Q. 13 - Precalculus Enhanced with Graphing Utilities

Question

In problems $11-14, \left(a\right)$ find the slope of the line and $\left(b\right)$intercept the slope. 

Step-by-Step Solution

Answer

(a) The slope of the line is $m=-\frac{1}{3}$.

(b) The line makes an angle $\mathrm{\pi }-{\tan }^{-1}\left(\frac{1}{3}\right)$ with the positive $x-$axis.

Step 1. Given information.

We have the graph given below:

Part (a). Step 1. Find slope.

Two are are given in given graph:  

$\left(-2,2\right),\left(1,1\right)$

Slope between two points: $\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$$\begin{gathered} \left(x_1,\hspace{0.17em}{y}_1\right)=\left(-2,\hspace{0.17em}2\right),\hspace{0.17em}\left(x_2,\hspace{0.17em}{y}_2\right)=\left(1,\hspace{0.17em}1\right) \\ m=\frac{1-2}{1-\left(-2\right)} \\ m=-\frac{1}{3} \end{gathered}$$

Part (b). Step 1. Intercept the slope.

Slope of the line is given by:  

$$\begin{gathered} m=\tan \theta \\ -\frac{1}{3}=\tan \theta \\ \theta ={\tan }^{-1}\left(-\frac{1}{3}\right) \\ \theta =\pi -{\tan }^{-1}\left(\frac{1}{3}\right) \end{gathered}$$

Therefore, the line makes an angle $\mathrm{\pi }-{\tan }^{-1}\left(\frac{1}{3}\right)$ with the positive $x-$axis.