Question

show that a triangle with vertices at the points (-1,-5),(0,2), and (3,-2) is a right triangle.
the slope of the segment through (-1,-5) and (3,-2) is 4.
(type an integer or a simplified fraction.)
find the product of the slopes of the two segments meeting at the point (-1,-5).
the product of the slopes of the two segments meeting at the point (-1,-5) is 21/4.
(type an integer or a simplified fraction.)
find the product of the slopes of the two segments meeting at the point (0,2).
the product of the slopes of the two segments meeting at the point (0,2) is -28/3.
(type an integer or a simplified fraction.)
find the product of the slopes of the two segments meeting at the point (3,-2).
the product of the slopes of the two segments meeting at the point (3,-2) is -1.
(type an integer or a simplified fraction.)
since the product of the slopes of the two segments meeting at the point is , the triangle with vertices at the points (-1,-5),(0,2), and (3,-2) is a right triangle.
(type an integer or a simplified fraction.)

Explanation:

Step1: Calculate the slope between two points

The formula for slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\).
For points \((-1,-5)\) and \((0,2)\), \(m_1=\frac{2-(-5)}{0 - (-1)}=\frac{2 + 5}{0+1}=7\).
For points \((0,2)\) and \((3,-2)\), \(m_2=\frac{-2 - 2}{3-0}=\frac{-4}{3}\).
For points \((-1,-5)\) and \((3,-2)\), \(m_3=\frac{-2-(-5)}{3-(-1)}=\frac{-2 + 5}{3 + 1}=\frac{3}{4}\).

Step2: Check the product of slopes

We know that for two perpendicular lines with slopes \(m_a\) and \(m_b\), \(m_a\times m_b=-1\).
\(m_2\times m_3=\frac{-4}{3}\times\frac{3}{4}=-1\)

Answer:

Since the product of the slopes of the two segments meeting at the point \((3,-2)\) is \(-1\), the triangle with vertices at the points \((-1,-5)\), \((0,2)\), and \((3,-2)\) is a right - triangle.