# 6.5: Rectangular to Polar Conversions

Difficulty Level: At Grade Created by: CK-12
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Practice Rectangular to Polar Conversions

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You are looking at a map of your state with a rectangular coordinate grid. It looks like this

You realize that if you convert the the coordinates of Yourtown (labelled with a "YT") to polar coordinates, you can more easily see the distance between the Capitol at the origin and Yourtown. Can you make this conversion from rectangular to polar coordinates? Keep reading, and at the end of this Concept, you'll be able to perform the conversion.

### Guidance

When converting rectangular coordinates to polar coordinates, we must remember that there are many possible polar coordinates. We will agree that when converting from rectangular coordinates to polar coordinates, one set of polar coordinates will be sufficient for each set of rectangular coordinates. Most graphing calculators are programmed to complete the conversions and they too provide one set of coordinates for each conversion. The conversion of rectangular coordinates to polar coordinates is done using the Pythagorean Theorem and the Arctangent function. The Arctangent function only calculates angles in the first and fourth quadrants so π\begin{align*}\pi\end{align*} radians must be added to the value of θ\begin{align*}\theta\end{align*} for all points with rectangular coordinates in the second and third quadrants.

In addition to these formulas, r=x2+y2\begin{align*}r = \sqrt{x^2 + y^2}\end{align*} is also used in converting rectangular coordinates to polar form.

#### Example A

Convert the following rectangular coordinates to polar form.

P(3,5)\begin{align*}P (3, -5)\end{align*}

Solution: For P(3,5) x=3\begin{align*}P (3, -5) \ x = 3\end{align*} and y=5\begin{align*}y = -5\end{align*}. The point is located in the fourth quadrant and x>0\begin{align*}x > 0\end{align*}.

\begin{align*}r &= \sqrt{x^2 + y^2} && \theta = \ \arctan \frac{y}{x}\\ r &= \sqrt{(3)^2 + (-5)^2} && \theta = \tan^{-1} \left ( -\frac{5}{3} \right )\\ r &= \sqrt{34} && \theta \approx - 1.03\\ r &\approx 5.83\end{align*}

The polar coordinates of \begin{align*}P (3, -5)\end{align*} are \begin{align*}P (5.83, -1.03)\end{align*}.

#### Example B

Convert the following rectangular coordinates to polar form.

\begin{align*}Q (-9, -12)\end{align*}

Solution: For \begin{align*}Q (-9, -12) \ x = -9\end{align*} and \begin{align*}y = -5\end{align*}. The point is located in the third quadrant and \begin{align*}x < 0\end{align*}.

\begin{align*}r & = \sqrt{x^2 + y^2} && \theta = \arctan \frac{y}{x} + \pi \\ r & = \sqrt{(-9)^2 + (-12)^2} && \theta = \tan^{-1} \left (\frac{-12}{-9} \right ) + \pi \\ r & = \sqrt{225} && \theta \approx 4.07 \\ r & = 15\end{align*}

The polar coordinates of \begin{align*}Q (-9, -12)\end{align*} are \begin{align*}Q (15, 4.07) \end{align*}

#### Example C

Convert the following rectangular coordinates to polar form.

\begin{align*}Q (2, 7)\end{align*}

Solution: For \begin{align*}Q (2, 7) \ x = 2\end{align*} and \begin{align*}y = 7\end{align*}. The point is located in the first quadrant and \begin{align*}x > 0\end{align*}.

\begin{align*}r & = \sqrt{x^2 + y^2} && \theta = \arctan \frac{y}{x} + \pi \\ r & = \sqrt{(2)^2 + (7)^2} && \theta = \tan^{-1} \left (\frac{7}{2} \right ) + \pi \\ r & = \sqrt{54} && \theta \approx 74.05 \\ r &\approx 7.35\end{align*}

The polar coordinates of \begin{align*}Q (2, 7)\end{align*} are \begin{align*}Q (7.35, 74.05) \end{align*}

### Vocabulary

Polar Coordinates: A set of polar coordinates are a set of coordinates plotted on a system that uses the distance from the origin and angle from an axis to describe location.

Rectangular Coordinates: A set of rectangular coordinates are a set of coordinates plotted on a system using basis axes at right angles to each other.

### Guided Practice

1. Write the following rectangular point in polar form: \begin{align*}A(-2, 5)\end{align*} using radians

2. Write the following rectangular point in polar form: \begin{align*}B(5, -4)\end{align*} using radians

3. Write the following rectangular point in polar form: \begin{align*}C(1, 9)\end{align*} using degrees

Solutions:

1. For \begin{align*}A (-2, 5) x = -2\end{align*} and \begin{align*}y = 5\end{align*}. The point is located in the second quadrant and \begin{align*}x < 0\end{align*}. \begin{align*}r = \sqrt{(-2)^2 + (5)^2} = \sqrt{29} \approx 5.39, \ \theta = \ \arctan \frac{5}{-2} + \pi = 1.95.\end{align*} The polar coordinates for the rectangular coordinates \begin{align*}A(-2,5)\end{align*} are \begin{align*}A(5.39,1.95)\end{align*}

2. For \begin{align*}B (5,-4) x = 5\end{align*} and \begin{align*}y = -4\end{align*}. The point is located in the fourth quadrant and \begin{align*}x > 0\end{align*}. \begin{align*}r = \sqrt{(5)^2 + (-4)^2} = \sqrt{41} \approx 6.4, \ \theta = \tan^{-1} \left (\frac{-4}{5} \right ) \approx -0.67\end{align*} The polar coordinates for the rectangular coordinates \begin{align*}B(5, -4)\end{align*} are \begin{align*}A(6.40, -0.67)\end{align*}

3. \begin{align*}C(1, 9)\end{align*} is located in the first quadrant. \begin{align*}r = \sqrt{1^2 + 9^2} = \sqrt{82} \approx 9.06, \ \theta = \tan^{-1} \frac{9}{1} \approx 83.66.\end{align*}

### Concept Problem Solution

To convert these rectangular coordinates into polar coordinates, first use the Pythagorean Theorem:

\begin{align*}r = \sqrt{(4)^2 + (2)^2} = \sqrt{20} \approx 4.47\end{align*}

and then use the tangent function to find the angle:

\begin{align*}\theta = \arctan \frac{4}{2} = 63.43^\circ\end{align*}

The polar coordinates for Yourtown are \begin{align*}YT(4.47,63.43^\circ)\end{align*}

### Practice

Write the following points, given in rectangular form, in polar form using radians where \begin{align*}0\leq \theta \leq 2\pi\end{align*}.

1. (1,3)
2. (2,5)
3. (-2,3)
4. (2,-1)
5. (3,2)
6. (4,5)
7. (-1,2)
8. (-3,3)
9. (-2,5)
10. (1,-4)
11. (5,2)
12. (1,6)

For each equation, convert the rectangular equation to polar form.

1. \begin{align*}x=5\end{align*}
2. \begin{align*}2x-3y=5\end{align*}
3. \begin{align*}2x+4y=2\end{align*}
4. \begin{align*}(x-1)^2+y^2=1\end{align*}
5. \begin{align*}(x+3)^2+(y+3)^2=18\end{align*}
6. \begin{align*}y=7\end{align*}

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Sep 26, 2012
Last Modified:
Mar 23, 2016
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