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

( 2 , 4 π 3 ) (2,\frac{4\pi}{3}) ( 2 , 3 4 π )

Convert the point to polar coordinates. ( − 1 , − 3 ) (-1,-\sqrt{3}) ( − 1 , − 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> )

Here we're asked to convert the point given in rectangular coordinates into polar coordinates, and the point that we're given is the negative one negative square root of 3. So let's start with step number 1 and plot our point. Now it may not be easy to locate the square root of 3 on your graph, but you can always type this into your calculator and you would get that this is equal to about negative 1.73. So with that in mind, graphing this point at negative 1, negative 1.73, we end up right about here in quadrant 3. Now we can move on to step number 2 in calculating r. Now in calculating r, I get that r is equal to that x value negative one squared plus that y value negative root three squared. Now in actually squaring these values, this ends up giving me the square root of 1 +3, which is really just the square root of 4, which we know is just 2. So my r value here is 2, and we can go ahead and move on to step number 3 in finding theta. Now looking at the location of our point here, I see that it's located in quadrant 3. Now because it's located in quadrant 3, that tells me that I need to take the inverse tangent of y over x and then add pi to it. So let's go ahead and do that here. Now here theta will be equal to the inverse tangent of my y value negative root 3 over negative 1. And remember we're going to add pi after we take that inverse tangent. Now taking the inverse tangent here ends up giving me the inverse tangent of the square root of 3, still adding that pi to the end, and the inverse tangent of the square root of 3 is simply pi over 3. So pi over 3 plus pi, making sure to add that factor to get our point located in the right quadrant. This ends up giving me 4pi over 3 for my angle theta. Now I my point in polar coordinates, 24pi over 3, and this is my final answer. Let me know if you have any questions. Thanks for watching.