Lesson Objectives
• Learn how to convert lines between rectangular and polar form
• Learn how to graph a line on the polar grid

## Converting Lines Between Rectangular and Polar Form

### Graphs of Polar Equations

A polar equation will have r and θ as the variables vs the typical x and y that we see with a rectangular equation. Let's begin by thinking about the polar equation of a line in standard form: $$ax + by=c$$ Plug in for x and y: $$x=r \cdot \text{cos}\hspace{.1em}θ$$ $$y=r \cdot \text{sin}\hspace{.1em}θ$$ $$a(r \cdot \text{cos}\hspace{.1em}θ) + b(r \cdot \text{sin}\hspace{.1em}θ)=c$$ $$a \cdot r \cdot \text{cos}\hspace{.1em}θ + b \cdot r \cdot \text{sin}\hspace{.1em}θ=c$$ Factor out the r: $$r(a \cdot \text{cos}\hspace{.1em}θ + b \cdot \text{sin}\hspace{.1em}θ)=c$$ Solve for r: $$r=\frac{c}{a \cdot \text{cos}\hspace{.1em}θ + b \cdot \text{sin}\hspace{.1em}θ}$$ Let's look at an example.
Example #1: Find the polar equation and sketch the graph. $$y=x - 4$$ Let's write this in standard form: $$x - y=4$$ Plug into the formula given: $$r=\frac{c}{a \cdot \text{cos}\hspace{.1em}θ + b \cdot \text{sin}\hspace{.1em}θ}$$ Here, a is 1, b is -1, and c is 4: $$r=\frac{4}{1 \cdot \text{cos}\hspace{.1em}θ - 1 \cdot \text{sin}\hspace{.1em}θ}$$ $$r=\frac{4}{\text{cos}\hspace{.1em}θ - \text{sin}\hspace{.1em}θ}$$ Note: You can achieve the same equation by simply substituting r cos θ for x and r sin θ for y and solving for r. $$x - y=4$$ $$r \hspace{.1em}\text{cos}\hspace{.1em}θ - r \hspace{.1em}\text{sin}\hspace{.1em}θ=4$$ $$r(\hspace{.1em}\text{cos}\hspace{.1em}θ - \hspace{.1em}\text{sin}\hspace{.1em}θ)=4$$ $$r=\frac{4}{\hspace{.1em}\text{cos}\hspace{.1em}θ - \hspace{.1em}\text{sin}\hspace{.1em}θ}$$ To sketch the graph of a polar equation we use a different strategy based on the equation given. To sketch the graph of a line in polar form, we can create a table of values just like we did with the rectangular counterpart. We would plug in for θ and get a value for r. Alternatively, we can grab points in rectangular form and convert them over to polar form. If we think about the equation in rectangular form, notice that the y-intercept would occur at (0, -4) on the rectangular plane. If we think about this with polar coordinates, this would be (4, 270°). Additionally, our x-intercept would occur at (4, 0) on the rectangular plane. If we think about this with polar coordinates, this would be (4, 0°). Example #2: Find the polar equation and sketch the graph. $$y=-2x - 6$$ Let's write this in standard form: $$2x + y=-6$$ Plug into the formula given: $$r=\frac{c}{a \cdot \text{cos}\hspace{.1em}θ + b \cdot \text{sin}\hspace{.1em}θ}$$ Here, a is 2, b is 1, and c is -6: $$r=\frac{-6}{2 \cdot \text{cos}\hspace{.1em}θ + 1 \cdot \text{sin}\hspace{.1em}θ}$$ $$r=-\frac{6}{2 \cdot \text{cos}\hspace{.1em}θ + \text{sin}\hspace{.1em}θ}$$ To sketch the graph of our equation, let's consider the x-intercept of (-3, 0), which is (3, 180°) when written with polar coordinates. Additionally, we can consider the y-intercept of (0, -6), which is (6, 270°) when written with polar coordinates. In some cases, we will be asked to convert a polar equation to a rectangular equation. These types of problems are more difficult. Let's take a look at a few examples.
Example #3: Sketch the given equation on the polar grid and find the rectangular equation. $$θ=\frac{2π}{3}$$ A line of the form: $$θ=k$$ Is a line that passes through the origin. To graph this type of equation, no matter what the r-value, our line will coincide with that angle. $$\frac{2π}{3}=120°$$ For r-values that are positive, our line will coincide with the angle 120° on the polar grid, for r-values that are negative, we will go backwards and our line will coincide with the angle 300°: Recall that negative r-values tell us to find our given angle and then walk backwards.
Let's now convert this to rectangular form. By definition: $$\text{tan}\hspace{.1em}θ=\frac{y}{x}$$ Let's take the tangent of each side: $$\text{tan}\hspace{.1em}θ=\text{tan}\left(\frac{2π}{3}\right)$$ Let's replace tan θ with y/x: $$\frac{y}{x}=\text{tan}\left(\frac{2π}{3}\right)$$ Evaluate tan(2$π$/3): $$\text{tan}\left(\frac{2π}{3}\right)=-\sqrt{3}$$ Let's replace this in our problem. $$\frac{y}{x}=-\sqrt{3}$$ From here, we can solve for y by multiplying both sides by x: $$y=-x\sqrt{3}$$ Example #4: Sketch the given equation on the polar grid and find the rectangular equation. $$r=\text{csc}(θ + 45°)$$ Let's begin by writing cosecant in terms of sine: $$\text{csc}\hspace{.1em}θ=\frac{1}{\text{sin}\hspace{.1em}θ}$$ $$r=\frac{1}{\text{sin}(θ + 45°)}$$ Let's multiply both sides by sin(θ + 45°): $$r \cdot \text{sin}(θ + 45°)=1$$ Let's now use our sum identity for sine: $$r[\text{sin}\hspace{.1em}θ \hspace{.1em}\text{cos}\hspace{.1em}45° + \text{cos}\hspace{.1em}θ \hspace{.1em}\text{sin}\hspace{.1em}45°]=1$$ Let's replace the sin 45° and the cos 45°: $$\text{sin}(45°)=\frac{\sqrt{2}}{2}$$ $$\text{cos}(45°)=\frac{\sqrt{2}}{2}$$ $$r\left[\text{sin}\hspace{.1em}θ \hspace{.1em}\frac{\sqrt{2}}{2}+ \text{cos}\hspace{.1em}θ \hspace{.1em}\frac{\sqrt{2}}{2}\right]=1$$ Distribute the r to each term and factor out the $\frac{\sqrt{2}}{2}$: $$\frac{\sqrt{2}}{2}(r \hspace{.1em}\text{sin}\hspace{.1em}θ + r \hspace{.1em}\text{cos}\hspace{.1em}θ)=1$$ Replace r cos θ with x and r sin θ with y: $$\frac{\sqrt{2}}{2}(y + x)=1$$ Multiply both sides by $\frac{2}{\sqrt{2}}$: $$y + x=\frac{2}{\sqrt{2}}$$ Rationalize the Denominator: $$y + x=\frac{2}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}$$ $$y + x=\frac{2\sqrt{2}}{2}$$ $$y + x=\sqrt{2}$$ We can switch around x and y to write this line in standard form: $$x + y=\sqrt{2}$$ We can also write this in slope-intercept form by subtracting x away from each side: $$y=-x + \sqrt{2}$$ To sketch the graph of this equation on the polar grid, we could return to the equation and create a table of values. We can choose a value for θ and find the associated r-value. We only need to find two points to graph our line. An easier method is to look at the equation in rectangular form and grab the intercepts if possible. These can then be changed over to polar form, plotted, and then our line can be graphed:
y-intercept: $(0, \sqrt{2})$
x-intercept: $(\sqrt{2}, 0)$
If we convert these two points over to polar form, we get our two points as: $$(\sqrt{2}, 90°)$$ $$(\sqrt{2}, 0°)$$ We can always grab a third point as a check, but this isn't necessary. Let's look at the equation in polar form: $$r=\frac{1}{\text{sin}(θ + 45°)}$$ From inspecting our equation, think about a θ value of 105°. This would be added to 45° and produce 150°. We know the sine of 150° is 1/2 and dividing 1 by 1/2 will give us 2. $$(2, 105°)$$ Let's sketch our graph: ### Horizontal and Vertical Lines in Polar Form

We will also come across horizontal and vertical lines in polar form.
Let's begin with a horizontal line. $$y=b$$ Replace y with r sin θ: $$r \hspace{.1em}\text{sin}\hspace{.1em}θ=b$$ Solve for r: $$r=\frac{b}{\text{sin}\hspace{.1em}θ}$$ Let's now think about a vertical line. $$x=a$$ Replace x with r cos θ: $$r \hspace{.1em}\text{cos}\hspace{.1em}θ=a$$ Solve for r: $$r=\frac{a}{\text{cos}\hspace{.1em}θ}$$ Let's look at a few examples.
Example #5: Find the polar equation and sketch the graph. $$y=3$$ When we see that we have a horizontal line, we can use our formula or just plug in r sin θ for y. Either way, we will obtain: $$r=\frac{3}{\text{sin}\hspace{.1em}θ}$$ How can we graph this equation? We know the y-intercept would occur at (0, 3) on the rectangular plane. On the polar plane, this would be (3, 90°). We can plot this point and then draw a horizontal line. Example #6: Find the polar equation and sketch the graph. $$x=-3$$ When we see that we have a vertical line, we can use our formula or just plug in r cos θ for x. Either way, we will obtain: $$r=-\frac{3}{\text{cos}\hspace{.1em}θ}$$ How can we graph this equation? We know the x-intercept would occur at (-3, 0) on the rectangular plane. On the polar plane, this would be (3, 180°). We can plot this point and then draw a vertical line. #### Skills Check:

Example #1

Convert each to polar form.

$$y=\frac{x}{3}$$

A
$$θ=\frac{π}{6}$$
B
$$\text{csc}\hspace{.1em}θ=3$$
C
$$\text{cot}\hspace{.1em}θ=1$$
D
$$θ=\frac{π}{4}$$
E
$$\text{cot}\hspace{.1em}θ=3$$

Example #2

Convert each to rectangular form. $$r=-4\hspace{.1em}\text{sec}\left(θ + \frac{π}{4}\right)$$

A
$$y=-3$$
B
$$y=-\frac{x\sqrt{3}}{3}$$
C
$$y=-x\sqrt{3}- 4$$
D
$$y=-\sqrt{x}- 3$$
E
$$y=x + 4\sqrt{2}$$

Example #3

Convert each to rectangular form. $$θ=\frac{π}{3}$$

A
$$y=x\sqrt{3}$$
B
$$y=3x$$
C
$$y=x$$
D
$$y=-\sqrt{x}$$
E
$$y=\frac{x\sqrt{3}}{3}$$         