Convert Points Between Polar and Rectangular Coordinates: Study with Video Lessons, Practice Problems & Examples

To convert polar coordinates (r, θ) to rectangular coordinates (x, y), use the equations x = r × cos(θ) and y = r × sin(θ). For example, for polar coordinates (5, π/3), x = 5 × cos(π/3) = 5/2 and y = 5 × sin(π/3) = 5√3/2. Conversely, to convert rectangular coordinates (x, y) to polar coordinates, calculate r using r = √(x² + y²) and θ using θ = tan⁻¹(y/x), adjusting for the quadrant. This process emphasizes the relationship between these coordinate systems through trigonometric functions and the Pythagorean theorem.

Convert Points from Polar to Rectangular

Hey, everyone. Now that we're familiar with both polar and rectangular coordinates, we're going to have to convert between the two and take a point given in polar coordinates as $r\theta$ converting it into its rectangular $xy$ equivalent. This all just comes back down to triangles, something that you are already an expert on. So here, I'm going to break down for you exactly how to take a point given in polar coordinates and convert it into rectangular coordinates just using this triangle here. Let's go ahead and get started. Now, looking at the point that we have on our graph here given in polar coordinates $5\pi \frac{3}{4}$, we can go ahead and form a triangle with this, giving our triangle a hypotenuse of 5 and an inner angle $\theta$ of $\frac{\pi }{3}$. Then our side lengths here are $x$ and $y$ the same way that we've seen before on the unit circle. But here, our hypotenuse is no longer just 1 the same way that it was there. Now we can dive a bit deeper here because if I set up a cosine equation here, I would see that the cosine of $\theta$, of course, is equal to the adjacent side over the hypotenuse. And specifically for this triangle, that would tell me that the cosine of that inner angle $\frac{\pi }{3}$ same thing for a sine function. Setting up my sine function, sine of $\theta$, is equal to the opposite side, which in this case is $y$, over that hypotenuse value of 5. Now I can go ahead and solve each of these for $x$ and $y$. So looking at my cosine equation here, if I go ahead and multiply both sides by 5 to cancel that out on the right side, I end up getting that 5 times the cosine of $\frac{\pi }{3}$ is equal to $x$. Then doing the same thing for my sine equation, again multiplying both sides by 5 here, having that cancel on the right side, I see that 5 times the sine of $\frac{\pi }{3}$ is equal to $y$. Now if I actually multiply these out, I see that I get an $x$ value of $\frac{5}{2}$ and a $y$ value of $\frac{5\sqrt{3}}{2}$. So looking at the point that I started with, $5\pi \frac{3}{4}$ in polar coordinates, I now have my point in rectangular coordinates as $\frac{5}{2},\frac{5\sqrt{3}}{2}$. Now this will work for any point given in polar coordinates when converting to rectangular. My $x$ value is always going to be equal to $r$ times the cosine of $\theta$ and $y$ will always be equal to $r$ times the sine of $\theta$. This probably looks really familiar to you because from the unit circle, we saw that $x$ was equal to the cosine of $\theta$ and $y$ was equal to the sine of $\theta$. Now we just have this $r$ value to account for, but we're still doing the same exact thing, just that $r$ is no longer always equal to 1. But now that we know how to find these rectangular points given a polar coordinate, let's go ahead and get some practice here and work through some examples together. Here, we're given a point in polar coordinates negative $3\pi \frac{6}{1}$. Now here, we want to go ahead and plot this point on our polar coordinate system and then convert it into rectangular coordinates. So let's go ahead and start by plotting this on our polar coordinate system. Negative 3 $\pi \frac{6}{1}$. I locate my angle $\theta$ first $\frac{\pi }{6}$ along this line. But since I have a negative $r$ value of 3, I'm going to count in the opposite direction, But since I have a negative $r$ value of 3, I'm going to count in the opposite direction 3 units, and I end up with my point right here. Now that I have that point plotted, let's go ahead and convert this into rectangular coordinates. Now I know that my $x$ value is going to be equal to $r$ times the cosine of $\theta$. So $x$ equals $r\times \cos (\theta )$. Here, plugging in my values for $r$ and $\theta$, I get negative three times the cosine of $\frac{\pi }{6}$. Now the cosine of $\frac{\pi }{6}$ is $\frac{\sqrt{3}}{2}$. So this is negative three times $\frac{\sqrt{3}}{2}$. Now I have my $x$ value. Let's find our $y$ value. $y$ is going to, of course, be equal to $r\times \sin (\theta )$. So when we plug these values in here, that ends up giving me a negative three times the sine of that angle $\frac{\pi }{6}$. Now the sine of $\frac{\pi }{6}$ is $\frac{1}{2}$, so this is equal to negative three times $\frac{1}{2}$, which I can rewrite as being negative $\frac{3}{2}$. So now I have my $x$ and $y$ values, negative $\frac{3\sqrt{3}}{2}$ and negative $\frac{3}{2}$. This is my point in rectangular coordinates. Now in this problem, we're also asked to go ahead and plot this on the $xy$ plane, which is actually going to be easier to do if we have these in their decimal form. So this negative three $\sqrt{3}\frac{2}{1}$ is going to be about negative 2.6, and negative $\frac{3}{2}$ is, of course, just negative 1.5. So I can go ahead and use those decimals to plot that on my rectangular coordinate system. Now actually locating that $x$ value negative 2.6 and then going down to my $y$ value of negative 1.5, I end up right here in quadrant 3 of my rectangular coordinate system. Now here we see that these points are located in the exact same spot, which makes sense because these are just the exact same point but represented in different ways, 1 in polar and one in rectangular. Now let's look at one final example here. Here we have the point given in polar coordinates, 0 negative $\pi \frac{6}{1}$. Now again, here we want to first plot this on our polar coordinate system. So locating my angle negative $\pi \frac{6}{1}$, which will end up being right along this line, Since I know that my $r$ value is 0, I actually just stay right here at the pole, what we think of as being the origin in rectangular coordinates. So here's my point $b$. Now that we have that plotted on that polar coordinate system, let's go ahead and calculate our $x$ and $y$ values and get this point in rectangular coordinates. Now I know that $x$ is going to be equal to $r\times \cos (\theta )$, which here my $r$ value is 0. So this is 0 times $r\times \cos (\theta )$. But 0 times anything is just 0, so this ends up giving me an $x$ value of 0. Then for $y$, if I also do the same thing here, $r\times \sin (\theta )$, plugging in my values of $r$ and $\theta$, I get 0 times $r\times \sin (\theta )$, which, of course, again, $0\times \text{anything}$ is still just 0. So I again get a $y$ value of 0. Then that gives me my point in rectangular coordinates $0,0$, which we know is located at the origin, which I can go ahead and plot right here. So anytime we have an $r$ value of $0$, that's always going to end up being the point 0 in rectangular coordinates. And we again see that this is located in the exact same spot as it was on our polar coordinate system. Again, these are the same exact point represented in different ways. Now that we know how to convert points from polar to rectangular, let's continue practicing. Thanks for watching and I'll see you in the next one.

Convert the point to rectangular coordinates.
$(-2,-\frac{\pi}{4})$

• A. $(\sqrt{2},-\sqrt2)$
• B. $(-1,1)$
• C. $(-\sqrt{2},\sqrt{2})$
• D. $(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$

Answer: (C)

Convert the point to rectangular coordinates.
$(4,\frac{\pi}{6})$

• A. $(2\sqrt{3},2)$
• B. $(4\sqrt{3},4)$
• C. $(2,2\sqrt{3})$
• D. $(2,\sqrt{3})$

Answer: (A)

Convert the point to rectangular coordinates.
$(-3,0)$

• A. $(-3,0)$
• B. $(0,-3)$
• C. $(0,0)$
• D. $(3,0)$

Answer: (A)

Convert the point to rectangular coordinates.
$(0,\frac{7\pi}{4})$

• A. $(0,\frac74)$
• B. $(-4,0)$
• C. $(0,0)$
• D. $(0,7)$

Answer: (C)

Convert Points from Rectangular to Polar

Hey, everyone. We just learned how to take points given in polar coordinates and convert them to rectangular coordinates, but we also need to be able to do the opposite. Take points given in rectangular coordinates and convert them to polar coordinates. Now, looking at this point given in rectangular coordinates here, we see this sort of suspiciously familiar-looking triangle that we've solved time and time again. And that's all this is. It all just comes back down to triangles. So here, we're going to keep using everything that we already know, specifically working with the Pythagorean theorem and the tangent function in order to convert points from their rectangular coordinates into polar coordinates. Now here, I'm going to walk you through how to do that step by step. So let's go ahead and get started.

Now, looking at the point that we have here in rectangular coordinates, 3, 4, we're able to form this triangle. And from this rectangular coordinate point, 3, 4, I know that this side length x is equal to 3 and my side length y is equal to 4. Since I already know x and y, it's just left to find r and theta, which will give me my point in polar coordinates.

Now let's start by finding r. Now, looking at my r value here, this is the hypotenuse of my triangle, but I have my other two side lengths already. So I can find that hypotenuse by simply using the Pythagorean theorem, which tells me that c2=a2+b2, or specifically for this triangle, r2=x2+y2. Now actually plugging in our values here, I get that r2=32+42. Now adding those together gives me 25, so I have r2=25, or completely solving that for r, simply r is equal to 5.

Now that I have that r value, let's turn to finding theta. Now, looking at theta here, I know that I have my opposite side and my adjacent side, which means that I can set up a tangent function because I know that the tangent of theta is equal to the opposite over the adjacent side. Or specifically for this triangle, for a point given in rectangular coordinates, the tangent of my angle theta is equal to y over x. Now, again, just plugging in the values that I have here, I end up getting that the tangent of theta is equal to that y value of 4 over that x value of 3. Now, in order to actually solve for theta here, I need to take the inverse tangent. So here, theta will be equal to the inverse tangent of 4 over 3. Now, if we actually plug that inverse tangent into our calculator, we end up getting that theta is equal to about 53 degrees, and I now have both my r and my theta value. Now, I originally started with my point in rectangular coordinates, 3, 4, and now I have my point in polar coordinates, 5, 53 degrees.

Now these are the general equations that we are going to use to convert points from rectangular to polar coordinates, but we have to be really careful here because we end up taking an inverse tangent. Remember that this function is only defined over the intervals contained in quadrant 1 and quadrant 4. And even though our point here was located in quadrant 1, it won't always be. So I actually have a step-by-step process for you that will work no matter where your point is located. So let's go ahead and work through these examples together.

Now, first, we're given this point here, -4, 0, and I want to go ahead and plot this point and then convert it into polar coordinates by following these steps here. Now starting with step 1, we just want to go ahead and plot this point on our rectangular coordinate system. Now, locating this point, -4, 0, I end up right here on my x-axis for this first point a. Now with step 1 done, we move on to step 2 in calculating r. Now this is the same equation that we saw above, but just already solved for r, having taken the square root on both sides. So here, plugging in my values, I get that r is equal to the square root of (-4)^2 + 0^2. Now this ends up being the square root of 16, which is simply equal to 4. So I have an r value here of 4. Now with r calculated, we move on to finding theta. Now, looking at the location of our point, and looking at where this point a is on my x-axis, because it's on an axis, I know that theta is going to be one of my quadrantal angles. And thinking specifically about where this point is located, thinking about my angles here going from 0 to pi radians or 180 degrees, I know that my angle here is simply pi. So that gives me my value of theta pi, and I now have my point in polar coordinates, 4, π.

Now let's move on to our next example here. We are specifically given the point -1, 3. Now we're going to restart our steps here, starting from step number 1, where we're going to start out by plotting our point. Now plotting my point here is going to be more useful to know that the square root of 3 is about 1.73 as a decimal. So plotting this point at -1, 1.73 ends me up right here in quadrant 2 for point b. Now with that point plotted, we want to go ahead and find r as we have before, taking the square root of x squared + y squared. Now plugging these values in here, I end up getting the square root of (-1)^2 + (3)^2. Now actually adding those together gives me the square root of 4 or simply 2. So I have my r value of 2.

Now all that's left is to find theta. Again, we're looking at the location of our point, and I see that my point b is located in quadrant 2. So it's not on an axis. It's not in quadrant 1 or 4 as my original example was, but it is in quadrant 2. So what do we do now? Well, we're going to start off the same way that we did for our original angle that was in quadrant 1 by taking the inverse tangent of y over x. So let's start there. Theta is equal to the inverse tangent of my y value 3 over my x value negative 1. Now this simplifies to the inverse tangent of the negative square root of 3, which from my knowledge of the unit circle, or by simply typing this into a calculator, I will end up getting a negative π/3 for that inverse tangent of negative root 3. But we're not done yet because let's think about where this point is located or where this angle is located. The negative square root of 3 is here in quadrant 4, and that is not where my point b is located, so that wouldn't make any sense. Now, in order to make sure that this is located in the right place, because our point is located here in quadrant 2, we need to go ahead and add pi to that angle that we got from the inverse tangent. So this negative π/3, I need to add π to it. This ends up giving me 2π/3, which thinking about where that angle is located, that is located in quadrant 2 as it should be. So here, I have my final value of theta, 2π/3, and I have this point now in polar coordinates, 2, 2π/3.

Now, when converting points from rectangular to polar coordinates, we're going to start out the same way. We're going to plot our point on our graph, and we're going to find r by using this equation here. Then when finding theta, we need to pay attention to the location of our point, and then we're good to go. Now that we know how to convert points from rectangular to polar, let's continue practicing. Thanks for watching, and I'll see you in the next one.

Convert Points from Rectangular to Polar Example 1

Here, we're asked to convert the point given in rectangular coordinates into polar coordinates, and the point that we're given here is (3, -3). So let's go ahead and get started with our steps so that we can get this to polar coordinates. Now, step 1 tells us to go ahead and plot our point. So here on my rectangular coordinate system, I'm going to plot (3, -3), and I end up right down here in quadrant 4.

Now moving on to step 2, and actually finding our r value by taking the square root of x squared plus y squared. Plugging in my values here, 3 and -3, this gives me that $r$ is equal to the square root of $3^2+(-3{)}^2$. Now $3^2$ and $(-3{)}^2$ are both 9, so this gives me the square root of 9+9 or the square root of 18. Now we can simplify this further using our radical rules to get that $r$ is equal to $3\sqrt{2}$.

Now, let's go ahead and move on to step number 3 in finding $\theta$. Now here, we want to pay attention to where our point is located, and it is located in quadrant 4, so that tells me that I want to find $\theta$ by taking the inverse tangent of $\frac{y}{x}$. Setting that up and plugging my values in, I get that $\theta$ will be equal to the inverse tangent of my y value -3 over my x value 3. This simplifies to give me the inverse tangent of -1. If I plug this into my calculator, $\theta$ value is $-\frac{\pi }{4}$. So this is my final answer here, $3\sqrt{2},-\frac{\pi }{4}$.

Something that you might be thinking about here is if you chose to calculate your inverse tangent of $\frac{y}{x}$ differently and perhaps not using a calculator, but rather the unit circle, you may have thought that this angle is also $7\frac{\pi }{4}$. That's true. If I represented this point as $3\sqrt{2},7\frac{\pi }{4}$, that would still be correct. The reason that we got $-\frac{\pi }{4}$ is because of the restrictions placed on our inverse tangent function. But remember that there are always multiple ways to represent the same point in polar coordinates. So if you were to have $3\sqrt{2},7\frac{\pi }{4}$, that would still be correct. Even if you tried to take that point back to rectangular coordinates, you would still end up with this point, (3, -3). So let me know if you have any questions. Thanks for watching, and I'll see you in the next one.

Convert the point to polar coordinates.
$(0,5)$

• A. $(0,0)$
• B. $(5,0)$
• C. $(-5,\frac{\pi}{2})$
• D. $(5,\frac{\pi}{2})$

Answer: (D)

Convert the point to polar coordinates.
$(-2,2)$

• A. $(2\sqrt{2},\frac{3\pi}{4})$
• B. $(2\sqrt{2},-\frac{\pi}{4})$
• C. $(2\sqrt{2},\frac{\pi}{4})$
• D. $(-2\sqrt{2},\frac{3\pi}{4})$

Answer: (A)

Convert the point to polar coordinates.
$(1,1)$

• A. $(1,\frac{\pi}{4})$
• B. $(\sqrt{2},\frac{\pi}{4})$
• C. $(\sqrt{2},\frac{5\pi}{4})$
• D. $(\sqrt{2},-\frac{\pi}{4})$

Answer: (B)

Convert the point to polar coordinates.
$(-1,-\sqrt{3})$

• A. $(2,\frac{\pi}{3})$
• B. $(2,\frac{7\pi}{6})$
• C. $(2,\frac{4\pi}{3})$
• D. $(-2,\frac{4\pi}{3})$

Answer: (C)

Here’s what students ask on this topic:

How do you convert polar coordinates to rectangular coordinates?

To convert polar coordinates (r, θ) to rectangular coordinates (x, y), use the following equations:

$x=r\times \cos \left(\theta \right)$

$y=r\times \sin \left(\theta \right)$

For example, if you have polar coordinates (5, π/3), you can find the rectangular coordinates as follows:

$x=5\times \cos \left(\frac{\pi }{3}\right)=\frac{5}{2}$

$y=5\times \sin \left(\frac{\pi }{3}\right)=\text{mfrac}$

How do you convert rectangular coordinates to polar coordinates?

To convert rectangular coordinates (x, y) to polar coordinates (r, θ), follow these steps:

1. Calculate r using the Pythagorean theorem:

$r=\sqrt{x^2+y^2}$

2. Calculate θ using the tangent function:

$\theta =\tan ⁻¹\left(\frac{y}{x}\right)$

Adjust θ based on the quadrant where the point (x, y) is located. For example, for the point (3, 4):

$r=\sqrt{3{+}^4}=5$

$\theta =\tan ⁻¹\left(\frac{4}{3}\right)\approx 53°$

What are the equations for converting polar coordinates to rectangular coordinates?

The equations for converting polar coordinates (r, θ) to rectangular coordinates (x, y) are:

$x=r\times \cos \left(\theta \right)$

$y=r\times \sin \left(\theta \right)$

These equations use the trigonometric functions cosine and sine to find the x and y coordinates from the given r and θ values.

What are the equations for converting rectangular coordinates to polar coordinates?

The equations for converting rectangular coordinates (x, y) to polar coordinates (r, θ) are:

$r=\sqrt{x^2+y^2}$

$\theta =\tan ⁻¹\left(\frac{y}{x}\right)$

These equations use the Pythagorean theorem to find r and the inverse tangent function to find θ. Adjust θ based on the quadrant where the point (x, y) is located.

How do you find the angle θ when converting from rectangular to polar coordinates?

To find the angle θ when converting from rectangular coordinates (x, y) to polar coordinates, use the equation:

$\theta =\tan ⁻¹\left(\frac{y}{x}\right)$

After calculating θ, adjust it based on the quadrant where the point (x, y) is located:

• Quadrant I: θ is as calculated.
• Quadrant II: Add π to θ.
• Quadrant III: Add π to θ.
• Quadrant IV: Add 2π to θ if θ is negative.

This ensures θ is correctly positioned in the polar coordinate system.