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

( 2 2 , 3 π 4 ) (2\sqrt{2},\frac{3\pi}{4}) ( 2 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 4 3 π )

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

In this problem, we're asked to convert our point given in rectangular coordinates into polar coordinates. And the point that we're given here is negative 22. So let's get started with our steps and go ahead and plot this point. Now plotting my point here at negative two two, I end up here in quadrant 2. So we can move on to our step 2 and calculate our r value. Now here r is going to be equal to the square root of that x value negative 2 squared plus that y value 2 squared. Now this will end up giving me the square root of 4+4orthesquarerootof8. Now here we can use our radical rules in order to simplify this further to get that this is equal to 2 root 2 for our r value. Now with r calculated, we can go ahead and move on to calculating theta. Now here, we want to pay attention to the location of our point as we plotted it in step 1. And it's specifically located here in quadrant 2. Now whenever I have a point located in quadrant 2, I need to go ahead and take the inverse tangent of y over x, but I also need to add pi to it in order to account for what quadrant it's actually located in. So let's go ahead and calculate this value. Now I'm plotting and calculating this. I get that theta is equal to the inverse tangent of my y value 2 over my x value negative 2. Now this simplifies to the inverse tangent of negative one, which is equal to negative pi over 4. Now, if I stopped here forgetting that I needed to add this pi in, that would give me a point that is located here in quadrant 4, which is not where my actual point is. So that's why I need to add that factor of pi. So adding pi to this, making sure that I don't forget that step, if I add pi to that negative pi over 4, I end up getting that this is equal to 3 pi over 4 for my value of theta. Then my point in polar coordinates is located at 2 root 2, 3 pi over 4, and this is our final answer. Let me know if you have questions.

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

( 2 2 , 3 π 4 ) (2\sqrt{2},\frac{3\pi}{4}) ( 2 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 4 3 π )

( 2 2 , − π 4 ) (2\sqrt{2},-\frac{\pi}{4}) ( 2 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , − 4 π )

( 2 2 , π 4 ) (2\sqrt{2},\frac{\pi}{4}) ( 2 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 4 π )

( − 2 2 , 3 π 4 ) (-2\sqrt{2},\frac{3\pi}{4}) ( − 2 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 4 3 π )

9. Polar Equations

Convert Points Between Polar and Rectangular Coordinates

Trigonometry

9. Polar Equations

Convert Points Between Polar and Rectangular Coordinates

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Multiple Choice

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

$(2\sqrt{2},\frac{3\pi}{4})$

$(2\sqrt{2},-\frac{\pi}{4})$

$(2\sqrt{2},\frac{\pi}{4})$

$(-2\sqrt{2},\frac{3\pi}{4})$

Verified step by step guidance

To convert a point from Cartesian coordinates (x, y) to polar coordinates (r, θ), we use the formulas: r = \sqrt{x^2 + y^2} and θ = \arctan\left(\frac{y}{x}\right).

For the point (-2, 2), calculate the radius r: r = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}.

Next, calculate the angle θ using the arctangent function: θ = \arctan\left(\frac{2}{-2}\right) = \arctan(-1).

Since the point (-2, 2) is in the second quadrant, the angle θ should be adjusted to reflect this. The reference angle for \arctan(-1) is \frac{\pi}{4}, so in the second quadrant, θ = \pi - \frac{\pi}{4} = \frac{3\pi}{4}.

Thus, the polar coordinates for the point (-2, 2) are (2\sqrt{2}, \frac{3\pi}{4}).

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