Chapter 7, Problem 44SGA

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:

  $7x+3y-8=0$

Calculation:

The equation is in standard form, so the value of $A,B$ and $C$ is $7,3$ and $-8$ respectively. First evaluate $\sqrt{A^2+B^2}$ as shown below.

  $\begin{array}{c}\sqrt{A^2+B^2}=\sqrt{{\left(7\right)}^2+{\left(3\right)}^2}\end{array}$

  $\begin{array}{c}=\sqrt{58}\end{array}$

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.

  $\begin{array}{c}\frac{7x}{\sqrt{58}}+\frac{3y}{\sqrt{58}}-\frac{8}{\sqrt{58}}=0\end{array}$

  $\begin{array}{c}\frac{7\sqrt{58}}{58}x+\frac{3\sqrt{58}}{58}y-\frac{4\sqrt{58}}{29}=0\left[\text{Multiply and divide the equation by }\sqrt{58}\right]\end{array}$

Compare the above equation with the general normal form equation, $x\cos \varphi +y\sin \varphi -p=0$ to get $\cos \varphi =\frac{7\sqrt{58}}{58},\sin \varphi =\frac{3\sqrt{58}}{58}\text{ and }p=\frac{4\sqrt{58}}{29}$ .

Since sine and cosine are positive so $\varphi$ is in first quadrant. Now evaluate $\varphi$ as shown below.

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

  $\begin{array}{c}=\frac{\frac{3\sqrt{58}}{58}}{\frac{7\sqrt{58}}{58}}\end{array}$

  $\begin{array}{c}tan\varphi =\frac{3}{7}\end{array}$

  $\begin{array}{c}\varphi \approx 23°\end{array}$

Thus the value of $p$ and $\varphi$ is $\frac{4\sqrt{58}}{29}$ and $23°$ respectively.