In this video, you will learn how to do a reflection over a horizontal or vertical line, such as a reflection over the line x=-1.

Let’s use triangle ABC with points A(-6,1), B(-5,5), and C(-5,2).

Apply a reflection over the line x=-3

Since the line of reflection is no longer the x-axis or the y-axis, we cannot simply negate the x- or y-values. This is a different form of the transformation.

Let’s work with point A first. Since it will be a horizontal reflection, where the reflection is over x=-3, we first need to determine the distance of the x-value of point A to the line of reflection. We’ll be using the absolute value to determine the distance.

Since point A is located three units from the line of reflection, we would find the point three units from the line of reflection from the other side. The y-value will not be changing, so the coordinate point for point A’ would be (0, 1)

Repeat for points B and C. In the end, we found out that after a reflection over the line x=-3, the coordinate points of the image are:

A'(0,1), B'(-1,5), and C'(-1, 2)

Vertical Reflection

Apply a reflection over the line y=-1

The procedure to determine the coordinate points of the image are the same as that of the previous example with minor differences that the change will be applied to the y-value and the x-value stays the same.

In the end, we would have
A'(-6,-2), B'(-5,-7), and C'(-5, -3)

Reflection Over a Horizontal or Vertical Line

Video-Lesson Transcript

Let’s say we want to reflect this triangle over this line. A line x = -3 rather than the x-axis or the y-axis.

So, we have

r_{x = -3} \triangle{ABC}

This line is called x = -3 because anywhere on this line x = -3 and it doesn’t matter what the y value is.

Graphically, this is the same as reflecting over the y-axis.

We’re just going to treat it like we are doing reflecting over the y-axis. Let’s see how far away it is.

Point B is 2 units from the line x = -3, so we’re going 2 units to the right of it. Keep the same height. And we have B^\prime.

Same thing for points A and C.

C is 2 units away so we’re going to move 2 units horizontally and we get C^\prime.

Point A is 3 units from the line so we go 3 units to the right and we end up with A^\prime.

The reflection of triangle ABC will look like this.

Probably it’s best to do this graphically then get the coordinates from it.

A^\prime (0, 1)
B^\prime (-1, 5)
C^\prime (-1, 2)

Similarly, let’s reflect this over a vertical line.

r_{y = -1} \triangle {ABC}

This line represents y = -1 because anywhere on this line is y = -1, it doesn’t matter what the x value is.

We’ll treat this the same way as we treat everything so far in reflection.

Point A is 2 spots away from the y axis so we’ll go 2 spots below it. And we end up with A^\prime.

C is 3 spots above the line so we’ll go 3 spots below it. We end up with C^\prime.

Point B is 6 spots away so we’ll go 6 spots below y = -1 line.

Now, we can draw a triangle.

We’re able to get the new coordinates graphically by graphing it.

As we look at it, we can now figure out the coordinates.

Looking at the graph, we’ll see that:

A^\prime (-6, -2)
C^\prime (-5, -3)
B^\prime (-5, -7)

A quick sketch easily helps us figure out the coordinates for the image if we reflect triangleABC over the line y = -1.