Reflections in the x-y Plane
Transcript
Now we'll talk about a bit of an odd topic, reflections in the x-y plane. Questions about reflections of points are not very common, they appear mostly in the more advanced questions. But if we understand a few simple principles, these rare and challenging questions become quite easy. First, we need to review a couple ideas from pure geometry.
Suppose we reflect a point over a line and then draw a segment between the original point and its reflected image. So we have an original point, reflected image and we draw a little segment, that green segment connecting them. Of course the original and the reflected are equidistant from the line, but more than that, the mirror line is the perpendicular bisector of that segment.
That's a big idea that a mirror acts as a perpendicular bisector. As you may remember from the geometry module, every point on the perpendicular bisector of a segment, is equidistant from the two endpoints of the segment. This means that every point on the mirror line is equidistant from the original point and the reflection. These properties are true for all reflections, so these are really big ideas.
So let's talk about reflections over the x-axis. If we reflect a point in the x-y plane over the x-axis, the original point and the reflected image will have the same x-coordinate, will be on the same vertical line. The y-coordinate has equal absolute values and opposite signs. So we just take the y-coordinate, if it's positive we make it negative, if it's negative we make it positive, very simple.
Now, reflections over the y-axis, same thing really. If we reflect a point over the y-axis, the original point and the reflected image have the same y-coordinate. They lie on the same horizontal line, so here are two points on the same horizontal line. The x-coordinates have equal absolute values and opposite positive or negative signs.
So, if we have a negative x value it becomes positive, if we have a positive x value it becomes negative. So in each case the two points are reflections of each other. (-2, 5) is a reflection of (2, 5), and (2, 5) is a reflection of (-2, 5). Now, a little harder, we'll think about the line y = x. This line has a slope of 1 and a y-intercept of 0, it's a very special line.
It makes an angle of 45 degrees with the x- and y-axes. As is obvious from he equation, this line is the set of all points in the x-y plane for which the x- and y-coordinates are identical. There are no points from which the x- and y- coordinates are identical that are not on this line.
And incidentally a little piece of trivia, all the points above this line have a y-coordinate bigger than the x-coordinate. All the points below the line have an x-coordinate bigger than the y-coordinate, that's also something that can be helpful. Let's talk about reflections over this line. When we reflect a point in the x-y plane over the line y = x, the image has the x- and y-coordinates switched.
So here, (2, 5) and (5, 2) are reflected images of each other over the line y = x. In other words, we swap the place of the x-coordinate and the y-coordinate, that's the effect of reflecting over this particular line. Similarly, (2, -4) and (-4, 2), those would be reflections of each other over the line y = x.
(-1, 7) and (7, -1) also reflections over the line y = x and (-3, -5), (-5, -3) also reflections of the line y = x. If we pick any pair of points like these, then pick any point on the line y = xy, would be equidistance from this point, because it's a point on the mirror line. And as we said above any point on the mirror line is equidistant from the original point and its reflected point.
Thus, for example, (1, 7) and (7,1) and (k, k) would form an isosceles triangle, for any value of k, positive or negative. Because (k,k) has to be a point on the line y = x, a point on the mirror line. Any point on the mirror line is equidistant from original point and it's reflection. And obviously these two points (1, 7) and (7, 1) are reflection of the line y = x.
So here we can get a visual of it, and we can actually see these various isosceles triangle. So here I picked just a few example points but you get the idea, any point on that line. The (k, k) could be anywhere on that line and we'd get isosceles triangles. Although it's a rare topic on the test, we can also mention the line y = -x.
This line has a slope of -1 and a y-intercept of 0, like y = x, it makes an angle of 45 degrees with the x- and y-axes. Also, like y=x, this line has some special reflection properties. What happens when we reflect a point over the line y = -x? So here we have some examples. (2, 5) gets reflected to (-5, -2), and (4, 2) gets reflected to (2, -4).
So the reflection of (2, 5) is (-5, 2), the reflection of (2, -4) is (4, -2). The x- and y- coordinates are switched and each is given the opposite plus and minus sign. So any positive number becomes negative and any negative number becomes positive, and they are switched in their order, so that's exactly what happens.
This can be a harder pattern to see but if two points have been switched and opposite-signed coordinates from each other, for example, (-3, 5) and (-5, 3). Then any point on the line y = -x will be equidistant from both of them, again, any point on the mirror line will be equidistant from both points. For example, (-3, 5) (-5, 3) and (12, -12) have to form an isosceles triangle.
Again, the point in the mirror line (12, -12) has to be equidistant from those two points. Here is a practice question, pause the video and then we will talk about this. Now this would be an example of a very, very hard question on a test. Ordinarily, this would be particularly hard especially without a diagram. Now of course you could Probably skip something you get a ballpark idea but it would be hardly verified exactly.
Well, we notice though that points J and K are reflections over the line y = -x because we switched the x- and y-coordinates and we've made the two positives negative, (5, 2) and (-2, -5). They have to be reflections of the line y = x, so any point on the line y = -x would be equidistant from them. And so what's a point on the line y = -x with a y-coordinate of 4 well has to be x = -4.
We get the point (-4, 4), so -4 is the x-coordinate of L. In summary, we reflect over the x-axis, we keep the same x, we're on the same vertical line, we get opposite side y-coordinates. If we are reflected with the y-axis, we still have the same y-coordinate, we stand in the same horizontal line, we got opposite sign x-coordinates, those are the really easy cases.
If we reflect over the line y = x, the 45 degree angle line, then we switch the x and the y-coordinates. When we're reflect over the line y = -x, we switch the x and y coordinates and we make each the opposite, positive or negative sign. And in all these cases, the mirror line is always the perpendicular bisector of the segment between the original point and its reflected image, that's an important geometry fact.
And most important, any point on the mirror line is equidistant from the original point and its reflected image.
Read full transcriptSuppose we reflect a point over a line and then draw a segment between the original point and its reflected image. So we have an original point, reflected image and we draw a little segment, that green segment connecting them. Of course the original and the reflected are equidistant from the line, but more than that, the mirror line is the perpendicular bisector of that segment.
That's a big idea that a mirror acts as a perpendicular bisector. As you may remember from the geometry module, every point on the perpendicular bisector of a segment, is equidistant from the two endpoints of the segment. This means that every point on the mirror line is equidistant from the original point and the reflection. These properties are true for all reflections, so these are really big ideas.
So let's talk about reflections over the x-axis. If we reflect a point in the x-y plane over the x-axis, the original point and the reflected image will have the same x-coordinate, will be on the same vertical line. The y-coordinate has equal absolute values and opposite signs. So we just take the y-coordinate, if it's positive we make it negative, if it's negative we make it positive, very simple.
Now, reflections over the y-axis, same thing really. If we reflect a point over the y-axis, the original point and the reflected image have the same y-coordinate. They lie on the same horizontal line, so here are two points on the same horizontal line. The x-coordinates have equal absolute values and opposite positive or negative signs.
So, if we have a negative x value it becomes positive, if we have a positive x value it becomes negative. So in each case the two points are reflections of each other. (-2, 5) is a reflection of (2, 5), and (2, 5) is a reflection of (-2, 5). Now, a little harder, we'll think about the line y = x. This line has a slope of 1 and a y-intercept of 0, it's a very special line.
It makes an angle of 45 degrees with the x- and y-axes. As is obvious from he equation, this line is the set of all points in the x-y plane for which the x- and y-coordinates are identical. There are no points from which the x- and y- coordinates are identical that are not on this line.
And incidentally a little piece of trivia, all the points above this line have a y-coordinate bigger than the x-coordinate. All the points below the line have an x-coordinate bigger than the y-coordinate, that's also something that can be helpful. Let's talk about reflections over this line. When we reflect a point in the x-y plane over the line y = x, the image has the x- and y-coordinates switched.
So here, (2, 5) and (5, 2) are reflected images of each other over the line y = x. In other words, we swap the place of the x-coordinate and the y-coordinate, that's the effect of reflecting over this particular line. Similarly, (2, -4) and (-4, 2), those would be reflections of each other over the line y = x.
(-1, 7) and (7, -1) also reflections over the line y = x and (-3, -5), (-5, -3) also reflections of the line y = x. If we pick any pair of points like these, then pick any point on the line y = xy, would be equidistance from this point, because it's a point on the mirror line. And as we said above any point on the mirror line is equidistant from the original point and its reflected point.
Thus, for example, (1, 7) and (7,1) and (k, k) would form an isosceles triangle, for any value of k, positive or negative. Because (k,k) has to be a point on the line y = x, a point on the mirror line. Any point on the mirror line is equidistant from original point and it's reflection. And obviously these two points (1, 7) and (7, 1) are reflection of the line y = x.
So here we can get a visual of it, and we can actually see these various isosceles triangle. So here I picked just a few example points but you get the idea, any point on that line. The (k, k) could be anywhere on that line and we'd get isosceles triangles. Although it's a rare topic on the test, we can also mention the line y = -x.
This line has a slope of -1 and a y-intercept of 0, like y = x, it makes an angle of 45 degrees with the x- and y-axes. Also, like y=x, this line has some special reflection properties. What happens when we reflect a point over the line y = -x? So here we have some examples. (2, 5) gets reflected to (-5, -2), and (4, 2) gets reflected to (2, -4).
So the reflection of (2, 5) is (-5, 2), the reflection of (2, -4) is (4, -2). The x- and y- coordinates are switched and each is given the opposite plus and minus sign. So any positive number becomes negative and any negative number becomes positive, and they are switched in their order, so that's exactly what happens.
This can be a harder pattern to see but if two points have been switched and opposite-signed coordinates from each other, for example, (-3, 5) and (-5, 3). Then any point on the line y = -x will be equidistant from both of them, again, any point on the mirror line will be equidistant from both points. For example, (-3, 5) (-5, 3) and (12, -12) have to form an isosceles triangle.
Again, the point in the mirror line (12, -12) has to be equidistant from those two points. Here is a practice question, pause the video and then we will talk about this. Now this would be an example of a very, very hard question on a test. Ordinarily, this would be particularly hard especially without a diagram. Now of course you could Probably skip something you get a ballpark idea but it would be hardly verified exactly.
Well, we notice though that points J and K are reflections over the line y = -x because we switched the x- and y-coordinates and we've made the two positives negative, (5, 2) and (-2, -5). They have to be reflections of the line y = x, so any point on the line y = -x would be equidistant from them. And so what's a point on the line y = -x with a y-coordinate of 4 well has to be x = -4.
We get the point (-4, 4), so -4 is the x-coordinate of L. In summary, we reflect over the x-axis, we keep the same x, we're on the same vertical line, we get opposite side y-coordinates. If we are reflected with the y-axis, we still have the same y-coordinate, we stand in the same horizontal line, we got opposite sign x-coordinates, those are the really easy cases.
If we reflect over the line y = x, the 45 degree angle line, then we switch the x and the y-coordinates. When we're reflect over the line y = -x, we switch the x and y coordinates and we make each the opposite, positive or negative sign. And in all these cases, the mirror line is always the perpendicular bisector of the segment between the original point and its reflected image, that's an important geometry fact.
And most important, any point on the mirror line is equidistant from the original point and its reflected image.
