Glide Reflection and Reflection Geometry: Understanding Plane Transformations

Apart from rotations and translations, there is a third type of rigid motion called reflections. In a plane, reflections act similar to mirrors. However, unlike a mirror, a figure is reflected across a line. In this lesson, the formal definition and properties of reflections will be developed.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Explore

Reflecting a Triangle

In the following applet, the vertices of

△ A B C

can be moved. Also, the slope of the line

ℓ

can be set by moving the slider point. Once everything is set, reflect

△ A B C

across line

ℓ .

Is there any relationship between

\overline{A A^{'}}

and

ℓ ?

If so, do

\overline{B B^{'}}

and

ℓ

have the same relationship? What about

\overline{C C^{'}}

and

ℓ ?

Discussion

Properties of a Reflection

As can be checked in the previous exploration, after a point

A

is reflected across a line

ℓ,

the segment connecting

A

with its image is perpendicular to

ℓ .

Additionally, the line

ℓ

intersects

\overline{A A^{'}}

at its midpoint.

A point and its reflection across a line

From these two facts, it can be concluded that

ℓ

is the perpendicular bisector of

\overline{A A^{'}} .

With this information in mind, reflections can be defined properly.

Discussion

Reflections

A reflection is a transformation in which every point of a figure is reflected across a line. The line across the points are reflected is called the line of reflection and acts like a mirror.

Triangle being reflected across a movable line

More precisely, a reflection across a line

ℓ

maps every point

A

in the plane onto its image

A^{'}

such that one of the following statements is satisfied.

If $A$ is on the line $ℓ,$ then $A$ and $A^{'}$ are the same point.
If $A$ is not on the line $ℓ,$ then $ℓ$ is the perpendicular bisector of $\overline{A A^{'}} .$

Segment AA' intersects line ell perpendicularly, and line ell bisects segment AA'. Points B and B' coincide.

Like rotations and translations, reflections are rigid motions because they preserve the side lengths and angle measures. However, reflections can change the orientation of the preimage.

Example

Reflecting a Polygon

The principal of Jefferson High wants to build a physics lab by the chemistry lab. The plan, seen from the sky, is that the new building looks like a reflection of the chemistry lab through the walkway that connects the soccer field with the library.

Perform a reflection to the chemistry lab across the walkway in order to draw the physics lab.

Answer

Hint

To reflect the chemistry lab, reflect each corner of the building. The physics lab is the quadrilateral formed by the images. Remember that the image of a point that is on the line of reflection is the same point.

Solution

For simplicity, start by labeling each corner of the quadrilateral and the walkway.

To reflect $A B C D$ across $ℓ,$ a reflection can be performed on each vertex, one at a time. For example, to reflect $A$ is a good start. To do so, follow the definition of reflections. First, draw a line perpendicular to $ℓ$ passing through $A .$

Then $A^{'}$ can be plotted as the point on line $m$ where its distance to $ℓ$ is the same as the distance from $A$ to $ℓ .$

The same steps can be applied to reflect vertices $B,$ $C,$ and $D .$ Notice that because both $C$ and $D$ are on the line $ℓ,$ their images will maintain their same point locations, respectively.

Finally, the image of $A B C D$ under a reflection across the line $ℓ$ is the quadrilateral formed by $A^{'},$ $B^{'},$ $C^{'},$ and $D^{'} .$ This quadrilateral represents the physics lab.

Discussion

Reflections Performed by Hand

Reflections can be performed by hand with the help of a straightedge and a compass.

To reflect $△ A B C$ across the line $ℓ,$ follow these three steps.

Draw an Arc Centered at

A

that Intersects

ℓ

at Two Points

Place the compass tip on vertex

A

and draw an arc that intersects the line

ℓ

at two different points. Let

P

and

Q

be these points.

Drawing an Arc centered at A that intersects ell at two points

Draw Two Arcs Centered at the Intersection Points

With the same compass setting, place the compass tip on

P

and draw an arc. Keeping the same setting, place the compass tip on

Q

and draw a second arc. The intersection point of these two arcs is the image of

A .

Drawing arcs centered at P and Q that intersect each other

Notice that both arcs need to be drawn on the side of

ℓ

not containing vertex

A .

Repeat the Previous Steps for the Other Vertices

To reflect $B$ and $C,$ repeat the two previous steps.

The image of $△ A B C$ after the reflection is the triangle formed by $A^{'},$ $B^{'},$ and $C^{'} .$

Notice that the first two steps are the same as the first two steps to construct a line perpendicular to a given line through a point.

Discussion

Reflections in the Coordinate Plane

In the coordinate plane, there is a particular relationship between the coordinates of a point and those of its image after a reflection across certain lines worth considering. These lines are the coordinate axes and lines

y = x

and

y = - x .

Investigate each relationship by using the following applet.

Applet to investigate the coordinates of a point after a reflection across the coordinate axes and the lines y=x and y=-x

Drawn from diagram, the following relations can be determined.

The image of $(a, b)$ under a reflection across the $x -$ axis is $(a, - b) .$
The image of $(a, b)$ under a reflection across the $y -$ axis is $(- a, b) .$
The image of $(a, b)$ under a reflection across the line $y = x$ is $(b, a) .$
The image of $(a, b)$ under a reflection across the line $y = - x$ is $(- b, - a) .$

Example

Identifying the Line of Reflection

Up to this point, how to perform a reflection, when already given the line of reflection, has been understood. Now, consider a case if given a figure and its image under a reflection. How can the line of reflection be found? This question is answered in the following example.

While visiting a museum, Tearrik saw a painting containing the word MATH and two pentagons. The picture caught his attention. There is some sort of reflection but he wants to know for sure.

A painting showing a pentagon and its reflection an unknown certain line

Tearrik analyzed the painting. He determined that the picture was made by performing a reflection. Show how Tearrik figured that out and draw the line of reflection used in the picture.

Answer

Hint

The line of reflection is the perpendicular bisector of any segment connecting a point to its image.

Solution

When a point is reflected across a line, its image is such that the line of reflection is the perpendicular bisector of the segment connecting the point to its image. Therefore, to find the line of reflection of the painting, start by drawing a segment that connects a vertex and its image. For instance, draw $\overline{C C^{'}} .$

Next, construct the perpendicular bisector of $\overline{C C^{'}} .$ That will represent the line of reflection used to make the painting.

Notice that drawing only one segment that connects a point to its image is enough to find the line of reflection. Before performing the reflection, the painting looked as follows.

Pop Quiz

Practicing Reflections

In the following applet, there are two possible requests.

Reflect $△ A B C$ across the given line.
Draw the line of reflection used to map $△ A B C$ onto $△ A^{'} B^{'} C^{'} .$

To reflect $△ A B C,$ place points $A^{'},$ $B^{'},$ and $C^{'}$ where they should be after the reflection is applied. To draw the line of reflection, place the two points, so they lie on the line of reflection.

Performing random reflections to random triangles

Example

Performing Reflections

In previous lessons, the composition of rotations and translations were studied. Now it is time to learn about the composition of reflections. The first case to consider is when the lines of reflection are parallel.

Using a drone, Kevin took a photo of the roof of his house, Main Street, and Euclid Sreet — they are parallel streets.

A pircufe of two vertical parallel streets: Main St. and Euclid St. The streets are separated by a median strip. On the left side of Main St., there are three buildings arranged from top to bottom: the Bank, Kevin's House, and the Bakery.

Perform the following transformations to Kevin's house.

Reflect Kevin's house across the midline of Main Street.
Reflect the obtained image across the midline of Euclid Street.

Draw both reflections over the original photo. Is there a single transformation that maps Kevin's house onto the final image?

Answer

Hint

When a figure is reflected, its orientation is flipped. What happens if the figure is reflected a second time? Draw the segments connecting the vertices of Kevin's house to the corresponding vertices of the final image. Look for a relationship between these segments.

Solution

To reflect Kevin's house across the midline of Main Street, start by drawing lines perpendicular to the line of reflection passing through each vertex of the house. Label each vertex and line for added clarity.

The image of $A$ after the reflection is a point $A^{'}$ on $ℓ_{2}$ such that $A^{'}$ and $A$ are equidistant from the midline. The images of $B,$ $C,$ and $D$ can be located in a similar way. By connecting $A^{'},$ $B^{'},$ $C^{'}$ and $D^{'},$ the image of Kevin's house after performing the first reflection can be obtained.

Applying a similar reasoning, the second reflection can be performed. After the second reflection is performed, the image of Kevin's house lies to the right of Euclid Street.

To determine whether there is a single transformation that maps Kevin's house to the final image, label the vertices of the initial polygon and its images.

From the previous diagram, the following conclusions can be drawn.

Both $A B C D$ and $A^{''} B^{''} C^{''} D^{''}$ have the same orientation, in contrast to $A B C D$ and $A^{'} B^{'} C^{'} D^{'} .$
The corresponding sides of $A B C D$ and $A^{''} B^{''} C^{''} D^{''}$ are parallel.
The vectors $A A^{''},$ $B B^{''},$ $C C^{''},$ and $D D^{''}$ are all parallel, congruent, and have the same direction.

Notice that the third statement corresponds to the definition of translations. Consequently,

A B C D

can be mapped onto

A^{''} B^{''} C^{''} D^{''}

using a single transformation, a translation.

Discussion

Reflections Across Parallel Lines

In the previous example, it was concluded that the composition of reflections in parallel lines gives the same result as a translation. This conclusion is not an isolated fact. Actually, there is a theorem that guarantees this result.

Rule

Reflections in Parallel Lines Theorem

The composition of two reflections across parallel lines is a single translation. Furthermore, the translation vector is perpendicular to both parallel lines, and its magnitude is twice the distance between the parallel lines.

Reflection of a triangle across two parallel lines

In the diagram, $△ A B C$ is first reflected across $ℓ$ and then the image is reflected across $m .$ Equivalently, the following statements hold true.

Triangle $A^{''} B^{''} C^{''}$ is a translation of triangle $A B C$ along $v .$
Vector $v$ is perpendicular to $ℓ$ and $m .$
The magnitude of $v$ is $2 d,$ where $d$ is the distance between $ℓ$ and $m .$

Proof

The proof will be developed focusing the attention on vertex $C$ and its images, but the conclusions are true for all the vertices. First, start reflecting $△ A B C$ across $ℓ .$ By definition of reflection, the line $ℓ$ is the perpendicular bisector of $\overline{C C^{'}} .$ Let $P_{1}$ be the intersection point of $ℓ$ and $\overline{C C^{'} .}$

Next, perform the reflection of $△ A^{'} B^{'} C^{'}$ across line $m .$ This time, the line $m$ is the perpendicular bisector of $\overline{C^{'} C^{''}} .$ Let $P_{2}$ be the intersection point of $m$ and $\overline{C^{'} C^{''} .}$

Since $ℓ$ and $m$ are parallel lines and $C C^{'}$ is perpendicular to $ℓ,$ by the Perpendicular Transversal Theorem, $C C^{'}$ is perpendicular to $m .$ Also, $C^{'} C^{''}$ is perpendicular to $m .$ Consequently, $C C^{'}$ and $C^{'} C^{''}$ are parallel vectors with a common point. Therefore, these vectors belong to the same line.

The points $C,$ $C^{'},$ and $C^{''}$ are collinear.

Due to collinearity, it can be concluded that

\overline{C C^{''}}

is perpendicular to

ℓ

and

m .

In addition, by the Segment Addition Postulate, the length of

\overline{C C^{''}}

can be rewritten in terms of the lengths of

\overline{C C^{'}}

and

\overline{C^{'} C^{''}} .

C C^{''} = C C^{'} + C^{'} C^{''} = (C P_{1} + P_{1} C^{'}) + (C^{'} P_{2} + P_{2} C^{''})

Remember that, by definition of reflection,

C P_{1} = P_{1} C^{'}

and

C^{'} P_{2} = P_{2} C^{''} .

Substitute these values into the equation.

C C^{''} = C P_{1} + P_{1} C^{'} + C^{'} P_{2} + P_{2} C^{''}

▼

Substitute values and evaluate

SubstituteII

$C P_{1} = P_{1} C^{'}$ , $P_{2} C^{''} = C^{'} P_{2}$

C C^{''} = P_{1} C^{'} + P_{1} C^{'} + C^{'} P_{2} + C^{'} P_{2}

AddTerms

Add terms

C C^{''} = 2 P_{1} C^{'} + 2 C^{'} P_{2}

FactorOut

Factor out $2$

C C^{''} = 2 (P_{1} C^{'} + C^{'} P_{2})

Segment Addition Postulate

C C^{''} = 2 P_{1} P_{2}

Finally, notice that

P_{1} P_{2}

is the distance between the lines, that is,

P_{1} P_{2} = d .

That way, it has been shown that

\overline{C C^{''}}

is perpendicular to

ℓ

and

m

and

C C^{''}

is twice the distance between the lines.

Generalizing

Applying the same reasoning it can be concluded that

\overline{A A^{''}}

and

\overline{B B^{''}}

are perpendicular to

ℓ

and

m .

\overline{A A^{''}} ⊥ ℓ \overline{B B^{''}} ⊥ ℓ \overline{C C^{''}} ⊥ ℓ and and and \overline{A A^{''}} ⊥ m \overline{B B^{''}} ⊥ m \overline{C C^{''}} ⊥ m

From these relations, it is obtained that

\overline{A A^{''}},

\overline{B B^{''}},

and

\overline{C C^{''}}

are parallel segments. Additionally, all these segments have the same length.

A A^{''} B B^{''} C C^{''} = 2 d = 2 d = 2 d

Last but not least, notice that

A A^{''},

B B^{''},

and

C C^{''}

all have the same direction, which is the same as pointing from the first line to the second line. Consequently,

△ A^{''} B^{''} C^{''}

is a translation of

△ A B C

along

v,

where

v

is either

A A^{''},

B B^{''},

C C^{''} .

Extra

Different Positions of

C,

C^{'},

and

C^{''}

Notice that the vertices of the preimage can be positioned differently relatively to the lines of reflection. Here, the proof was developed for

C

positioned not too far to the left of

ℓ,

which resulted in

C^{'}

lying between the two lines. However, there are several other possibilities.

The same is true for vertices

A

and

B .

Nevertheless, the claim is true for any case. One interesting fact here is that the translation vector always points in the same direction as pointing from the first line to the second line.

In the diagram above, $△ A B C$ was first reflected across $m$ and then reflected across $ℓ .$

Discussion

Reflections Across Intersecting Lines

Additionally, there is also a theorem for the case where the lines of reflection intersect each other.

Rule

Reflections in Intersecting Lines Theorem

The composition of two reflections across intersecting lines is a single rotation. Additionally, the center of rotation is the point of intersection of the lines, and the angle of rotation is twice the measure of the acute or right angle formed by the lines.

Reflection of a triangle across two intersecting lines

In the diagram, $△ A B C$ is first reflected across $ℓ$ and then the image is reflected across $m .$ The same result is obtained when $△ A B C$ is rotated by an angle of $2 α^{\circ}$ about point $P .$

Proof

Let $P$ be the intersection point between the lines $ℓ$ and $m .$ To prove that $△ A^{''} B^{''} C^{''}$ is a rotation of $△ A B C,$ the following two facts will be proved.

The point $P$ is the same distance from a vertex of $△ A B C$ as it is from the corresponding vertex of $△ A^{''} B^{''} C^{''} .$ That is, $P A = P A^{''},$ $P B = P B^{''},$ and $P C = P C^{''} .$
The angles $∠ A P A^{''},$ $∠ B P B^{''},$ and $∠ C P C^{''}$ have all a measure of $2 α,$ where $α$ is the acute angle formed by the lines of reflection.

Distance From $P$ to Each Vertex

By definition of reflection, the line $ℓ$ is the perpendicular bisector of the segments $\overline{A A^{'}},$ $\overline{B B^{'}},$ and $\overline{C C^{'}} .$

Since

P

is on

ℓ,

the Perpendicular Bisector Theorem guarantees that

P

is equidistant from the endpoints of

\overline{A A^{'}},

\overline{B B^{'}},

and

\overline{C C^{'}} .

P A P B P C = P A^{'} = P B^{'} = P C^{'}

Similarly, the line

m

is the perpendicular bisector of

\overline{A^{'} A^{''}},

\overline{B^{'} B^{''}},

and

\overline{C^{'} C^{''}} .

Therefore,

P

is equidistant from the endpoints of these segments.

In consequence, $P A^{'} = P A^{''},$ $P B^{'} = P B^{''},$ and $P C^{'} = P C^{''} .$ Finally, the Transitive Property of Equality can be used to obtain the first part of the proof.

That way, it has been shown that $P$ is the same distance from a vertex of $△ A B C$ as it is from the corresponding vertex of $△ A^{''} B^{''} C^{''} .$

Angle of Rotation

Here, it will be shown that $m ∠ A P A^{''}$ is $2 α,$ where $α$ is the acute angle formed by the lines. Let $Q$ and $T$ be the intersection points between $\overline{A A^{'}}$ and $ℓ,$ and $\overline{A^{'} A^{''}}$ and $m$ respectively.

Next, consider the right triangles $P Q A$ and $P Q A^{'} .$ Notice that their hypotenuses $\overline{P A}$ and $\overline{P A^{'}}$ are congruent as well as their legs $\overline{A Q}$ and $\overline{A^{'} Q} .$

Therefore, $△ P Q A$ and $△ P Q A^{'}$ are congruent thanks to the Hypotenuse Leg Theorem. This congruence implies that $∠ A P Q$ and $∠ Q P A^{'}$ have the same measure.

$m ∠ A P Q = m ∠ Q P A^{'} (I)$

Similarly, by the Hypotenuse Leg Theorem, $△ P T A^{'}$ and $△ P T A^{''}$ are congruent. Consequently, $∠ A^{'} P T$ and $∠ T P A^{''}$ are congruent.

$m ∠ A^{'} P T = m ∠ T P A^{''} (II)$

Now, applying the Angle Addition Postulate,

m ∠ A P A^{''}

can be rewritten as the sum of

m ∠ A P A^{'}

and

m ∠ A^{'} P A^{''} .

m ∠ A P A^{''} = m ∠ A P A^{'} + m ∠ A^{'} P A^{''}

Once more, thanks to the Angle Addition Postulate, each of the two angles on the right-hand side can be rewritten in terms of the angle measures involved in Equations (I) and (II).

m ∠ A P A^{''} = m ∠ A P A^{'} + m ∠ A^{'} P A^{''}

SubstituteII

$m ∠ A P A^{'} = m ∠ A P Q + m ∠ Q P A^{'}$ , $m ∠ A^{'} P A^{''} = m ∠ A^{'} P T + m ∠ T P A^{''}$

m ∠ A P A^{''} = m ∠ A P Q + m ∠ Q P A^{'} + m ∠ A^{'} P T + m ∠ T P A^{''}

▼

Substitute

m ∠ Q P A^{'}

for

m ∠ A P Q

and simplify

Substitute

$m ∠ A P Q = m ∠ Q P A^{'}$

m ∠ A P A^{''} = m ∠ Q P A^{'} + m ∠ Q P A^{'} + m ∠ A^{'} P T + m ∠ T P A^{''}

AddTerms

Add terms

m ∠ A P A^{''} = 2 m ∠ Q P A^{'} + m ∠ A^{'} P T + m ∠ T P A^{''}

▼

Substitute

m ∠ A^{'} P T

for

m ∠ T P A^{''}

and simplify

Substitute

$m ∠ T P A^{''} = m ∠ A^{'} P T$

m ∠ A P A^{''} = 2 m ∠ Q P A^{'} + m ∠ A^{'} P T + m ∠ A^{'} P T

AddTerms

Add terms

m ∠ A P A^{''} = 2 m ∠ Q P A^{'} + 2 m ∠ A^{'} P T

FactorOut

Factor out $2$

m ∠ A P A^{''} = 2 (m ∠ Q P A^{'} + m ∠ A^{'} P T)

Finally, notice that

m ∠ Q P A^{'} + m ∠ A^{'} P T

is the angle formed by the lines of reflection, that is,

m ∠ Q P A^{'} + m ∠ A^{'} P T = α .

Consequently, the measure of

∠ A P A^{''}

is twice the measure of the angle formed by the lines.

$m ∠ A P A^{''} = 2 α$

Applying a similar reasoning, it can be shown that

m ∠ B P B^{''}

and

m ∠ C P C^{''}

are also equal to

2 α .

This completes the proof of the fact that

△ A^{''} B^{''} C^{''}

is a rotation of

△ A B C .

Discussion

Combining Translations and Reflections

To this point, a few compositions of rigid motions have been analyzed.

The composition of rotations is a rotation or a translation.
The composition of translations is a translation.
The composition of reflections is a translation or a rotation.

In these cases, the composition of two rigid motions can be presented as a single transformation. However, not every composition of two rigid motions can be expressed as a single transformation. Such is the case for glide reflections.

Concept

Glide Reflection

A glide reflection is a transformation that combines a translation and a reflection across a line parallel to the translation vector. Since a glide reflection is a composition of rigid motions, it is also a rigid motion.

Thanks to the fact that the line of reflection and the translation vector are parallel, a glide reflection could instead be a reflection followed by a translation. That is, the image does not depend on the order of the transformations.

Example

Performing a Glide Reflection

Consider a glide reflection defined by the line

y = - x

and the vector

v = ⟨ 2, - 2 ⟩ .

Apply this glide reflection to

P (- 4, 1) .

What are the coordinates of

P^{'} ?

Hint

The order in which the transformations are applied does not affect the image. When a point $(a, b)$ is reflected across the line $y = - x,$ the image has coordinates $(- b, - a) .$

Solution

Algebraic Solution

The given glide reflection is the composition of a translation along $v$ and a reflection across $y = - x .$ Start by plotting the point $P$ along with the line $y = - x$ and the vector $v = ⟨ 2, - 2 ⟩ .$

Remember, the order in which the transformations are applied does not affect the image. To begin, the translation can be performed first. To translate $P$ along $v = ⟨ 2, - 2 ⟩,$ add $2$ to the $x -$ coordinate of $P$ and $- 2$ to the $y -$ coordinate of $P .$

Point	$(a, b) \to (a + v_{1}, b + v_{2})$	Image
$P (- 4, 1)$	$(- 4, 1) \to (- 4 + 2, 1 - 2)$	$(- 2, - 1)$

Next, reflect the point $(- 2, - 1)$ across the line $y = - x .$ To do that, the coordinates are swapped and their signs are changed.

Point	$(a, b) \to (- b, - a)$	Image
$(- 2, - 1)$	$(- 2, - 1) \to (- (- 1), - (- 2))$	$(1, 2)$

Consequently, the coordinates of

P^{'}

are

(1, 2) .

Solution

Geometric Solution

Start by plotting the point $P$ along with the line $y = - x$ and the vector $v = ⟨ 2, - 2 ⟩ .$

Since the order in which the transformations defining the glide reflection are applied does not affect the image, the translation can be performed first. To translate

P,

draw the vector

v = ⟨ 2, - 2 ⟩

such that

P

is the tail of the vector. The tip of the vector is the translation of

P .

Next, reflect the obtained image across

y = - x .

To do it, draw the line perpendicular to

y = - x

passing through

(- 2, - 1) .

Then,

P^{'}

is the point on this line such that

(- 2, - 1)

and

P^{'}

are equidistant from

y = - x .

Consequently, the coordinates of

P^{'}

are

(1, 2) .

Closure

Reflections as Rigid Motions

Although there are four types of rigid motions — rotations, translations, reflections, and glide reflections — any rigid motion can be seen as a composition of some reflections. That is the case due to the two theorems previously mentioned in this lesson. Interact with the applet to review each rigid motion.

Flowchart showing that any rigid motion is the composition of reflections

Therefore, there is no need for more transformations other than reflections when talking about rigid motions. The minimum number of reflections that need to be composed can be determined by looking at the orientation of a figure and its image closely.

Since translations and rotations keep the orientation of a figure, two reflections are needed.
Since the glide reflections combine translations and reflections, three reflections are needed.

In general, any rigid motion is the composition of either one, two, or three reflections. Before moving on from this lesson, keep in mind that reflections are everywhere, so look around and identify some reflections. No, not just the one in the mirror. Think about reflections in nature or a favorite hobby!

External credits: Ethan Conley

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Catch-Up and Review

Answer

Hint

Solution

Answer

Hint

Solution

Answer

Hint

Solution

Proof

Generalizing

Extra

Proof

Distance From P to Each Vertex

Angle of Rotation

Hint

Solution

Solution

Distance From $P$ to Each Vertex