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Reflections in the x-y Plane
Now we’ll talk about a bit of an odd topic, reflections in the x-y plane. Questions about reflections and 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 of 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, a 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 the segment is equidistant from the two end points 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.
It 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, 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 each, in each case, the two points are reflections of each other. Negative 2, 5 is a reflection of 2, 5, and 2, 5 is a reflection of negative 2, 5. Now a little harder, we’ll think about the line y equals x. This line has a slope of one and a y-intercept of zero.
It’s a very special line. It makes an angle of 45 degrees with the x and y-axes. As is obvious from the equation, this line is a set of all points in the x-y plane for which the x and y coordinates are identical. There are no points for 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 equals 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 equals 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, negative 4 and negative 4, 2, those will be reflections of each other over the line y equals x.
Negative 1, 7 and 7, negative 1, also reflections over the line y equals x. And negative 3, negative 5, negative 5, negative 3, also reflections over the line y equals x. If we pick any pair of points like these, then pick any point on the line x-y.
Then any point on the point, the line y equals y would be equidistant from those points 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, 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 equals x, a point on the mirror line. Any point on the mirror line is equidistant from a point and it original point and its reflection, and obviously, these two points 1, 7 and 7, 1 are reflections over the line y equals x. So here we can get a visual of it, and we can actually see these various isosceles triangles.
Here, I’ve 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 equals negative x.
This line has a slope of negative 1 and a y-intercept of 0. Like y equals x, it makes an angle of 45 degrees with the x and y-axes. Also, like y equals x, this line has some special reflection properties. What happens when we reflect a point over the line y equals negative x. So here we have some examples, 2, 5 gets reflected to negative 5, negative 2 and 4, 2 gets reflected to 2, negative 4.
So the reflection of 2, 5 is negative 5, 2. The reflection of 2, negative 4 is the, is 4, negative 2. The x-co, and y-coordinates are switched, and each is given the opposite plus or minus sign. So any positive number becomes negative, any negative number becomes positive, and they are switched in their order.
So that’s exactly what happens. This gonna be a harder pattern to see, but if two points have been switched an opposite-signed x and y and have, have switched an opposite-signed x and y-coordinates from each other. For example, negative 3, 5 and negative 5, 3, then any point on the line y equals negative x will be equidistant from both of them.
Again, any point on the mirror line will be equidistant from both points. For example, negative 3, 5, negative 5, 3, and 12, negative 12 have to form an isosceles triangle. Again, the point on the mirror line, 12, negative 12, has to be equidistant from those two points. Here’s a practice question, pause the video then we’ll talk about this.
Now this would be an example of a very, very hard question on the test. Ordinarily, this would be particularly hard especially without a diagram. Now, now of course you could probably sketch something and you get a ballpark idea, but it would be hard to verify it exactly. Well, we notice though that points J and K are reflections over the line y equals negative x because we switched the x and y-coordinates, and we’ve made the two positives negative, 5, 2 and negative 2, negative 5.
They have to be reflections over the line y equals x, so any point on the line y equals negative x would be equidistant from that. And so, what’s a point on the line y equals negative x with a y-coordinate of 4? Well it has to be x equals negative 4. We get the point negative 4, 4.
So negative 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-signed y-coordinates. If we reflect over the y-axis, we still have the same y-coordinate. We stay in the same horizontal line. We get opposite-signed x-coordinates.
Those are the really easy cases. If we reflect over the line y equals x. The 45 degree angle line, then we switch the x and the y-coordinates. When we reflect over the line y equals negative x, we switch the 8, 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 se, the perpendicular bisector of the segment between the original point and it’s reflected image, that’s an important geometry fact.
And most important, any point on the mirror line is equidistant from the original point and it’s reflected image.
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