# Lesson Explainer: Reflections Mathematics • 11th Grade

In this explainer, we will learn how to reflect points and shapes in given reflection lines.

There are many different ways we can transform an object without altering its size or shape. We will focus on one of these specific types of transformations, called reflections.

Reflections can be thought of as mirror images. When we look in a mirror, we see a virtual image of ourselves and our surroundings, as if we were on the other side of the mirror. While it is not within the scope of this explainer to explain this optical process, we note that the virtual image of a real object obtained from a reflection in a mirror is geometrically described as the image of the real object by the geometrical transformation called reflection.

An important feature of reflection is that the real object and its virtual image are at the same distance from the mirror. We have all experienced this when moving toward a mirror or away from it; the reflected image of ourselves moves as well, closer to or further away from the mirror.

In geometry, and looking at reflections in a plane, we represent mirrors with a line called the mirror line. The reflection of a given point through line gives point , called its image by this reflection, such that line is the perpendicular bisector of line segment .

To construct geometrically, we first construct a line perpendicular to that passes through by drawing an arc centered at that cuts at and . Then, using the same radius, we draw arcs centered at and that intersect at , on the other side of the mirror line with respect to point . As , quadrilateral is a rhombus, which means that its diagonals are perpendicular and bisect each other. We, therefore, have that bisects . In other words, , where is the intersection point between and ; that is, and are at the same perpendicular distance from the mirror line.

We can use this to define the reflection of any object through a line formally as follows.

### Definition: Reflection through a Line

A reflection is a transformation that preserves the perpendicular distances of all points from the mirror line.

The reflection of a given point through line is the point such that line is the perpendicular bisector of line segment .

Let’s now see an example of using this definition to correctly identify the image of the reflection of a point through a line.

### Example 1: Reflecting a Point in a Line

Five points are plotted on the grid in the figure below.

When reflecting point across the dotted line, which of the four other points will it map onto?

1. Point
2. Point
3. Point
4. Point

We first recall that a reflection is a transformation that preserves the perpendicular distances of all points from the mirror line. In particular, we say that is the image of in the reflection through if is the perpendicular bisector of .

We can find the image of in this reflection by drawing a line perpendicular to the dotted line passing through .

We see that point is at the same perpendicular distance as from the dotted line. Hence, the answer is option D: point is the image of in the reflection through the dotted line.

We can apply the definition of reflection to more than just points. For example, a straight line or a shape is a geometric object consisting of points. This means that we can reflect these objects through a line by reflecting all of their points through the line.

To see an example of this, let’s reflect a line segment through a line .

We want to reflect every point on through line . We can start with point . We find its image by sketching a line perpendicular to through such that bisects . We can do this with a compass and straightedge. We first trace a circle centered at that intersects at two distinct points, say, and .

We then trace arcs of congruent circles centered at these points of intersection such that they intersect. It is important that we keep the radius of the circles constant for all three circles. We label the point of intersection of the circles and the point of intersection between and  .

We can then note that is a rhombus, so its diagonals and are perpendicular bisectors of each other.

This construction of will make , and will also bisect . Thus, is the image of after a reflection through .

We can follow the same process to find the image of any point of after a reflection through .

We can note that for any point , its image after a reflection through must lie on , since the perpendicular distances from are preserved. This means that we can reflect a line segment through a line by reflecting the endpoints of the line segment and then connecting their images with a line segment.

This actually extends even further. Since polygons are defined by their sides, we can reflect any polygon through a line by just reflecting the endpoints of its sides. In other words, we can reflect a polygon through a line by reflecting its vertices.

Let’s now see an example of correctly identifying the reflection of a line segment through a given mirror line.

### Example 2: Reflecting a Line Segment in a Horizontal Line

Which of the following represents the image of after a reflection in ?

We begin by recalling that a reflection is a transformation that preserves the perpendicular distances of all points from the mirror line. In particular, we can recall that we can reflect a line segment through a line by reflecting its endpoints. So, the image of after a reflection through will be , where and are the images of and after the reflection respectively.

Since and remain constant in each of the options, we can use the diagrams to reflect and to construct ourselves. We need the perpendicular distance of each point from to stay constant. We can do this using a construction; however, in this case, is horizontal, so the lines perpendicular to will be the given vertical lines. We can sketch the lines perpendicular to that pass through and as follows.

To reflect and through , we mark the points and that make the perpendicular bisector of and .

Therefore, the reflection of through is .

This matches the diagram given in option A.

In our next example, we will determine which diagram correctly reflects a triangle through a line.

### Example 3: Reflecting a Rectangle in a Line

Which of the following represents the image of triangle after a reflection in ?

We start by recalling that a reflection preserves the perpendicular distances of all points from the mirror line. In particular, we can recall that we can reflect a polygon through a line by reflecting its vertices. So, the image of triangle after a reflection through will be triangle , where , , and are the images of , , and after the reflection respectively.

Since the positions of triangle and line are constant in each of the options, we can just reflect the triangle through the line and then match our drawing with the correct option. To reflect a point through , we want to draw a line perpendicular to through the point such that bisects the line segment between the point and its image.

Let’s start by reflecting through . We will do this using a construction. We start by tracing a circle centered at that intersects at two distinct points.

While keeping the radius the same, we then trace congruent circles centered at both points of intersection. We then label the point of intersection between the circles .

This construction makes the perpendicular bisector of , so must be the image of after reflection through .

We follow the same process to find and .

Connecting the images of the points will then give us the reflection of triangle through .

We can see that this is the same triangle given in option C.

Before we move on to our next example, we can note some useful properties of reflections. We start by noting that reflecting line segments gives a congruent line segment. We can show this by noting that if we have a line segment that is parallel to the mirror line , then its reflection will form a rectangle .

So, . We can think of this as sketching the line segment on a piece of paper and then folding the paper along the mirror line to find ; the points and their respective images will lie on top of each other.

If is not parallel to , then we can extend so that it intersects the mirror line at a point .

Since lies on the perpendicular bisector of and , it is equidistant from the endpoints of these line segments. So, , and . Since reflects through to be , we have the following.

We have

We know that , so we can equate the right-hand sides of the equations to get

We also know that ; hence,

The mirror line is the perpendicular bisector of any line segment between a point and its image after reflection through the mirror line. We can think of this as folding a piece of paper along the mirror line; a point and its image will lie on top of each other after the fold. Thus, the image of a shape after reflection is congruent to the original shape, and, in particular, the angle measure is preserved.

We can also use this to consider what happens when we reflect parallel lines.

If we reflect any pair of parallel lines through a line , then their images will also be parallel. We can prove this by noting that the parallel lines can be thought of as the sides of a parallelogram, and we know that reflecting a parallelogram will give a congruent shape. Hence, the images of the lines must be parallel.

Let’s now see an example of using these properties to determine the lengths of the sides and the angle measures of a triangle using a given reflection of the triangle.

### Example 4: Solving Problems Related to a Reflected Triangle

In the following figure, is the image of by a reflection in .

1. Fill in the blanks: The length of , and the length of .
2. Fill in the blanks: is to , and is to .
3. Find the measure of .

We first recall that a reflection leaves angle measures and line segment lengths unchanged. In particular, we know that triangles and must be congruent.

Part 1

Since triangles and are congruent, then their corresponding sides are congruent. Thus,

This gives us that the length of and the length of .

Part 2

We note that for any point , its image, , is such that is the perpendicular bisector of .

Hence, is the perpendicular bisector of both and .

Since and are both perpendicular to the same line , we can conclude that they must be parallel.

Using the same property, we note that is the perpendicular bisector of , so is perpendicular to .

This gives us that is parallel to , and is perpendicular to .

Part 3

Since triangles and are congruent, then their corresponding angles are congruent. Thus,

It is worth reiterating that we can reflect any shape through a line by reflecting all of the points of the shape through the line. However, this is not usually necessary since we can often use the definitions of the shape to help us reflect it.

We have seen, for example, that reflecting a polygon gives a congruent polygon, and we only need to reflect the vertices to find the image of the polygon.

We can find a similar result if we reflect a circle through a line. Say we have a circle with center and a radius . When we reflect through , the length of is preserved. So, we get a circle centered at with radius .

Therefore, we can reflect a circle through a line by reflecting its center and then tracing a congruent circle centered at this point.

In our final example, we will determine which geometric property is correct by reflecting a circle through a line that intersects the circle.

### Example 5: Reflecting a Circle in a Line, Where the Line Intersects the Circle

Given a circle with center that intersects with at points and , draw an image of circle after a reflection in .

Which of the following statements is correct?

We first recall that we can reflect a circle with center through a line by reflecting through to get and then tracing a congruent circle centered at .

We can also note that the circle intersects at two distinct points and . We recall that when we reflect points on the mirror line, their images remain unchanged. So, is coincident with and is coincident with . We can also note that this means that , so option E is not correct.

We can reflect through line by tracing a circle centered at that intercepts at two distinct points and then, without changing the radius, tracing circles centered at these points of intersection. We then label the point of intersection between the circles . Finally, we trace a circle of radius centered at .

We can see in this instance that is not parallel to and that is not parallel to . We can show that the lengths of and do not have to be the same by noting that is twice the distance from to , whereas is just the distance between the points of intersection between and the circle.

This leaves option C. We know that for any point , its image, , after a reflection through is such that is the perpendicular bisector of for any point .

Applying this to , we have that is the perpendicular bisector of . Since is contained in , we must have that , which is option C.

Let’s finish by recapping some of the important points from this explainer.

### Key Points

• A reflection is a transformation that preserves the perpendicular distances of all points from the mirror line.
• The image of a point under reflection through a line is such that the line is the perpendicular bisector of the line segment between the point and its image. In particular, if , then its image under reflection through is coincident with .
• We can reflect a point through a line by tracing a circle centered at the point that intercepts at two distinct points and then, without changing the radius, tracing circles centered at these points of intersection. We then label the point of intersection between these circles so that is the perpendicular bisector of .
• We can reflect a line segment through a line by finding the images of the endpoints.
• We can reflect a polygon through a line by finding the images of the vertices.
• Reflecting a shape gives a congruent shape, so it preserves the lengths of the line segments and the measures of angles.
• Reflecting a pair of parallel lines will give another pair of parallel lines.
• We can reflect a circle through a line by finding the image of its center and tracing a congruent circle centered at this point.