Chapter 10.1, Problem 43E

To determine

Tofind:The lengths of the sides of triangle with vertices $\left(1,-3,-2\right)$ , $\left(5,-1,2\right)$ and $\left(-1,1,2\right)$ , and check the triangle is a right triangle or an isosceles or neither.

Answer to Problem 43E

The length of three sides of the triangle are $6$ , $2\sqrt{10}$ and $6$ . The given triangle is an isosceles triangle.

Explanation of Solution

Given information:

The vertices of the triangle are $\left(1,-3,-2\right)$ , $\left(5,-1,2\right)$ and $\left(-1,1,2\right)$ .

Calculation:

The vertices of the right triangle are $\left(1,-3,-2\right)$ , $\left(5,-1,2\right)$ and $\left(-1,1,2\right)$ .The length of the sides of triangle is the distance between the vertices.

Use the distance formula for two points $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ .

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

The distance between the points $\left(1,-3,-2\right)$ and $\left(5,-1,2\right)$ is,

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(5-1\right)}^2+{\left(-1-\left(-3\right)\right)}^2+{\left(2-\left(-2\right)\right)}^2}\\ =\sqrt{{\left(4\right)}^2+{\left(2\right)}^2+{\left(4\right)}^2}\\ =\sqrt{16+4+16}\\ =\sqrt{36}\\ =6\end{array}$

So, the length of the first side of triangle is $6$ .

The distance between the points $\left(5,-1,2\right)$ and $\left(-1,1,2\right)$ is,

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(-1-5\right)}^2+{\left(1-\left(-1\right)\right)}^2+{\left(2-2\right)}^2}\\ =\sqrt{{\left(-6\right)}^2+{\left(2\right)}^2+{\left(0\right)}^2}\\ =\sqrt{36+4+0}\\ =\sqrt{40}\\ =2\sqrt{10}\end{array}$

So, the length of the second side triangle is $2\sqrt{10}$ .

The distance between the points $\left(1,-3,-2\right)$ and $\left(-1,1,2\right)$ is,

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(-1-1\right)}^2+{\left(1-\left(-3\right)\right)}^2+{\left(2-\left(-2\right)\right)}^2}\\ =\sqrt{{\left(-2\right)}^2+{\left(4\right)}^2+{\left(4\right)}^2}\\ =\sqrt{4+16+16}\\ =\sqrt{36}\\ =6\end{array}$

So, the length of the third side of triangle is $6$ .

Therefore, the length of three sides of triangle are $6$ , $2\sqrt{10}$ and $6$ .

The lengths of two sides of triangle are equal. Therefore, the given triangle is an isosceles triangle.