Chapter 7, Problem 47SGA

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:

  $x-7y=-5$

Calculation:

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

  $\begin{array}{c}x-7y=-5\end{array}$

  $\begin{array}{c}x-7y+5=0\left[\text{Add }5\text{ on each side}\right]\end{array}$

Thus the value of $A,B$ and $C$ is $1,-7$ and $5$ respectively.

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

  $\begin{array}{c}\sqrt{A^2+B^2}=\sqrt{{\left(1\right)}^2+{\left(-7\right)}^2}\end{array}$

  $\begin{array}{c}=5\sqrt{2}\end{array}$

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.

  $\begin{array}{c}\frac{1}{-5\sqrt{2}}x-\frac{7}{-5\sqrt{2}}y+\frac{5}{-5\sqrt{2}}=0\end{array}$

  $\begin{array}{c}-\frac{\sqrt{2}}{10}x+\frac{7\sqrt{2}}{10}y-\frac{\sqrt{2}}{2}=0\left[\text{Multiply and divide the equation by }5\sqrt{2}\right]\end{array}$

Compare the above equation with the general normal form equation, $x\cos \varphi +y\sin \varphi -p=0$ to get $\cos \varphi =-\frac{\sqrt{2}}{10},\sin \varphi =\frac{7\sqrt{2}}{10}\text{and }p=-\frac{\sqrt{2}}{2}$ .

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

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

  $\begin{array}{c}=\frac{\frac{7\sqrt{2}}{10}}{-\frac{\sqrt{2}}{10}}\left[\text{Put the value of sine and cosine}\right]\end{array}$

  $\begin{array}{c}\tan \varphi =-7\end{array}$

  $\begin{array}{c}\varphi \approx 98°\end{array}$

Thus the value of $p$ and $\varphi$ is $\frac{\sqrt{2}}{2}$ and $98°$ respectively.