Chapter 7.6, Problem 22E

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:

  $x+y=1$

Calculation:

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

  $\begin{array}{c}x+y=1\end{array}$

  $\begin{array}{c}x+y-1=0\left[\text{Subtract 1 from each side}\right]\end{array}$

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

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

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

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

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.

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

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

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

Since cosine and sine both are negative so $\varphi$ must lie in the first quadrant. Now evaluate $\varphi$ as shown below.

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

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

  $\begin{array}{c}\tan \varphi =1\end{array}$

  $\begin{array}{c}\varphi \approx 45°\end{array}$

Thus the value of $p$ and $\varphi$ is $\frac{\sqrt{2}}{2}$ and $45°$ respectively.