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

Explanation:

Step1: Calculate the slope between \((-1,-5)\) and \((0,2)\)

Use the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Here \(x_1=-1,y_1 = - 5,x_2=0,y_2=2\).
\(m_1=\frac{2-(-5)}{0 - (-1)}=\frac{2 + 5}{0+1}=\frac{7}{1}=7\)

Step2: Calculate the slope between \((0,2)\) and \((3,-2)\)

Use the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Here \(x_1 = 0,y_1=2,x_2=3,y_2=-2\).
\(m_2=\frac{-2 - 2}{3-0}=\frac{-4}{3}\)

Step3: Calculate the slope between \((-1,-5)\) and \((3,-2)\)

Use the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Here \(x_1=-1,y_1=-5,x_2=3,y_2=-2\).
\(m_3=\frac{-2-(-5)}{3-(-1)}=\frac{-2 + 5}{3 + 1}=\frac{3}{4}\)

Step4: Check the product of slopes

We know that if two lines are perpendicular, the product of their slopes is \(-1\).
Check \(m_1\times m_2=7\times(-\frac{4}{3})=-\frac{28}{3}
eq - 1\)
Check \(m_2\times m_3=(-\frac{4}{3})\times\frac{3}{4}=-1\)

Answer:

The slope of the segment through \((-1,-5)\) and \((0,2)\) is \(7\). Since the product of the slopes of the line segments joining \((0,2)\) to \((3,-2)\) and \((3,-2)\) to \((-1,-5)\) is \(-1\), the triangle with vertices \((-1,-5)\), \((0,2)\) and \((3,-2)\) is a right - triangle.