Chapter 8.2, Problem 31PPS

To determine

The type of the triangle.

Answer to Problem 31PPS

Acute triangle.

Explanation of Solution

Given:

The given vertices are:

  $\begin{array}{l}\mathrm{X}\left(-7,-3\right)\\ \mathrm{Y}\left(-2,-5\right)\\ \mathrm{Z}\left(-4,-1\right)\end{array}$

Calculation:

The figure can be drawn as:

  

The length of the sides will be:

  $\begin{array}{l}\mathrm{X}\mathrm{Y}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}\\ \mathrm{X}\mathrm{Y}=\sqrt{{\left(-2-\left(-7\right)\right)}^2+{\left(-5-\left(-3\right)\right)}^2}\\ \mathrm{X}\mathrm{Y}=\sqrt{25+4}\\ \mathrm{X}\mathrm{Y}=\sqrt{29}\\ \\ \mathrm{Y}\mathrm{Z}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}\\ \mathrm{Y}\mathrm{Z}=\sqrt{{\left(-4-\left(-2\right)\right)}^2+{\left(-1-\left(5\right)\right)}^2}\\ \mathrm{Y}\mathrm{Z}=\sqrt{4+16}\\ \mathrm{Y}\mathrm{Z}=\sqrt{20}\\ \\ \mathrm{X}\mathrm{Z}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}\\ \mathrm{X}\mathrm{Z}=\sqrt{{\left(-4-\left(-6\right)\right)}^2+{\left(-1-\left(-3\right)\right)}^2}\\ \mathrm{X}\mathrm{Z}=\sqrt{9+4}\\ \mathrm{X}\mathrm{Z}=\sqrt{13}\end{array}$

According to Pythagorean Theorem,

  $\begin{array}{l}{\mathrm{c}}^2\stackrel{?}{=}{\mathrm{a}}^2+{\mathrm{b}}^2\\ {\left(\sqrt{29}\right)}^2\stackrel{?}{=}{\left(\sqrt{20}\right)}^2+{\left(\sqrt{13}\right)}^2\\ 29\stackrel{?}{=}20+13\\ 29<33\end{array}$

Hence, the triangle formed is an acute triangle.