Chapter 11.1, Problem 39E

To determine

To find :The length of the sides of the triangle with the given vertices and to determine whether the triangle is a right triangle, an isosceles triangle, or neither.

Answer to Problem 39E

The length of the sides of triangle are 6 , $\sqrt{40}$ and 6 and the triangle is isosceles.

Explanation of Solution

Given vertices: $\left(4,-1,-2\right),\left(8,1,2\right),\left(2,3,2\right)$

Formula used: The distance between the points $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ is

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$

Calculation:

Let ABC be the triangle with vertices $A\left(4,-1,-2\right),B\left(8,1,2\right),C\left(2,3,2\right)$

Then by using distance formula , length of the sides of the triangle are

  $\begin{array}{l}AB=\sqrt{{\left(8-4\right)}^2+{\left(1+1\right)}^2+{\left(2+2\right)}^2}=\sqrt{16+4+16}=6\\ BC=\sqrt{{\left(2-8\right)}^2+{\left(3-1\right)}^2+{\left(2-2\right)}^2}=\sqrt{36+4+0}=\sqrt{40}\\ CA=\sqrt{{\left(4-2\right)}^2+{\left(-1-3\right)}^2+{\left(-2-2\right)}^2}=\sqrt{4+16+16}=6\end{array}$

Since, the length of two sides are equal, therefore the triangle is isosceles triangle.