Chapter 9.4, Problem 18E

To determine

To write each equation in rectangular form

Answer to Problem 18E

The rectangular form of $6=r\cos \left[\theta -120°\right]$ is $-x+\sqrt{3}y-12=0$

Explanation of Solution

Given information

The given polar form of equation is $6=r\cos \left[\theta -120°\right]$

Formula used: $p=r\cos \left[\theta -\phi \right]$ and $0=x+y-p$

Calculation:

The polar form of the equation is $6=r\cos \left[\theta -120°\right]$

Now,

  $\begin{array}{l}6=r\cos \left(\theta -120°\right)\\ 6=r\left(\cos \theta \cos 120°+\sin \theta \sin 120°\right)\\ 6=r\left(-\frac{1}{2}\cos \theta +\frac{\sqrt{3}}{2}\sin \theta \right)\\ 6=-\frac{1}{2}r\cos \theta +\frac{\sqrt{3}}{2}r\sin \theta \left[\text{polar to rectangular }r\cos \theta =x,r\sin \theta =y\right]\\ 6=-\frac{1}{2}x+\frac{\sqrt{3}}{2}y\\ 12=-x+\sqrt{3}y\\ 0=-x+\sqrt{3}y-12\\ -x+\sqrt{3}y-12=0\end{array}$

Hence,

The rectangular form of $6=r\cos \left[\theta -120°\right]$ is $-x+\sqrt{3}y-12=0$