Chapter 8.1, Problem 23E

Solution

Classify the triangle with given vertices as scalene, isosceles, or equilateral.

Isosceles.

Given information:

Given vertices are:

  $\left(0,-3\right),\left(0,3\right),\left(3,0\right)$

Formula used:

• The distance between two points, $\left(x_1,y{1}_{}\right),\text{ and }\left(x_2,y{2}_{}\right)$, is given by:

$\sqrt{{\left(x_2-x_1\right)}^2+{\left(y{2}_{}-y{1}_{}\right)}^2}$

• A triangle is classified as
1. Scalene if length of all sides are different.
2. Isosceles if length of two sides are equal.
3. Equilateral if length of all sides are equal.

Calculation:

To classify the triangle with vertices given below as scalene, isosceles, or equilateral

  $\left(0,-3\right),\left(0,3\right),\left(3,0\right)$ , First calculate the length of the sides of the triangle by using the formula of the distance between two points $\left(x_1,y{1}_{}\right),\text{ and }\left(x_2,y{2}_{}\right)$ that is

  $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y{2}_{}-y{1}_{}\right)}^2}$

The length of the side with end points $\left(0,-3\right),\text{ and }\left(0,3\right)$ is calculated as:

  $\begin{array}{c}\sqrt{{\left(0-0\right)}^2+{\left(3-\left(-3\right)\right)}^2}=\sqrt{0+6^2}\\ =\sqrt{36}\\ =6\end{array}$

Similarly, the length of the side with end points $\left(0,-3\right),\text{ and }\left(3,0\right)$ is calculated as:

  $\begin{array}{c}\sqrt{{\left(3-0\right)}^2+{\left(0-\left(-3\right)\right)}^2}=\sqrt{{\left(3\right)}^2+3^2}\\ =\sqrt{9+9}\\ =\sqrt{18}\\ =3\sqrt{2}\end{array}$

And, the length of the side with end points $\left(0,3\right),\text{ and }\left(3,0\right)$ is calculated as:

  $\begin{array}{c}\sqrt{{\left(3-0\right)}^2+{\left(0-3\right)}^2}=\sqrt{{\left(3\right)}^2+{\left(-3\right)}^2}\\ =\sqrt{9+9}\\ =\sqrt{18}\\ =3\sqrt{2}\end{array}$

Thus, the length of the sides of the triangle in given figure is as

  $6$ , $3\sqrt{2}$ , and $3\sqrt{2}$

Since the length of two sides are equal, thus the triangle with given vertices is isosceles.