Chapter 7.6, Problem 29E

To determine

To find: the length of the normal segment $\left(p\right)$ and angle the normal segment makes with the positive $x$ axis $\left(\varphi \right)$ by writing the equation in normal form.

Explanation of Solution

Given information:

  $\frac{x}{20}+\frac{y}{24}=1$

Calculation:

Rearrange the equation in the standard form of line $\left(Ax+By+C=0\right)$ as shown below.

  $\begin{array}{c}\frac{x}{20}+\frac{y}{24}=1\end{array}$

  $\begin{array}{c}6x+5y=120\left[\text{Multiply each side by 120}\right]\end{array}$

  $\begin{array}{c}6x+5y-120=0\left[\text{Subtract 120 from each side}\right]\end{array}$

Thus the value of $A,B$ and $C$ is $6,5$ and $-120$ respectively.

First evaluate $\sqrt{A^2+B^2}$ as shown below.

  $\begin{array}{c}\sqrt{A^2+B^2}=\sqrt{{\left(6\right)}^2+{\left(5\right)}^2}\end{array}$

  $\begin{array}{c}=\sqrt{61}\end{array}$

Since $C$ is negative, divide each term of the standard equation by $+\sqrt{A^2+B^2}$ as shown below to determine the normal form of the equation.

  $\begin{array}{c}\frac{6}{\sqrt{61}}x+\frac{5}{\sqrt{61}}y-\frac{120}{\sqrt{61}}=0\end{array}$

  $\begin{array}{c}\frac{6\sqrt{61}}{61}x+\frac{5\sqrt{61}}{61}y-\frac{120\sqrt{61}}{61}=0\left[\text{Multiply and divide the equation by }\sqrt{61}\right]\end{array}$

Compare the above equation with the general normal form equation, $x\cos \varphi +y\sin \varphi -p=0$ to get $\cos \varphi =\frac{6\sqrt{61}}{61},\sin \varphi =\frac{5\sqrt{61}}{61}\text{and }p=\frac{120\sqrt{61}}{61}$ .

Since cosine is negative and sine is positive so $\varphi$ must lie in the second quadrant. Now evaluate $\varphi$ as shown below.

  $\begin{array}{c}\tan \varphi =\frac{\sin \varphi }{\cos \varphi }\end{array}$

  $\begin{array}{c}=\frac{\frac{5\sqrt{61}}{61}}{\frac{6\sqrt{61}}{61}}\left[\text{Put the value of sine and cosine}\right]\end{array}$

  $\begin{array}{c}\tan \varphi =\frac{5}{6}\end{array}$

  $\begin{array}{c}\varphi \approx 41°\end{array}$

Thus the value of $p$ and $\varphi$ is $\frac{120\sqrt{61}}{61}$ and $41°$ respectively.