Chapter 11.4, Problem 71E

To determine

To find the length of the sides of given triangle.

Answer to Problem 71E

  $AB=6,BC=5,AC=5$

Explanation of Solution

Given:

A vertices of triangle are $A(-1,3),B(-1,-3)$ and $C(3,0)$

Calculation:

Let us calculate the distances using distance formula-

The distance AB is

  $\begin{array}{l}AB=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}\\ AB=\sqrt{{(-1+1)}^2+{(3+3)}^2}\\ AB=\sqrt{36}\\ AB=6\end{array}$

The distance BC is

  $\begin{array}{l}BC=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}\\ BC=\sqrt{{(-1-3)}^2+{(-3-0)}^2}\\ BC=\sqrt{25}\\ BC=5\end{array}$

The distance AC is

  $\begin{array}{l}AC=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}\\ AC=\sqrt{{(3+1)}^2+{(0-3)}^2}\\ AC=\sqrt{25}\\ AC=5\end{array}$