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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 𝐴 through line ⃖⃗𝐿 gives point 𝐴′, called its image by this reflection, such that line ⃖⃗𝐿 is the perpendicular bisector of line segment 𝐴𝐴′.

To construct 𝐴′ geometrically, we first construct a line perpendicular to ⃖⃗𝐿 that passes through 𝐴 by drawing an arc centered at 𝐴 that cuts ⃖⃗𝐿 at 𝐡 and 𝐢. Then, using the same radius, we draw arcs centered at 𝐡 and 𝐢 that intersect at 𝐴′, on the other side of the mirror line with respect to point 𝐴. As 𝐴𝐡=𝐡𝐴′=𝐴′𝐢=𝐢𝐴, quadrilateral 𝐴𝐡𝐴′𝐢 is a rhombus, which means that its diagonals are perpendicular and bisect each other. We, therefore, have that 𝐿 bisects 𝐴𝐴′. In other words, 𝐴𝐻=𝐻𝐴′, where 𝐻 is the intersection point between 𝐴𝐴′ and ⃖⃗𝐿; that is, 𝐴 and 𝐴′ 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 𝐴 through line ⃖⃗𝐿 is the point 𝐴′ such that line ⃖⃗𝐿 is the perpendicular bisector of line segment 𝐴𝐴′.

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 𝑃 across the dotted line, which of the four other points will it map onto?

  1. Point 𝑄
  2. Point 𝑆
  3. Point 𝑇
  4. Point 𝑅

Answer

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 𝑃′ is the image of 𝑃 in the reflection through ⃖⃗𝐿 if ⃖⃗𝐿 is the perpendicular bisector of 𝑃𝑃′.

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

We see that point 𝑅 is at the same perpendicular distance as 𝑃 from the dotted line. Hence, the answer is option D: point 𝑅 is the image of 𝑃 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 𝐴𝐡 through a line ⃖⃗𝐿.

We want to reflect every point on 𝐴𝐡 through line ⃖⃗𝐿. We can start with point 𝐴. We find its image 𝐴′ by sketching a line perpendicular to 𝐿 through 𝐴 such that 𝐿 bisects 𝐴𝐴′. We can do this with a compass and straightedge. We first trace a circle centered at 𝐴 that intersects 𝐿 at two distinct points, say, 𝑋 and π‘Œ.

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 𝐴′ and the point of intersection between ⃖⃗𝐿 and 𝐴𝐴′ 𝐸.

We can then note that π΄π‘‹π΄β€²π‘Œ is a rhombus, so its diagonals 𝐴𝐴′ and π‘‹π‘Œ are perpendicular bisectors of each other.

This construction of 𝐴′ will make π΄π΄β€²βŸ‚βƒ–βƒ—πΏ, and ⃖⃗𝐿 will also bisect 𝐴𝐴′. Thus, 𝐴′ is the image of 𝐴 after a reflection through 𝐿.

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

We can note that for any point 𝐢∈𝐴𝐡, its image 𝐢′ after a reflection through ⃖⃗𝐿 must lie on 𝐴′𝐡′, since the perpendicular distances from ⃖⃗𝐿 are preserved. This means that we can reflect a line segment through a line ⃖⃗𝐿 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 ⃖⃗𝐿 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 𝐴𝐡 after a reflection in ⃖⃗𝐿?

Answer

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 𝐴𝐡 after a reflection through ⃖⃗𝐿 will be 𝐴′𝐡′, where 𝐴′ and 𝐡′ are the images of 𝐴 and 𝐡 after the reflection respectively.

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

To reflect 𝐴 and 𝐡 through 𝐿, we mark the points 𝐴′ and 𝐡′ that make ⃖⃗𝐿 the perpendicular bisector of 𝐴𝐴′ and 𝐡𝐡′.

Therefore, the reflection of 𝐴𝐡 through ⃖⃗𝐿 is 𝐴′𝐡′.

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 𝐴𝐡𝐢 after a reflection in ⃖⃗𝑀?

Answer

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 𝐴𝐡𝐢 after a reflection through ⃖⃗𝑀 will be triangle 𝐴′𝐡′𝐢′, where 𝐴′, 𝐡′, and 𝐢′ are the images of 𝐴, 𝐡, and 𝐢 after the reflection respectively.

Since the positions of triangle 𝐴𝐡𝐢 and line ⃖⃗𝑀 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 ⃖⃗𝑀, we want to draw a line perpendicular to ⃖⃗𝑀 through the point such that ⃖⃗𝑀 bisects the line segment between the point and its image.

Let’s start by reflecting 𝐴 through ⃖⃗𝑀. We will do this using a construction. We start by tracing a circle centered at 𝐴 that intersects ⃖⃗𝑀 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 𝐴′.

This construction makes ⃖⃗𝑀 the perpendicular bisector of 𝐴𝐴′, so 𝐴′ must be the image of 𝐴 after reflection through ⃖⃗𝑀.

We follow the same process to find 𝐡′ and 𝐢′.

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

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 𝐴𝐡 that is parallel to the mirror line ⃖⃗𝐿, then its reflection 𝐴′𝐡′ will form a rectangle 𝐴𝐡𝐡′𝐴′.

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

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

Since 𝑋 lies on the perpendicular bisector of 𝐴𝐴′ and 𝐡𝐡′, it is equidistant from the endpoints of these line segments. So, 𝑋𝐴=𝑋𝐴′, and 𝑋𝐡=𝑋𝐡′. Since 𝑋𝐡 reflects through ⃖⃗𝐿 to be 𝑋𝐡′, we have the following.

We have 𝑋𝐡=𝑋𝐴+𝐴𝐡,𝑋𝐡′=𝑋𝐴′+𝐴′𝐡′.

We know that 𝑋𝐡=𝑋𝐡′, so we can equate the right-hand sides of the equations to get 𝑋𝐴+𝐴𝐡=𝑋𝐴′+𝐴′𝐡′.

We also know that 𝑋𝐴=𝑋𝐴′; hence, 𝐴𝐡=𝐴′𝐡′.

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 𝐿, 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, △𝐴′𝐡′𝐢′ is the image of △𝐴𝐡𝐢 by a reflection in ⃖⃗𝐿.

  1. Fill in the blanks: The length of 𝐴′𝐢′=cm, and the length of 𝐴′𝐡′=cm.
  2. Fill in the blanks: 𝐴𝐴′ is to 𝐡𝐡′, and 𝐢𝐢′ is to ⃖⃗𝐿.
  3. Find the measure of ∠𝐴.

Answer

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

Part 1

Since triangles 𝐴𝐡𝐢 and 𝐴′𝐡′𝐢′ are congruent, then their corresponding sides are congruent. Thus, 𝐴′𝐢′=𝐴𝐢=4,𝐴′𝐡′=𝐴𝐡=6.cmcm

This gives us that the length of 𝐴′𝐢′=4cm and the length of 𝐴′𝐡′=6cm.

Part 2

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

Hence, ⃖⃗𝐿 is the perpendicular bisector of both 𝐴𝐴′ and 𝐡𝐡′.

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

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

This gives us that 𝐴𝐴′ is parallel to 𝐡𝐡′, and 𝐢𝐢′ is perpendicular to ⃖⃗𝐿.

Part 3

Since triangles 𝐴𝐡𝐢 and 𝐴′𝐡′𝐢′ are congruent, then their corresponding angles are congruent. Thus, π‘šβˆ π΄=π‘šβˆ π΄β€²=31.∘

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 𝑀 and a radius 𝑀𝐴. When we reflect 𝑀𝐴 through ⃖⃗𝐿, the length of 𝑀𝐴 is preserved. So, we get a circle centered at 𝑀′ with radius 𝑀𝐴.

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 𝑀 that intersects with ⃖⃗𝐿 at points 𝐴 and 𝐡, draw an image of circle 𝑀 after a reflection in ⃖⃗𝐿.

Which of the following statements is correct?

  1. 𝐴𝑀⫽𝐴𝑀′
  2. 𝐡𝑀⫽𝐡𝑀′
  3. π‘€π‘€β€²βŸ‚π΄π΅
  4. 𝑀𝑀′=𝐴𝐡
  5. 𝐴′𝐡′>𝐴𝐡

Answer

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

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

We can reflect 𝑀 through line ⃖⃗𝐿 by tracing a circle centered at 𝑀 that intercepts ⃖⃗𝐿 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 𝑀′. Finally, we trace a circle of radius 𝑀𝐴 centered at 𝑀′.

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

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

Applying this to 𝑀, we have that ⃖⃗𝐿 is the perpendicular bisector of 𝑀𝑀′. Since 𝐴𝐡 is contained in ⃖⃗𝐿, we must have that π‘€π‘€β€²βŸ‚π΄π΅, 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 ⃖⃗𝐿 is such that the line ⃖⃗𝐿 is the perpendicular bisector of the line segment between the point and its image. In particular, if π΄βˆˆβƒ–βƒ—πΏ, then its image 𝐴′ under reflection through ⃖⃗𝐿 is coincident with 𝐴.
  • We can reflect a point 𝐴 through a line ⃖⃗𝐿 by tracing a circle centered at the point that intercepts ⃖⃗𝐿 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 𝐴′ so that ⃖⃗𝐿 is the perpendicular bisector of 𝐴𝐴′.
  • 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.

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