Chapter 7, Problem 45SGA

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:

  $6x=4y-5$

Calculation:

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

  $\begin{array}{c}6x=4y-5\end{array}$

  $\begin{array}{c}6x-4y=-5\left[\text{Subtract }4y\text{ from each side}\right]\end{array}$

  $\begin{array}{c}6x-4y+5=0\left[\text{Add 5 on each side}\right]\end{array}$

Thus the value of $A,B$ and $C$ is $6,-4$ and $5$ respectively.

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

  $\begin{array}{c}\sqrt{A^2+B^2}=\sqrt{{\left(6\right)}^2+{\left(-4\right)}^2}\end{array}$

  $\begin{array}{c}=2\sqrt{13}\end{array}$

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.

  $\begin{array}{c}\frac{6}{-2\sqrt{13}}x-\frac{4}{-2\sqrt{13}}y+\frac{5}{-2\sqrt{13}}=0\end{array}$

  $\begin{array}{c}-\frac{6\sqrt{13}}{13}x+\frac{2\sqrt{13}}{13}y-\frac{5\sqrt{13}}{26}=0\left[\text{Multiply and divide the equation by }\sqrt{13}\right]\end{array}$

Compare the above equation with the general normal form equation, $x\cos \varphi +y\sin \varphi -p=0$ to get $\cos \varphi =-\frac{6\sqrt{13}}{13},\sin \varphi =\frac{2\sqrt{13}}{13}\text{ and }p=\frac{5\sqrt{13}}{26}$ .

Since sine is positive and cosine is negative so $\varphi$ belongs to second quadrant. Now evaluate $\varphi$ as shown below.

  $\begin{array}{c}\sin \varphi =\frac{2\sqrt{13}}{13}\end{array}$

  $\begin{array}{c}\varphi =180°-{\sin }^{-1}\frac{2\sqrt{13}}{13}\end{array}$

  $\begin{array}{c}=180°-34°\end{array}$

  $\begin{array}{c}\approx 146°\end{array}$

Thus the value of $p$ and $\varphi$ is $\frac{5\sqrt{13}}{26}$ and $146°$ respectively.