Chapter 7.2, Problem 64E

(a)

To determine

To prove: The slope $m$ of the line is given by $m=\tan \theta$ .

Explanation of Solution

Given information:

If $L$ is a line in the plane and $\theta$ is the angle formed by the line and the $x$ -axis as shown in the figure.

  

Proof:

Assume there exists two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ on the line $L$ .

Draw a line joining the points $\left(x_1,y_1\right)$ and $\left(x_2,y_1\right)$ . Now drop a perpendicular from point $\left(x_2,y_2\right)$ to that line.

The angle made by the two line joining the points $\left(x_1,y_1\right)$ , $\left(x_2,y_1\right)$ and line $L$ is also $\theta$ as the line joining the points $\left(x_1,y_1\right)$ , $\left(x_2,y_1\right)$ is parallel to $x$ -axis.

Now consider the tangent of the angle.

  $\tan \theta =\frac{y_2-y_1}{x_2-x_1}$

As the formula for the slope of line joining two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is equal to $\frac{y_2-y_1}{x_2-x_1}$ .

Hence, the slope of the line $m$ is given by $m=\tan \theta$ .

(b)

To determine

To prove: The expression $\tan \psi =\frac{m_2-m_1}{1+m_1{m}_2}$ .

Explanation of Solution

Given information:

Let $L_1$ and $L_2$ be two non parallel lines in the plane with slopes $m_1$ and $m_2$ , respectively. Let $\psi$ be the acute angle formed by the two lines.

  

Proof:

As the angle $\psi ={\theta }_2-{\theta }_1$ , use the identity $\tan \left(a-b\right)=\frac{\tan a-\tan b}{1+\tan a\tan b}$ .

Take the tangent of angle $\psi$ and simplify.

  $\begin{array}{c}\tan \left(\psi \right)=\tan \left({\theta }_2-{\theta }_1\right)\\ =\frac{\tan {\theta }_2-\tan {\theta }_1}{1+\tan \left({\theta }_1\right)\tan \left({\theta }_2\right)}\end{array}$

As the slope of the line $L_1$ is equal to $m_1$ and the angle of inclination of line $L_1$ with $x$ -axis is equal to ${\theta }_1$ . So, $m_1=\tan {\theta }_1$ .

As the slope of the line $L_2$ is equal to $m_2$ and the angle of inclination of line $L_2$ with $x$ -axis is equal to ${\theta }_2$ . So, $m_2=\tan {\theta }_2$ .

Substitute the values in the above expression.

  $\begin{array}{c}\tan \left(\psi \right)=\frac{\tan {\theta }_2-\tan {\theta }_1}{1+\tan \left({\theta }_1\right)\tan \left({\theta }_2\right)}\\ =\frac{m_2-m_1}{1+m_1{m}_2}\end{array}$

Hence, the expression $\tan \psi =\frac{m_2-m_1}{1+m_1{m}_2}$ is proved.

(c)

To determine

To find: The acute angle formed by the two lines $y=\frac{1}{3}x+1$ and $y=-\frac{1}{2}x-3$ .

Answer to Problem 64E

The acute angle formed by the two lines $y=\frac{1}{3}x+1$ and $y=-\frac{1}{2}x-3$ is $\frac{\pi }{4}$ .

Explanation of Solution

Given information:

The two lines are $y=\frac{1}{3}x+1$ and $y=-\frac{1}{2}x-3$ .

Calculation:

Compare both the equations of line with general equation $y=mx+c$ . This gives the slope of the first line $m_1$ is equal to $\frac{1}{3}$ and the slope of the second line $m_2$ is equal to $-\frac{1}{2}$ .

Substitute the slopes in the formula for angle between lines.

  $\begin{array}{l}\tan \psi =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|\\ \tan \psi =\left|\frac{-\frac{1}{2}-\frac{1}{3}}{1+\left(\frac{1}{3}\right)\left(-\frac{1}{2}\right)}\right|\\ \tan \psi =\left|\frac{\frac{-3-2}{6}}{1-\frac{1}{6}}\right|\\ \tan \psi =\left|\frac{-\frac{5}{6}}{\frac{5}{6}}\right|\end{array}$

Further simplify,

  $\begin{array}{c}\tan \psi =\left|\frac{-\frac{5}{6}}{\frac{5}{6}}\right|\\ \tan \psi =1\\ \psi ={\tan }^{-1}\left(1\right)\\ \psi =\frac{\pi }{4}\end{array}$

Therefore, the acute angle formed by the two lines $y=\frac{1}{3}x+1$ and $y=-\frac{1}{2}x-3$ is $\frac{\pi }{4}$ .

(d)

To determine

To prove: The slope of one is negative reciprocal of the slope of other, if they are perpendicular.

Explanation of Solution

Given information:

Two lines are perpendicular.

Proof:

As calculated in part(b), the angle between two lines is $\psi$ then $\tan \psi =\frac{m_2-m_1}{1+m_1{m}_2}$ .

Convert tangent into cotangent.

  $\begin{array}{l}\tan \psi =\frac{m_2-m_1}{1+m_1{m}_2}\\ \cot \psi =\frac{1+m_1{m}_2}{m_2-m_1}\end{array}$

As the angle between the two lines is $\frac{\pi }{2}$ .

Substitute $\frac{\pi }{2}$ for $\psi$ in the formula.

  $\begin{array}{c}\cot \psi =\frac{1+m_1{m}_2}{m_2-m_1}\\ \cot \frac{\pi }{2}=\frac{1+m_1{m}_2}{m_2-m_1}\\ 0=\frac{1+m_1{m}_2}{m_2-m_1}\\ 1+m_1{m}_2=0\end{array}$

Further simplify,

  $\begin{array}{c}1+m_1{m}_2=0\\ m_1{m}_2=-1\\ m_1=-\frac{1}{m_2}\end{array}$

Hence, the slope of one is negative reciprocal of the slope of other, if they are perpendicular.