Question

Find the perimeter of the triangle whose vertices are the following specified points in the plane.

Answer 1

Let A = (1,-5), B = (2,9) and C = (-6,-8). Solve first for AB, BC, and CA.

$\begin{gathered} \text{Solve for }\overline{AB} \\ \overline{AB}=\sqrt[]{(1-2)^2+(-5-9)^2} \\ \overline{AB}=\sqrt[]{(-1)^2+(-14)^2} \\ \overline{AB}=\sqrt[]{1+196} \\ \overline{AB}=\sqrt[]{197} \\ \overline{AB}\approx14.04\text{ units} \end{gathered}$$\begin{gathered} \text{Solve for }\overline{BC} \\ \text{ }\overline{BC}=\sqrt[]{(2-(-6))^2+(9-(-8))^2} \\ \text{ }\overline{BC}=\sqrt[]{(8)^2+(17)^2} \\ \text{ }\overline{BC}=\sqrt[]{(64)^{}+(289)^{}} \\ \text{ }\overline{BC}=\sqrt[]{353^{}} \\ \overline{BC}\approx18.79\text{ units} \end{gathered}$$\begin{gathered} \text{Solve for }\overline{CA} \\ \text{ }\overline{CA}=\sqrt[]{(-6-1)^2+(-8-(-5))^2} \\ \text{ }\overline{CA}=\sqrt[]{(-7)^2+(-3)^2} \\ \text{ }\overline{CA}=\sqrt[]{49+9} \\ \text{ }\overline{CA}=\sqrt[]{58} \\ \text{ }\overline{CA}\approx7.62\text{ units} \end{gathered}$

Now solve for Perimeter

$\begin{gathered} P=\overline{AB}+\overline{BC}+\overline{CA} \\ P=14.04+18.79+7.62 \\ P=40.45\text{ units} \end{gathered}$

Therefore, the perimeter of the triangle whose points in the plane are (1,-5), (2,9) and (-6,-8) is 40.45 units.