\), let \((x_1,y_1)=(3,-2)\) and \((x_2,y_2)=(0,2)\).

Step3: Substitute into the slope formula

$$
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$$

Answer:

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