Chapter 7.6, Problem 9CFU

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:

  $y=-3x+2$

Calculation:

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

  $\begin{array}{c}y=-3x+2\\ 3x+y-2=0\left[\text{Add }\left(3x-2\right)\text{ on each side}\right]\end{array}$

Thus the value of $A,B$ and $C$ is $3,1$ and $-2$ respectively.

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

  $\begin{array}{c}\sqrt{A^2+B^2}=\sqrt{{\left(3\right)}^2+{\left(1\right)}^2}\\ =\sqrt{9+1}\\ =\sqrt{10}\end{array}$

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.

  $\begin{array}{c}\frac{3}{\sqrt{10}}x+\frac{1}{\sqrt{10}}y-\frac{2}{\sqrt{10}}=0\\ 3\sqrt{10}x+\sqrt{10}y-2\sqrt{10}=0\left[\text{Multiply and divide the equation by }\sqrt{10}\right]\end{array}$

Compare the above equation with the general normal form equation, $x\cos \varphi +y\sin \varphi -p=0$ to get $\cos \varphi =3\sqrt{10},\sin \varphi =\sqrt{10}\text{ and }p=2\sqrt{10}$ .

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

  $\begin{array}{c}\tan \varphi =\frac{\sin \varphi }{\cos \varphi }\\ =\frac{\sqrt{10}}{3\sqrt{10}}\left[\text{Put the value of sine and cosine}\right]\\ \tan \varphi =\frac{1}{3}\\ \varphi \approx 198°\end{array}$

Thus the value of $p$ and $\varphi$ is $2\sqrt{10}$ and $198°$ respectively.