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 $square^{circ}$ angle.
(type an integer or a simplified fraction.)

Explanation:

Step1: Find the slopes of the sides of the triangle

Let \(A(-1,-5)\), \(B(0,2)\), \(C(3,-2)\)
The slope formula is \(m = \frac{y_2 - y_1}{x_2 - x_1}\)
Slope of \(AB\): \(m_{AB}=\frac{2 - (-5)}{0 - (-1)}=\frac{2 + 5}{0 + 1}=7\)
Slope of \(BC\): \(m_{BC}=\frac{-2 - 2}{3 - 0}=\frac{-4}{3}\)
Slope of \(AC\): \(m_{AC}=\frac{-2 - (-5)}{3 - (-1)}=\frac{-2 + 5}{3 + 1}=\frac{3}{4}\)

Step2: Check if the product of two slopes is \(- 1\)

Since \(m_{AB}\times m_{BC}=7\times(-\frac{4}{3})=-\frac{28}{3}
eq - 1\)
\(m_{AB}\times m_{AC}=7\times\frac{3}{4}=\frac{21}{4}
eq - 1\)
\(m_{BC}\times m_{AC}=(-\frac{4}{3})\times\frac{3}{4}=-1\)

Answer:

The triangle is a right - triangle because the product of the slopes of \(BC\) and \(AC\) is \(-1\), which means the two sides \(BC\) and \(AC\) are perpendicular.