Question

Triangle ABC has vertices located at A( 0, 2), B (2, 5), and C (−1, 7).

Answer 1

Answer: See below

Step-by-step explanation:

Triangle ABC has vertices located at A( 0, 2), B (2, 5), and C (−1, 7)

A)

$\begin{aligned}&A B=\sqrt{(2-0)^{2}+(5-2)^{2}}=\sqrt{13} \\&B C=\sqrt{(-1-2)^{2}+(7-5)^{2}}=\sqrt{13} \\&C A=\sqrt{(-1-0)^{2}+(7-2)^{2}}=\sqrt{26} \\&A B=\sqrt{13}, B C=A B=\sqrt{13}, C A=\sqrt{26}\end{aligned}$

B) Slope of AB line = m₁=3/2

Slope of BC line = m₂=-2/3

Slope of CA line = m₃=-5

C) $\mathrm{AB}=\mathrm{BC} \& \mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{CA}^{2}$

Therefore, the triangle is a isosceles right angle triangle