Question

the vertices of triangle abc are: a(-3,4) b(-8,10) c(3,0)
complete the steps below to prove whether triangle abc is an isosceles right triangle.
find the slopes of the following:
\\( \overline { a b } = \\)
\\( \overline { b c } = \\)
\\( \overline { a c } = \\)
find the lengths of the following:
\\( \overline { a b } = \\)
\\( \overline { b c } = \\)
\\( \overline { a c } = \\)
(b) prove whether triangle abc is an isosceles right triangle. use your answers from parts a and b to support your answer.

Explanation:

Step1: Calculate the slopes

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\).
For \(\overline{AB}\) with \(A(-3,4)\) and \(B(-8,10)\):
\(m_{AB}=\frac{10 - 4}{-8+3}=\frac{6}{-5}=-\frac{6}{5}\)
For \(\overline{BC}\) with \(B(-8,10)\) and \(C(3,0)\):
\(m_{BC}=\frac{0 - 10}{3 + 8}=\frac{-10}{11}=-\frac{10}{11}\)
For \(\overline{AC}\) with \(A(-3,4)\) and \(C(3,0)\):
\(m_{AC}=\frac{0 - 4}{3+3}=\frac{-4}{6}=-\frac{2}{3}\)

Step2: Calculate the lengths

The distance formula is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
For \(\overline{AB}\):
\(d_{AB}=\sqrt{(-8 + 3)^2+(10 - 4)^2}=\sqrt{(-5)^2+6^2}=\sqrt{25 + 36}=\sqrt{61}\)
For \(\overline{BC}\):
\(d_{BC}=\sqrt{(3 + 8)^2+(0 - 10)^2}=\sqrt{121+100}=\sqrt{221}\)
For \(\overline{AC}\):
\(d_{AC}=\sqrt{(3 + 3)^2+(0 - 4)^2}=\sqrt{36 + 16}=\sqrt{52}=2\sqrt{13}\)

Answer:

Slopes: \(\overline{AB}:-\frac{6}{5}\), \(\overline{BC}:-\frac{10}{11}\), \(\overline{AC}:-\frac{2}{3}\)
Lengths: \(\overline{AB}:\sqrt{61}\), \(\overline{BC}:\sqrt{221}\), \(\overline{AC}:2\sqrt{13}\)