Common Core Geometry Unit 5 Lesson 10 Reflections in the Coordinate Plane

Hello and welcome to another common core geometry lesson by E math instruction. My name is Kirk weiler, and today we'll be doing unit 5 lesson number ten on reflections in the coordinate plane. All right, so unit 5, all about the coordinate plane, the tools of coordinate geometry like the slope formula, the midpoint formula, the distance formula. And then the last few lessons in this unit, what we're doing is we're moving back, and we're taking a look at some of the transformations that we worked with in unit two, but strictly now in the coordinate plane. Today, we're going to be looking at reflections in the coordinate plane. And we'll be combining that somewhat with rotations in the coordinate plane. So if you haven't already worked with rotations in the coordinate plane, either by watching our video or by having it in class, some of the work that we do today is going to be a little bit challenging and maybe inaccessible. So make sure you've had rotations in the coordinate plane before you challenge what we do today. But let's jump right into it. All right, and we'll be able to review some properties about reflections, including even in this first problem. So let's take a look at what exercise one asks us to do. On the grid below, a three comma 7 has been plotted. Find its image a prime after reflecting it in each of the following lines. Great. So what we're going to be doing is we're going to be reflecting that point that's plotted in the X axis, then in the Y axis, and then something that we hadn't done back in unit two. We're going to reflect it in some vertical and horizontal lines. IE the line X equals 6 and the line Y equals four. But we'll get to those in a minute. Why don't you take a second because these are relatively easy, even though we haven't done them for a while, and see if you can find the image of a of a after a reflection in the X axis, and then one in the Y axis. All right, let's do it. Now, when we're reflecting points in vertical and horizontal lines, it's actually relatively easy because a reflection should always be flipped across the line in a perpendicular way, okay? And what I mean by that is if we take a look at the point a, right? And we want to reflect it in the X axis. Well, we want to come into the X axis, all right? And obviously that ends up being 7 units. And then we want to go 7 units. Pass the X axis. Right? The key here is, of course, that that is a perpendicular kind of situation. In other words, we're not going to get a prime down here. We're going to take it and flip it over the X axis so that when I connect a to its image, a prime, that line segment is perpendicular to this line segment, and is bisected by the X axis. That, of course, gives us a prime at three comma negative 7. And one of the rules that's very easy to sort of remember or memorize is that when you reflect a point like a across the X axis, somewhat ironically, the Y coordinate gets negated, but the X coordinate doesn't change at all. Look at that. The X coordinate stayed at three, but the Y coordinate went from 7 to negative 7. Again, that probably makes some sense. Let's now take a look at the reflection across the Y axis. In this case, a happens to be three units from the Y axis in this direction. So when we reflect it across, it should be three units away from the Y axis. In that direction, right? Again, that horizontal line now is perpendicular to the vertical Y axis. And again, the vertical Y axis has bisected that line segment, putting me at an a prime value of negative three positive 7. Of course, this carries on the trend that we saw in letter a, which is when we reflect across the Y axis, the X coordinate ends up getting negated, changing sign, but the Y coordinate doesn't change at all. Do you need to memorize those rules, not likely because they're pretty easy to visualize, and quite frankly, once we get into lines, vertical lines other than the X and the Y axis, the rules, kind of go out the window. You could still develop rules for it, but they would be kind of tough to memorize. So what I'd like to do is I'm going to erase kind of all the ink that's up there right now. And then let's start taking a look at reflections across those other two lines. Let me just get rid of all of this really fast. Wonderful. So letter C asks us to reflect the .3 comma 7 in the line X equals 6. Now it's key that when you see X equals 6, you know number one that it's a line. But number two, it's a vertical line. All right? And that's important because I want to come over here and I'm going to actually draw the vertical line X equals 6. And I think I'll go with it in red just so it pops. All right, so oh, that just didn't work at all. Let me try that again. That's because I was using the wrong edge. Let's try it again. And there. Right? That is the line X equals 6. Now, once we have that plotted, it's actually pretty easy than to do exactly what we did with the Y and the X axis, which is to take a, go perpendicular into the line X equals 6 come perpendicular out of the line X equals 6. And of course, what will happen because we have to go the same distance on both sides is that we end up having an a prime like that. Again, this line segment being perpendicular to that vertical line X equals 6. And what happens is we're at an a prime value of 9 comma 7. Just stick in red here. A prime of 9 comma 7. Certainly the Y coordinate didn't change, but the X coordinate did. All right, why don't you go ahead and try to reflect it across this horizontal line, Y equals four. All right, let's go through it. Again, I'm going to actually draw that line Y equals four. The horizontal line that is. Eh, I guess I'll leave that on for now. Again, it's really important that when you see something like Y equals four, you know it's a horizontal line. So you can draw it quickly, and then just like we did before, we can take point a, we can walk it into that horizontal line a certain distance that's specifically three units again, then come out of it by three units, right? So we come into it, and it's three units we go out of it three units, and we end up being an a prime value of three comma one. All right. So not hard. It's all about counting, recognizing when you have a vertical line versus a horizontal line. Again, if you want, you can memorize rules for reflecting across the X and the Y axis, they can be helpful. They can speed problems up. But if you've got a piece of graph paper there, it's not absolutely necessary. All right, let's move on to some more challenging reflections. And specifically reflections in slanted lines. Now again, the key of doing reflections in slanted lines is really understanding what a reflection is. So a little review on that, even though we've been already doing it in the previous example, a little more of a review on it in exercise number two. Let's take a look. In the diagram at the right, point a has been reflected across line M, such that its image a prime is shown. Points a and a prime have been connected to form a a prime segment AA prime, which intersects line M at point B letter a what is true about a a prime and M so what's true about segment a a prime compared to line or segment M what's true about those? Well, hopefully, you said that they were get me ink back here. Hopefully you said they are perpendicular to each other, so you can easily write that all in symbols. If you'd like AA prime is perpendicular to M the other piece is what's true about AB versus a prime B prime. They're the same length, right? Which you can either say like this. Or if you want to get fancy with the symbol, you can say that a, B is congruent to a prime B again, the idea is pretty simple, right? If we've got a point, any point, and we've got a line, and we want to reflect this point in that line, right? Then the idea is that we have to, if you will kind of construct a perpendicular line, right? Through that one, boy, that doesn't look perfect. Then we have to take this distance and this distance, and that's where our image point will be, even though my drawing is absolutely horrible. Okay. Now the two lines that we are most likely to have to reflect in that are slanted. Are the lines Y equals X and the lines Y equals negative X so before we even do reflections in those two lines, let's actually examine them again, okay? We looked at these two lines a little bit earlier in this unit, but let's take a look at them again and exercise number three. Okay, I'm going to extend the page a little bit. Okay. On the grid at the right, plot the lines Y equals X and Y equals negative X states and points that must lie on both lines and also state the slopes of both lines. So we reviewed specifically how to plot these two things, okay? So what I'd like you to do is I'd like you to spend a little bit of time drawing Y equals X, Y equals negative X if you need to, you could definitely just simply write down some points that you know must be on both of those lines and then use those points to help you graph them, and also I'd like to know what the slope of both lines are. I'm going to go grab a ruler while you work on this. There is my brother. All right, let's talk about it. Now, I would encourage you to simply know the graphs of these two lines. There are extremely helpful lines they should be pretty easy. But if you don't know their graphs, it's simple enough to say, well, the line Y equals X has to contain all the points where literally Y is equal to X that's why it's called Y equals X even points like zero comma zero. So that gives me points like this as zero zero one one two two three three four four, and negative one negative one, et cetera. And very quickly, I can then draw the graph of this line. Great. All right, so here is my line. Y equals X. All right. Y equals negative X is also a very important line, and it's got points where the Y and the X coordinates are the same size, same absolute value, but our opposite in sign. For instance, that would include points like one comma negative one, two comma negative two, three comma negative three, let's do somewhere in the first coordinate is negative, negative one, comma positive one. So it includes these points. All right, so let's get that thing graphed. And there's the line. Y equals negative X. Okay. Now, one thing that's kind of interesting about these two lines are their slopes. And remember, slope is rise divided by run. Okay? You can also think about it in terms of this Y equals MX plus B but that actually might be a little bit tricky for people because it almost looks like there's no number of multiplying X here. And there's no number multiplying negative X there. All right? Or sorry, no number multiplying X here. But if you think about it right, at least on this first line Y equals X, we have a run of one and a rise of one each time. So it's slope. Is one over one. Now, of course, that's just equal to one. And therefore, we can really think about this as Y equals one times X if we want to think about it in terms of Y equals MX plus B on the other hand, Y equals negative X, right? For every one unit that we run, we fall one unit, right? One negative one. So we can think about its slope as being negative one divided by one. Or just a slope of negative one. But what's kind of cool is looking at them sort of in this form. Because what that should show you is that these two lines are themselves perpendicular to each other. Remember perpendicular lines will have slopes that are negative reciprocals, meaning that you take the slope, flip it, and negate it. But if I took this slope and I flipped it and I negated it, I would get that slope. Right? So these two lines are perpendicular to each other because they have negative reciprocal slopes. That'll be a little bit important in just a moment. So let's actually take a look at some reflections in these two lines. All right, exercise number four. On the grid to the right, the point negative 6 two has been plotted, along with the lines Y equals X and Y equals negative X find the image of this point after a reflection across Y equals X and a reflection across Y equals negative X all right, let's take a look. So let's first together do a reflection across the line. Y equals X okay. So I want to take this point and I run over and reflect it across this line. Now, what I know is that to do that, I need to come into this line perpendicular and go out to of it the same distance. Well, since this line's got a slope of one, what that means is I have to come in at a slope of negative one. I literally have to come in parallel to the line Y equals negative X, okay? And literally, I would kind of do this. Right? That's coming in perpendicular. If you will, I'm coming in along the diagonals. Of those squares. I'm going over one down one, over one down, one, but I'm just kind of doing that. The question is, how many diagonals did I go through? Well, I can count them. I've got one, two, three, four. I went in four diagonals, so I'm going to come out one, two, three, four, and I'm going to be right down here. And where is that at? That's at a point a prime of two comma negative 6. All right, simple enough, right? I just kind of come in along the diagonal by four. Or 5 diagonals, one, two, three, four. Sorry, four. And I come out by four. Why don't you do the same with the line Y equals negative X? Ready? Let's do it. Okay, all right, get rid of this. Not really even sure why that's there. But let me actually switch to red. In this case, when I'm reflecting across Y equals negative X, I have to come in perpendicular to it, which means I've got to have a slope of positive one. I've got to be parallel to Y equals X now. But I'm going to do kind of the same trick. I'm going to count my diagonals. So that's going to be one, two, and I hits the line one, two, and that's where my a prime must be. And what's that now? That's now at negative two, positive 6. Okay. Now, I'm not big on memorizing rules of rotation or translation or whatnot. But these two reflections arise so often that I do think that it's helpful to actually sort of memorize what happens to a point X comma Y when you reflect it across the line Y equals X and what happens to a point X comma Y when you reflect it over a line Y equals negative X just those two lines, just those two. Not much of anything else. So let's see an exercise 5. It says using your results from above generate rules for these two reflections. So when I took negative 6 comma two and I reflected it in the line Y equals X, what happened to those X and Y coordinates? It should be fairly obvious. Well, they were just flip flopped, right? The negative 6 went into the Y, the positive two went into the X so I can say, when I reflect in the line Y equals X, X comma Y, just becomes Y comma X, you just switch them. Just switch the two coordinates. And I would suggest memorizing that. Likewise, a little bit trickier. When I reflected in the line Y equals negative X, what happened to the two coordinate points, or not the two coordinate points, but the X and the Y coordinates. Well, it looks like they were switched again, but then they had their signs changed. Notice the negative 6 is now sitting in the Y as a positive 6, and that positive two is now sitting in the X as a negative two. So when we reflect in Y equals negative X, X comma Y becomes negative Y comma negative X we both switch and we negate. Switch plus negate. All right? And again, I would suggest memorizing those as we move forward in this particular lesson, we've only got a few more problems left. I'm going to actually use the fact that you know these, okay? So when we reflect across the line Y equals X, if I don't have a piece of graph paper, I'm just going to say, oh, we just flip flop the X and Y coordinates. And when we reflect across the line Y equals negative X, I'm going to say, oh, we flip flop the X and Y coordinates, and then we negate where we change the sign, right? That's what negating really is. We change the sign on both of them. All right. Let me actually step out of the way for a moment and give you a little bit of time with this particular slide. All right, let's move on. Okay. Exercise number 6. If negative 5 three is reflected across the line Y equals X, then rotate it 90° counterclockwise about the origin, which of the following would be its final image point. Now, if you've already memorized, you know, the reflection across Y equals X and a rule for rotating around 90°, then you don't need this. It just don't need it. Okay? If you haven't memorized them, then it takes a little bit more work. Now again, that one is so easy that for me, if I've got negative 5 three, which is sitting right here, right? And I reflect it across the line Y equals X and then that's going to be three comma negative 5. And that's going to be down here. Now, of course, you could always, you could always draw on Y equals X and then count those diagonals and then come back out and end up having the point right there. Now, once you have that point plotted, I tend not to memorize the rotation rules. Because to me, it's all you switch this and then you negate that one, but not this one. You know, it gets to be kind of confusing in my own head. This is definitely one. Where I would now say, okay, you know, I've got this, I don't even need that ruler necessarily this time. I group these things together. Okay. Got to remember that that's the point I care about. I'm going to rotate it 90° counterclockwise, and now that point that used to be at three negative 5 has now been mapped up to the point. 5 comma whoops, what happened there? And I didn't actually plot it at three negative 5. There it is. It actually ends up being at 5 comma three. Sorry about that. I'm going to rotate it back now and it's not going to stay there. There we go. So my mistake in plotting on the first one. Again, three negative 5 actually would have been down here and then when it rotated up, it ended up up there. All right, so what was that? The correct choice, choice one. All right. And again, this illustrates why it's sort of nice to at least have some of the rules memorized. Again, I could take negative 5 three and I can plot that line Y equals X really quick. And then I could go one, two, three, four, one, two, three, four, and kind of get down there. And then take it and rotate it up to here. It's all good. It's fine. But this is such an easy rule, switching the X and the Y that for me, it's a nice one to have memorized. All right, let's keep going. Now, in all likelihood, just making sure. In all likelihood, those are the most difficult reflections you're going to do. But you could end up having to reflect a point. In a slanted line, other than Y equals X and Y equals negative X well, one of the things that I like about these problems is that they still get at the fundamental essence of what a reflection is. So let's take a look at that in this problem. On the grid shown, line N with equation Y equals two X minus three is graphed, along with the point R at negative two three. Okay, great. If R was reflected over line N to produce its image R prime, what would the slope of R R prime B explain? So if I took this point and I reflected it across this line and then I connected it with a line segment, what would the slope of that be? Now you might think this is awfully unfair because you don't know where R R prime lands. But strangely enough, you can tell me what the slope of it's going to be. So pause the video now and think about this a bit. All right, well let's talk about it. I'm going to do a little sketch right here of the problem that we have going on. We've got this. And we've got this Y equals two X minus three in them. We've got this point R that's sitting right about here. Right, now I'm going to reflect it in this line. I don't know if it ends up being down in here or not, but it's going to be somewhere around there. And what I know is that our R prime must be perpendicular to Y equals two X minus three. Which means it must have a slope that is the negative reciprocal. The negative reciprocal of the slope of this line. But the slope of this line, because it's already written in Y equals MX plus B form, the slope of this line is two, right? So this line is two or two over one. So the slope of R are prime. It's slope would have to be negative one half. Negative reciprocal slopes. But see, that then gives me the key on how I can actually figure out where R prime lands. See if you can use the fact that the slope should be negative one half of this line segment. To figure out where are our prime lens? Take a few minutes. All right, let's talk about it. Well, let's start drawing the line segment, right? Since it's got a slope of negative one half, it means that for a run of two, it's going to fall by one, run of two, four by one, right? Run of two, fall by one, run of two, fall by one, et cetera, okay? So what I can tell is that it's going to land right there. Okay? It's going to be at the point 6 comma negative one. Now, you might ask, well, why would it be at 6 comma negative one? Why did I just come out of here twice? With that run of two and drop of one, run of two drop of one. And that's because that's how many it took to get into the intersection with that line. That distance has to be the same as that distance. So you just sort of reproduce what you did to come in. If it had taken me three of these to get here, then I would have gone three of them out. But it only took me two, right? So that means that this distance then ends up being the same as this distance, we get that perpendicularity happening. All right, and it allows us then to figure out the coordinates of R prime. It's quite a challenge. Quite a challenge because you have to keep a lot of different things in your head, most importantly, the fact that this line, the line of reflection will be perpendicular to the segment that connects the original point to its image. All right. Let me get out of the way for a second. Copy down anything you need, and then we'll wrap up. Let's do it. All right. So today we looked at reflections in the coordinate plane. Reflections in horizontal and vertical lines like the X axis, the Y axis or other vertical and horizontal lines are relatively easy. They tend to all be about just kind of counting the squares or counting the distance, not too bad. Reflection and slanted lines becomes much more difficult. I would highly suggest memorizing those rules for the reflections in Y equals X that would be just changing the X and the Y coordinates and Y equals negative X, which is changing the X and Y coordinates. And then negating both of them, changing their signs. And the most challenging thing we looked at was reflecting in a slanted line other than Y equals X and Y equals negative X and that still has to do with the fact that you've got that perpendicularity between the segment and the line that you're reflecting across. That's the key, right? That and those negative reciprocal slopes. Okay. In the next lesson, we're going to look a little bit at translations in the coordinate plane and how to put all of this stuff together because quite frankly translations are pretty easy all by themselves. They don't need a whole lesson. For now, though, I'd like to thank you for joining me for another common core geometry lesson by E math instruction. My name is Kirk weiler, and until next time, keep thinking. And keep solving problems.