Chapter 9.4, Problem 30E

To determine

To write: The polar form of the equation of the line that passes through the given rectangular coordinates

Answer to Problem 30E

The polar form of the given rectangular coordinate is $r=\frac{7}{\left(2\cos \theta +3\sin \theta \right)}$

Explanation of Solution

Given information:

The rectangular coordinate is $\left(4,-1\right)\text{ and }\left(-2,3\right)$

Calculation:

Here, $x_1=4,x_2=-2,y_1=-1\text{ and }{y}_2=3$

The slope $m$ of the line is,

  $\begin{array}{l}m=\frac{3-\left(-1\right)}{-2-4}\\ m=\frac{4}{-6}\\ m=-\frac{2}{3}\end{array}$

Use the point slope form of the equation of the line,

  $\begin{array}{l}y=m\left(x-x_1\right)+y_1\\ y=-\frac{2}{3}\left(x-4\right)-1\\ 3y=-2\left(x-4\right)-1\left[\text{multiply both sides by 3}\right]\\ 3y=-2x+8-1\\ 2x+3y=7\left[\text{add }2x\text{ to both sides}\right]\\ 2r\cos \theta +3r\sin \theta =7\left[\text{substitute }r\cos \theta =x\text{ and }r\sin \theta =y\right]\\ r\left(2\cos \theta +3\sin \theta \right)=7\left[\text{factor out }r\right]\\ r=\frac{7}{\left(2\cos \theta +3\sin \theta \right)}\left[\text{divide both side by}\left(2\cos \theta +3\sin \theta \right)\right]\end{array}$

Therefore, the polar form of the given rectangular coordinate is $r=\frac{7}{\left(2\cos \theta +3\sin \theta \right)}$