Chapter 9.4, Problem 31E

To determine

To write the polar form of the equation of line.

Answer to Problem 31E

The polar form of the equation of line is $\frac{0.4}{\sqrt{1.6561}}=r\left(\theta +51\right)$

Explanation of Solution

Given information:

The line passes through points with polar coordinates $\left(3,\frac{\pi }{4}\right)\text{ and }\left(2,\frac{7\pi }{6}\right)$

Formula used:

Polar to Cartesian conversion formula.

  $x=r\cos \theta ,y=r\sin \theta$

Calculation:

First convert the polar coordinates into Cartesian coordinates.

Compare the given polar coordinates to $\left(r,\theta \right)$

  $\begin{array}{l}x=r\cos \theta \\ x=3\cos \left(\frac{\pi }{4}\right)\\ x=\frac{3}{\sqrt{2}}\\ y=r\sin \theta \\ y=3\sin \left(\frac{\pi }{4}\right)\\ y=\frac{3}{\sqrt{2}}\end{array}$

Similarly for the second coordinate.

  $\begin{array}{l}x=r\cos \theta \\ x=2\cos \left(\frac{7\pi }{6}\right)\\ x=-\sqrt{3}\\ y=r\sin \theta \\ y=2\sin \left(\frac{7\pi }{6}\right)\\ y=-1\end{array}$

The two Cartesian coordinates are $\left(\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}}\right)\text{ and }\left(-\sqrt{3},-1\right)$

Form the equation of line passing through $\left(\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}}\right)\text{ and }\left(-\sqrt{3},-1\right)$ .

  $\begin{array}{l}\left(y-y_1\right)=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)\\ \left(y-\frac{3}{\sqrt{2}}\right)=\frac{-1-\frac{3}{\sqrt{2}}}{-\sqrt{3}-\frac{3}{\sqrt{2}}}\left(x-\frac{3}{\sqrt{2}}\right)\\ \left(y-\frac{3}{\sqrt{2}}\right)=\frac{-\sqrt{2}-3}{-\sqrt{6}-3}\left(x-\frac{3}{\sqrt{2}}\right)\\ \left(y-\frac{3}{\sqrt{2}}\right)=\frac{-4.414}{-5.449}\left(x-\frac{3}{\sqrt{2}}\right)\\ \left(y-\frac{3}{\sqrt{2}}\right)=0.81\left(x-\frac{3}{\sqrt{2}}\right)\\ y-0.81x-0.4=0\\ 0.81x-y+0.4=0\end{array}$

To convert to normal form, you need to find the value of $\pm \sqrt{A^2+B^2}$

  $\begin{array}{l}\pm \sqrt{A^2+B^2}\\ =\pm \sqrt{1+{\left(0.81\right)}^2}\\ =\pm \sqrt{1.6561}\end{array}$

Since C is positive take $-\sqrt{1.6561}$

Rewrite the equation as

  $-\frac{0.81x}{\sqrt{1.6561}}+\frac{y}{\sqrt{1.6561}}-\frac{0.4}{\sqrt{1.6561}}=0$

Compare the equation to normal form that is $x\cos \phi +y\sin \phi -p=0$

Therefore,

  $p=\frac{0.4}{\sqrt{1.6561}},\cos \phi =-\frac{0.81}{\sqrt{1.6561}},\sin \phi =\frac{1}{\sqrt{1.6561}}$

Now

  $\begin{array}{l}\tan \phi =\frac{\sin \phi }{\cos \phi }\\ =\frac{\frac{1}{\sqrt{1.6561}}}{\frac{-0.81}{\sqrt{1.6561}}}\\ =\frac{-1}{0.81}\\ \phi =\arctan \left(\frac{-1}{0.81}\right)\\ \phi =-51\end{array}$

Now substitute the value of p and $\phi$ into polar form

  $\begin{array}{l}p=r\cos \left(\theta -\phi \right)\\ \frac{0.4}{\sqrt{1.6561}}=r\left(\theta -\left(-51\right)\right)\\ \frac{0.4}{\sqrt{1.6561}}=r\left(\theta +51\right)\end{array}$

The polar equation is $\frac{0.4}{\sqrt{1.6561}}=r\left(\theta +51\right)$