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Rotating shapes: center ≠ (0,0)

Sal is given a triangle on the coordinate plane and the definition of a rotation about an arbitrary point, and he manually draws the image of that rotation.

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  • male robot johnny style avatar for user Benjamin Lathrop
    Is there an easier mathematical way to do this, because this will take too long for tests.
    (37 votes)
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    • aqualine seed style avatar for user Mecarath
      Okay, it took me a while to figure out a pattern, but there is an easier way to do by graphing. Create a pretend origin by drawing a dotted line Y-axis and X-axis where the arbitrary point is at. Then rotate your paper literally counter clockwise or clockwise whatever degrees you need it. You will see the dotted "pretend origin" has rotated. The shape in question also has rotated. Now again draw another "pretend orirgin2" at the arbitrary point as if it didn't move, You will see the "shift" or translation between the first pretend origin and the second. Remember that shape in question? Translate that shape using the same "shift" between the first and second pretend origin. Saves you TIME!
      (3 votes)
  • piceratops ultimate style avatar for user Theodore Astell
    Here's my idea about doing this in a bit more 'mathematical' way: every rotation I've seen until now (in the '...about arbitrary point' exercise) has been of a multiple of 90°. Because the axes of the Cartesian plane are themselves at right angles, the coordinates of the image points are easily predictable: with a bit of experimentation, you could easily 'prove' to yourself that rotating (a, b) 90° would result in (-b, a) [by the way, this has some interesting consequences in trig]; rotation of 180° gives us (-a, -b) and one of 270° would bring us to (b, -a). Rotating (a, b) 360° would result in the same (a, b), of course.

    Now, since (a, b) are coordinates with respect to the origin, this only works if we rotate around that point. But it's easy to calibrate it if you want to specify another point, around which you want to rotate -- just make that point the new origin! Figure out the other points' coordinates with respect to your new origin [if the new origin had coordinates (x, y), then (a, b)'s new coordinates would be (a-x, b-y) ], do the transformations, and then translate everything back to coordinates with respect to the old origin.
    (8 votes)
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    • piceratops ultimate style avatar for user Christopher
      Yeah, I was thinking about that too (that is, doing this in a more mathematical way). Computer languages employ this shifting of the matrix to simplify transformations for the programmer, but I don't know if that is how the calculations are actually done "under the hood".
      (5 votes)
  • piceratops ultimate style avatar for user Sidarth Kar
    Is there some kind of formula for this, because I want to take less time and I'm pretty sure there must be a formula, Please?
    (5 votes)
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  • piceratops tree style avatar for user Amy Lee
    How are we supposed to get the question in Rotations 1 & 2 correct when we don't even have the scratchpad available. Sal has a program that he uses to write with ( black background), but we don't have that. I have to wing the questions and just hope for the best. How have you guys been doing the questions?
    (3 votes)
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    • marcimus pink style avatar for user Tyler
      It might help to just get out a piece of paper and a pencil when the scratchpad is not available. Most of the time, for harder problems, I prefer paper over the scratchpad anyways.

      Otherwise, you could always get a basic drawing program like MS paint (or a drawing app if you're on a tablet) and use that in place of the scratchpad.
      (5 votes)
  • aqualine ultimate style avatar for user Christine Pham
    Can we rotate it by using the patterns of the rotations and thinking of the point of rotation as the origin? I think if we redraw the Y-axis and X-axis with the point and used the patterns it would work.
    (3 votes)
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  • male robot hal style avatar for user Marioland
    Is there a formula that we can use for arbitrary points? it is hard to do the exercise after this video if one lacks a lot of visualization skills.
    (4 votes)
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  • piceratops ultimate style avatar for user ∫∫ Greg Boyle  dG dB
    Wouldn't it be easier to just translate the second and third point? For example, once you have rotated a point (e.g M to M'), why not just figure out the translation of M to M' and translate the other points accordingly? This is a rigid body so distances and angles will be preserved.
    (1 vote)
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    • piceratops ultimate style avatar for user Magistra H
      If you rotate around something (whether something close like a baseball bat for a dizzy race, or far like a hissing cobra you want to avoid but can't get far enough away from), you change not only your position but the way you're facing. It's different from if you move across the room without changing the way you face. If you merely translate the figure, you risk neglecting the change in "direction" that comes with a rotation.
      (4 votes)
  • male robot hal style avatar for user Cesare Fusari
    I'm a bit confused. If the initial image is 270° rotated, and we want to draw the image before rotation, I think we have to rotate the image by more 90° or -270°. This will take our image to its initial position.
    In this video Sal rotate the image by -90° that is the same rotation already applied to the image and ends up with an image that is rotated by -180° or 180° that is the same. Does someone note this too?
    (1 vote)
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    • piceratops ultimate style avatar for user ILoveToLearn
      The question might be worded a bit ambiguously, but it means that you are required to rotate the initial image 270*. It says:
      SAM is rotated 270*. Draw the image of this rotation on the interactive graph.
      Therefore, SAM has already been rotated the set amount (here 270) of degrees around a certain point. We need to draw the image of that pre-done rotation on the graph.
      I hope this helps you! Have a great day!
      (4 votes)
  • purple pi teal style avatar for user Don't read my bio
    Can we rotate it by using the patterns of the rotations and thinking of the point of rotation as the origin? I think if we redraw the Y-axis and X-axis with the point and used the patterns it would work.
    (2 votes)
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  • ohnoes default style avatar for user neil.nimmagadda
    How do you find the center of rotation when you have the original figure and the rotated figure when it is not the origin?
    (2 votes)
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    • starky ultimate style avatar for user Awesome Dude
      A point (a, b) rotated around a point (x, y) 90 degrees will transform to point (-(b-y) + x, (a-x) + y).
      A point (a, b) rotated around a point (x, y) 180 degrees will transform to point (-(a - x) + x, -(b - y) + y).
      A point (a, b) rotated around the origin 270 degrees will transform to point (b - y + x, -(a - x) + y).

      Use the formula above to figure out how do rotate points around any given origin... (a,b) represents the point, while (x,y) represents the origin given.
      (1 vote)

Video transcript

- [Voiceover] Triangle SAM, S-A-M, and this is one over here, S-A-M, is rotated 270 degrees, about the point four comma negative two. So, this is four comma negative two right over here. Draw the image of this rotation using the interactive graph. We have this little interactive graph tool where we can draw points or if we wanna put them in the trash we can put them there. The direction of rotation by a positive angle is counter-clockwise. So, this would be 270 degrees in the counter-clockwise direction. So, to help us think about that I've copied and pasted this on my scratch pad and we can draw through it and the first thing that we might wanna think about is if you rotate, I've talked about this in the previous video when we were rotating around the origin, if you rotate something by, last time we talked about a negative 270 degree rotation, but if we're talking about a 270 degree rotation. If you imagine a point right over here this would be 90 degrees, 180, and then that is 270 degrees. That is a 200 and 70 degree rotation. We're going in a counter-clockwise direction. You see that that is equivalent, that is equivalent to a 90 degrees, to a 90 degrees clockwise rotation, or a negative 90 degree rotation. And 90 degree rotations are a little bit easier to think about. So, let's just, instead of thinking of this in terms of rotating 270 degrees in the positive direction, in the counter-clockwise direction, let's think about, let's think about this, rotating this 90 degrees in the clockwise direction. They are going to be equivalent. But how do we do that? Well, like we've done in previous videos what I'm gonna do is I'm gonna take each point, let's say point, using a color you can see, point M, I'm having trouble switching color, point M right over here, and I'm gonna construct a right triangle between point M and our center of rotation where the hypotenuse of the right triangles is gonna be the line that connects these two. So, let me draw that. So, that's the hypotenuse. Now, let me draw a right triangle like this. So, it's gonna go, it's gonna be, like, it's gonna look like that, and then it's gonna look like, whoops, that wasn't to press my, the command button before. Okay, there you go. All right, so I have this right triangle. So, you imagine, imagine rotating this right triangle, this magenta one that I just constructed, by negative 90 degrees, by negative 90 degrees, or 90 degrees in the clockwise direction. Well, what's gonna happen to each of these sides? Well, this side right over here, if I rotate it by negative 90 degrees it is going to end up, it is going to end up, it is going to end up pointing up and how long is it going to be? Well, let's see, this one is one, two, three, four, five, six, seven units long, so it's still gonna be seven units long. One, two, three, four, five, six, seven, it's gonna put us right over here. It is going to put us right over there. And then this side, let me make this clear, this side that I'm highlighting in green which we see are four units long. It forms a right angle. It forms a right angle. So, if you were to rotate this, the entire triangle 90 degrees in the clockwise direction well, then this is gonna go four in this direction just like that. Just like that. And then your new triangle, the image of, I guess, what we just, this right triangle I've constructed if I rotate it by negative 90 degrees, 90 degrees in the clockwise direction, is going to look like this. So, after rotation, so what did I just do? I just took this, I just took this triangle that I just constructed, and I rotated it negative 90 degrees to get to this triangle right over here. And so, the point M, its image after the 90 degree rotation would be right over here and I could call that, I'll call that M prime. What I'm essentially going to do, I'm going to do that for point M, I'm going to do that for point A, and I'm gonna do that for point S, and I'm gonna connect the three, the M prime, the A prime, and the S prime, and then that will give me the image, that'll give me the vertices of my new triangle. The image of triangle SAM after rotation. So, let's keep going. Let's do this, let's do this now with point A. With point A then I can construct another right triangle and I could do it, I could do it several, I could do it several ways. Actually, let me just do it this way. Let me, let me draw it. I could, I could draw this. The hypotenuse we know is gonna go from here to, from point A to our center of origin. And I could draw a right triangle that's up here like this or I could draw a right triangle that is down here like this. Maybe I'll focus on this one. Let me draw this right triangle. So, this right triangle right like this. So, if I draw this right triangle like this. So, this side, let me do this in a color that I can see more clearly or that is mostly this side. No, you won't be able to see that. All right, I'mma stick with the purple. So, this side right over here that I am drawing in purple, if you rotate it clockwise in the 90 degree direction or negative 90 rotation, well, it's going to, instead of being straight horizontal, it's gonna be straight vertical. So, it's going to be, it's going to be, it's gonna overlap with what we had already drawn and how long is it? Well, right now it goes from X equals negative five to X equals four. So, it's nine long and you can count that. One, two, three, four, five, six, seven, eight, nine, or you could say it's two longer than what we had already drawn. So, let's see, if you, if you start Y equals negative two and you add nine you're gonna get to Y equals seven. So, this is gonna go all the way like this and then, and then let's see, this side, this side forms a right angle with it. So, if you, just gonna go like this, it's gonna be a right angle and how long is this? It's one, two, three, four, five units long. So, now we're gonna go one, two, three, four, five units in that direction and just like that we get to our A prime. We get, we get to, we get to our A prime and we can draw the hypotenuse of this triangle. So, let me see if I can draw it. Press, I can only draw the straight line. All right, it's gonna look like, gonna look like that. So, we know where the image of point M is after the rotation. We know where the image of point A is after rotation and we see how these triangles we constructed. We've just rotated them. We've just rotated them 90 degrees in the clockwise direction. Now, we just have to do this for point S and then we have the three points for our image. So, point S, once again we can construct a right triangle. Let's do that. So, hypotenuse, we'll connect point S to our center of rotation, and I can do this several ways. Let me think about the best way to do it. Well, let me construct this right triangle so it looks like this. And then looks like this. So, this top side, this top side right over here. If I were to rotate it, if I were to rotate it negative 90 degrees, 90 degrees in the clockwise direction, where does that put us? Well, once again it's gonna put us straight up and down and this length is it goes from X equals negative three to X equals four. So, it's seven long. So, it gets us to this point right over here. Now, this side, let me be careful when I just say this, and let me do it in a different color. This side over here forms a right angle and so it's, and if you see when you go from here you go down so here you're gonna go left and you see that it's three long. So, it's three, it's three long. This is a right angle right over here and that gets us to our S prime. That gets us to our S prime. We can now draw the hypotenuse. So, notice this whole, like, kind of crazy complex that we have drawn of these different triangles, I've just rotated the whole thing by negative 90 degrees. But this triangle that we have, that we were, we had just made when you rotated it maps to that one right over there. But now we have in our image our S prime, our A prime, and our M prime, and we can connect them. So, it will look like this. So, let's just connect them. Connect the dots. So, that, and that, and that. And so, you see, you might have lost our original triangle. I'm just gonna shade it in so you can see it clear although what we really care about is mapping the points. Mapping, actually, let me. I don't want you to think that we're mapping, that we're mapping everything in between. We're just mapping the points that define the triangle. So, our original triangle that we cared about was this thing that I'm gonna draw really bold. It was this thing. And I'm gonna draw it really, really bold. And you rotate it by negative 90 degrees which is equivalent to positive 270. You could've rotated it positive 270. You could've rotated it positive 270 to get to this other one right over here but it maps to this. This is its image. Now we just have to go back onto the exercise. Now we have to go back onto the exercise and put it in. So, let's remember these points. We have the point eight comma five. So, we have the point eight comma five in our image. We have the point nine comma seven, nine comma seven, and we have the point, and we have the point one comma five. So, we have the point one comma five and now I just have to finish it by going to this point again and we're all done. We've rotated it by 270 degrees which is equivalent to a negative 90 degree rotation. We can check that we indeed got it right.