Q. 130 - Precalculus Enhanced with Graphing Utilities

Question

If $\theta ,0<\theta <\mathrm{\pi }$, is the angle between the positive x-axis and a nonhorizontal, nonvertical line L, show that the slope m of L equals $\tan \theta$. The angle $\theta$ is called the inclination of L.

[Hint: See the illustration, where we have drawn the line M parallel to L and passing through the origin. Use the fact that M intersects the unit circle at the point $(\cos \theta , \sin \theta )$.]

Step-by-Step Solution

Answer

The slope of line M is $m=\tan \theta$ and line I and M are parallel so the slope of the line L is $m=\tan \theta$

Step 1. Given information

The given figure is 

Range of $\theta$ is$0<\theta <\mathrm{\pi }$

Step 2. The slope of line Line

Line M passes through origin and $\left(\cos \theta , \sin \theta \right)$

so$$\begin{gathered} \left(x_1,y_1\right)=(0,0) \\ \left(x_2,y_2\right)=\left(\cos \theta , \sin \theta \right) \end{gathered}$$

The slope of the line M 

$$\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{\sin \theta -0}{\cos \theta -0} \\ m=\tan \theta \end{gathered}$$

Line L and M are parallel so their slope will be the same

So the slope of line L is $m=\tan \theta$