Chapter 7.6, Problem 27E

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:

  $y-2=\frac{1}{4}\left(x+20\right)$

Calculation:

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

  $\begin{array}{c}y-2=\frac{1}{4}\left(x+20\right)\end{array}$

  $\begin{array}{c}4y-8=x+20\left[\text{Multiply each side by 4}\right]\end{array}$

  $\begin{array}{c}x-4y+28=0\left[\text{Add }\left(-4y+8\right)\text{ on each side}\right]\end{array}$

Thus the value of $A,B$ and $C$ is $1,-4$ and $28$ respectively.

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

  $\begin{array}{c}\sqrt{A^2+B^2}=\sqrt{{\left(1\right)}^2+{\left(-4\right)}^2}\end{array}$

  $\begin{array}{c}=\sqrt{17}\end{array}$

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

  $\begin{array}{c}\frac{1}{-\sqrt{17}}x-\frac{4}{-\sqrt{17}}y+\frac{28}{-\sqrt{17}}=0\end{array}$

  $\begin{array}{c}-\frac{\sqrt{17}}{17}x+\frac{4\sqrt{17}}{17}y-\frac{28\sqrt{17}}{17}=0\left[\text{Multiply and divide the equation by }\sqrt{17}\right]\end{array}$

Compare the above equation with the general normal form equation, $x\cos \varphi +y\sin \varphi -p=0$ to get $\cos \varphi =-\frac{\sqrt{17}}{17},\sin \varphi =\frac{4\sqrt{17}}{17}\text{and }p=\frac{28\sqrt{17}}{17}$ .

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

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

  $\begin{array}{c}=\frac{\frac{4\sqrt{17}}{17}}{-\frac{\sqrt{17}}{17}}\left[\text{Put the value of sine and cosine}\right]\end{array}$

  $\begin{array}{c}\tan \varphi =-4\end{array}$

  $\begin{array}{c}\varphi \approx 104°\end{array}$

Thus the value of $p$ and $\varphi$ is $\frac{28\sqrt{17}}{17}$ and $104°$ respectively.