# Video: Reflections on the Coordinate Plane

In this video, we will learn how to find the images of points, lines, and shapes after their reflection in the π₯- or π¦-axis on the coordinate plane.

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### Video Transcript

In this video, weβll learn how to find the images of points, lines, and shapes after their reflection in the π₯- or π¦-axis on the coordinate plane. When we reflect a shape in a line, we flip it. And there are a number of ways we can do this. We could reflect the shape in the π₯-axis. Thatβs this line. We could reflect it in the π¦-axis. Thatβs this line. Or we might even reflect it in a different vertical line. Those are of the form π₯ equals π, where π is a constant.

The vertical line π₯ equals negative one, for example, is this one. It passes through the π₯-axis at negative one. We could use a different horizontal line π¦ equals π, where π is a real constant. For example, the line π¦ equals two passes through the π¦-axis at two as shown. Finally, we might even look at reflecting in a diagonal line like π¦ equals π₯. Remember, all the π₯- and π¦-coordinates on this line are equal, or the line π¦ equals negative π₯ as shown.

We say that the line in which we reflect the shape is called the mirror line. And the reflection of the shape is called its image. And we use the prime symbol to represent the image of an object. For example, the reflection of point π΄ would be π΄ prime. And the image, the reflection, of shape π΄π΅πΆ would be π΄ prime π΅ prime πΆ prime. Finally, itβs important that we realize that an object and its image have the same size and theyβll be located the same distance from the mirror line on opposite sides. Letβs see what this might look like.

Which pair of triangles represents a reflection in the π₯-axis?

The π₯-axis is this horizontal line. We could call that the line with equation π¦ equal zero. Now, weβre looking for the pair of triangles which represent a reflection in this line. Now, when we reflect a shape, we flip it. The two triangles will have the same size, and theyβll be located the same distance from the π₯-axis but on opposite sides. So, letβs look at some of these pairs.

Weβll begin by looking at shape π΄ and π΅. For shape π΄ and π΅, this first vertex is two units away from our mirror line. For both shapes, this second vertex is five units away from the mirror line on opposite sides. And our third vertices are both three units away from the mirror line on opposite sides. We see that each of the points are the same distance from the π₯-axis on opposite sides. And the shape is flipped but otherwise unchanged. That is indeed a reflection in the π₯-axis. So, thatβs a good indication of us that the pair of triangles that represent the relevant reflection are π΄ and π΅. But letβs check and see whatβs happened between the other pairs.

Letβs look at shape π΄ and πΆ. Once again, comparing the relevant vertices of our shapes, we see that they are the same distance from the π¦-axis on opposite sides. And the shape is flipped over the π¦-axis but otherwise unchanged. In this case then, shapes π΄ and πΆ represent a reflection in the π¦-axis.

Now, what about shapes π΅ and πΆ? Well, there are two ways to describe this. We add in a diagonal line with the equation π¦ equals negative π₯. Now, we compare the vertices of π΅ and πΆ. This time, our vertices are the same distance from the line π¦ equals negative π₯ but on opposite sides. Otherwise, shapes π΅ and πΆ, apart from being flipped β remember, thatβs a reflection β remain the same shape and size. So, we could say that shapes π΅ and πΆ represent a reflection in the line π¦ equals negative π₯.

If we look carefully though, we can even say the shape has been rotated 180 degrees about the origin. Thatβs the point zero, zero. In this case though, the pair of triangles we were looking for were π΄ and π΅.

In our next example, weβll look at how to reflect a shape in the π¦-axis.

Find the images of π΄π΅πΆπ· after reflection in the π¦-axis.

We begin by identifying the location of our mirror line. Weβre told to reflect the shape in the π¦-axis. Thatβs this vertical line here. We might say this line has the equation π₯ equals zero. The images of π΄π΅πΆπ· are the vertices of our rectangle after it has been reflected. And we use prime notation, π΄ prime π΅ prime πΆ prime π· prime. Now, when we reflect π΄π΅πΆπ· in the line π₯ equals zero or the π₯-axis [π¦-axis], we know it ends up being flipped. Otherwise it should end up the exact same size and the exact same distance from our mirror line on opposite sides.

Now, a really sensible way to perform the reflection is to do it vertex by vertex. Letβs start with vertex πΆ. We measure the distance between vertex πΆ and the π¦-axis. And of course, thatβs the perpendicular distance. Thatβs one, two units. This means vertex πΆ prime must be two units away from the π¦-axis in the opposite direction. That has Cartesian coordinates negative two, one.

Letβs repeat this process with π·. Once again, itβs two units away from the π¦-axis. And so, π· prime will also be two units away from the π¦-axis in the opposite direction. It has Cartesian coordinates negative two, six. Weβll now look at vertex π΄. This time thatβs one, two, three, four, five, six, seven, eight units away from our mirror line. π΄ prime is, therefore, eight units away from the mirror line in the opposite direction. So, itβs at the point negative eight, six.

Now, notice that since the size of the shape remains unchanged, we could have simply measured the distance between π΄ and π· and then repeated that on the other side. Letβs see what that might look like for the side πΆπ΅. The length of this side is one, two, three, four, five, six units. And so, the distance between vertices π΅ prime and πΆ prime must also be six units.

And so, we have the image of π΄π΅πΆπ·. π΄ prime has coordinates negative eight, six. Remember, we go along the corridor and up the stairs. π΅ prime is negative eight, one. πΆ prime is negative two, one. And π· prime is negative two, six.

Now, in fact, we can generalize something from this question. Letβs look at the original coordinates. They are eight, six; eight, one; two, one; and two, six; respectively. Notice that the π¦-coordinate in each case remains completely unchanged. The π₯-coordinate is almost the same, but we notice that itβs the negative version of it. In fact, we can say that the reflection of a point π₯, π¦ across the π¦-axis is the point negative π₯, π¦.

In a similar way, when we reflect a point across the π₯-axis, the π₯-coordinate remains unchanged, and we get the negative version of the π¦-coordinate. So, the reflection of π₯, π¦ across the π₯-axis is π₯, negative π¦.

In our next example, weβll look at how to reflect a shape in a vertical line of the form π₯ equals π for real constants π.

Reflect the rhombus π΄π΅πΆπ· in the line π₯ equals two.

Letβs begin by identifying our mirror line. We know that lines of the form π₯ equals π are vertical lines; they pass through the point π, zero. In other words, they pass through the π₯-axis at π. Now, the line π₯ equals two then must also be a vertical line, but it must pass through the π₯-axis at two as shown. So, weβre going to reflect our rhombus in this line.

Letβs do that vertex by vertex. Weβll begin with vertex π΄. We measure the perpendicular distance of this vertex from the mirror line. Itβs one, two, three, four units. This means the image of π΄, remember we call that π΄ prime, must also be four units away from the line π₯ equals two but in the opposite direction. And so, π΄ prime is here. Itβs got coordinates negative two, six.

Letβs now measure the perpendicular distance of vertex π΅ from our mirror line. This time thatβs one, two, three, four, five, six units. This means that the image of π΅ will be six units away from the mirror line in the opposite direction. Thatβs here. We repeat this process for the remaining vertices. πΆ is one, two, three, four units away from the mirror line. And so, its image, πΆ prime, will also be four units away.

Finally, π· is two units away from the line π₯ equals two, which means that π· prime will be here. Itβs two units away from our line in the opposite direction. And so, weβve reflected the rhombus π΄π΅πΆπ· in the line π₯ equals two. Weβve given its vertices π΄ prime, π΅ prime, πΆ prime, and π· prime.

In our next example, weβll consider how we can find the position of a point after reflecting it in a given diagonal line.

What is the image of the point nine, eight under reflection in the straight line π¦ equals π₯?

These sorts of questions can be quite difficult to visualize. So, letβs plot our point and the line π¦ equals π₯ on a Cartesian plane. Here is our point nine, eight. Its π₯-coordinate is nine, and its π¦-coordinate is eight. The line π¦ equals π₯ is a diagonal line. Every point on this line has equal π₯- and π¦-coordinates. For example, it will pass through the point one, one; three, three; five, five; negative two, negative two; and so on. In fact, it looks a little something like this.

Weβre going to reflect our point in this line. So, we look at the perpendicular distance from our point to the line. We can see thatβs half of the diagonal of one square. The image of our point will be the exact same distance away from the line in the opposite direction. So, what are its coordinates? Well, theyβre eight, nine. It now has an π₯-coordinate of eight and a π¦-coordinates of nine. And so, weβve worked out the image of the point nine, eight under reflection in the line π¦ equals π₯. Itβs eight, nine.

But we can, in fact, generalize this. We take a point π₯, π¦. When we reflect it in the line π¦ equals π₯, it becomes π¦, π₯. In other words, the values of the π₯- and π¦-coordinates interchange. In general, we say that the reflection of π₯, π¦ across the line π¦ equals π₯ is the point π¦, π₯. And the reflection of π₯, π¦ across the line π¦ equals negative π₯ gives us the point negative π₯, negative π¦.

In our final example, weβll consider how we can reflect a full shape in the diagonal line π¦ equals π₯.

Reflect triangle π΄π΅πΆ in the line π¦ equals π₯.

Remember, the line π¦ equals π₯ is a straight diagonal line that passes through the origin. Thatβs the point zero, zero. Every point on this line has equal π₯- and π¦-coordinates. Itβs this line. Notice it passes through the point two, two; 10, 10; negative six, negative six; and so on. And so, weβre looking to reflect triangle π΄π΅πΆ in this line. And so, we recall that when we reflect a shape, itβs flipped, that each vertex of an original figure and its image, thatβs the reflection, are the same distance away from the line of reflection but out the other side.

So, we could count the perpendicular distance from each vertex to our line. Alternatively, we can quote this fact. The reflection of a point π₯, π¦ across the line π¦ equals π₯ gives us the point π¦, π₯. Now, point π΄ has coordinates negative four, 12; point π΅ has coordinates negative four, six; and point πΆ has coordinates two, six. Notice that the values of our π₯- and π¦-coordinates are interchanged. So, the image of π΄ will have coordinates 12, negative four; the image of π΅ will have coordinates six, negative four; and the image of πΆ, πΆ prime, has coordinates six, two. π΄ prime is therefore here, π΅ prime is here, and πΆ prime is here.

And so, weβve reflected triangle π΄π΅πΆ in the line π¦ equals π₯. Itβs the image π΄ prime π΅ prime πΆ prime as shown.

In this video, weβve learnt that a reflection in a mirror line flips the original shape. The image, which is the reflection of the original and is given using prime notation, will have the same shape and size as the original object. Each vertex will be an equal distance away from the mirror line but on the opposite side.

We also saw that we can generalize the image of certain points after being reflected in various lines. The reflection of the point π₯, π¦ across the π₯-axis is the point π₯, negative π¦. Across the π¦-axis, and we get the point negative π₯, π¦. Across the line π¦ equals π₯, we get π¦, π₯. And across the line π¦ equals negative π₯, we get negative π¦, negative π₯.