Compositions and Congruence

Follow the order

A sequence gives each motion the image left by the preceding motion. For example, start at $(1,2)$, move three units right, then reflect across the horizontal axis. The slide produces $(4,2)$; the reflection acts on that new point and produces $(4,-2)$.

A composition is the single mapping obtained by applying motions in a stated order. We write the order in words: first the slide, then the reflection. Use $P'$ after the first motion and $P''$ after the second. The second motion acts on $P'$, not on the original $P$.

Each translation, reflection, and rotation preserves every distance and every angle measure. Those are the accepted slide, fold, and turn foundations. Therefore any sequence of them preserves both: each step passes the same length and angle measure to the next step. This does not derive angle preservation from lengths.

Undo the last step first

To undo a journey, retrace its last part before its first part. A slide right $5$ followed by a slide up $2$ is undone by going down $2$, then left $5$. The inverse of a mapping returns every image point to its original point.

A reflection undoes itself because folding back pairs the same two points. An opposite turn retraces the same circle. A zero translation or identity rotation already fixes every point and is its own inverse. A half-turn also undoes itself.

Undo a composition by reversing the order and inverting each motion. The first undoing step must act on the final image to restore the intermediate image.

Two parallel mirrors make a slide

Two reflections in distinct parallel mirrors give a translation across twice the gap, directed from the first mirror toward the second. To see the doubling, first use $x=0$, then $x=3$. Point $(1,2)$ reflects to $(-1,2)$, then to $(7,2)$. Its horizontal change is $7-1=6$, twice the $3$-unit gap, toward the second mirror.

First justify using one gap everywhere. In the $3$-unit example, slide a perpendicular joining segment up or down along the mirrors until its first endpoint reaches any chosen point of the first mirror. A translation along a line leaves that line in place, as in Translations. Both mirrors have that same direction, so the slide leaves both mirror lines in place. Its second endpoint therefore remains on the second mirror.

The slide preserves the segment’s length and right angles and takes it to a straight segment. Its image is thus a $3$-unit segment perpendicular to the second mirror. The unique perpendicular foot proved in lesson 2.3 makes it exactly the joining perpendicular from the new first endpoint to that mirror.

Nothing depended on $3$ or on the mirrors being vertical. Start with any perpendicular joining segment of length $d$ and translate it along the mirrors to any other first foot. Both mirrors stay in place, and the unique-foot result identifies its image as that foot’s perpendicular joining segment. Distance preservation gives length $d$ again. Thus the positive gap $d$ is the same on every perpendicular.

Now follow both reflections along the perpendicular through any starting point. Choose signed positions with the first mirror at $0$ and the second at $d$, positive toward the second mirror. If the point starts at position $s$, the first fold gives position $-s$. The second fold places it equally far on the other side of $d$:

$s''=2d-(-s).$

$s''=s+2d.$

Nothing depended on the chosen starting position. All these perpendiculars have the same direction because the mirrors are parallel, and the position along the mirrors is unchanged. Every point therefore moves the same distance $2d$ in the same direction. No quadrilateral theorem is needed.

Reversing the mirrors reverses the translation direction. If the mirrors coincide, reflecting twice gives the identity, a zero translation.

Two intersecting mirrors make a turn

Intersecting mirrors keep their intersection fixed and together turn the plane through twice their directed angle. For example, take a first mirror directed horizontally right and a second mirror directed $35^\circ$ counterclockwise from it. A point whose ray from the intersection has direction $20^\circ$ first reflects to direction $-20^\circ$. Its second reflected direction is $90^\circ$, equally far from the second mirror’s $35^\circ$ direction. The total change is $90-20=70$ degrees counterclockwise.

To generalize that calculation, choose a ray along the first mirror and measure signed directions from it, counterclockwise positive. Let the second mirror’s chosen ray have direction $\alpha$, and a point’s ray have direction $\theta$. The first fold negates the direction. The second fold puts that ray equally far across $\alpha$:

$\theta'=-\theta.$

$\theta''=2\alpha+\theta.$

Thus every off-center ray turns through $2\alpha$. Both reflections preserve its distance from the intersection, which lies on both mirrors and stays fixed. This meets the rotation definition for every point. Choosing the opposite ray along a mirror changes the doubled turn by a full turn, so the final image is unchanged.

Reversing the mirror order reverses the directed turn. When the mirrors are perpendicular, both orders give a half-turn: $180^\circ$ clockwise and counterclockwise have the same final image. These reflections commute, meaning either order gives the same mapping. Reversing order need not change a composition.

A glide follows its mirror

A reflection followed by a nonzero slide along its mirror produces a glide reflection. For a horizontal mirror, the reflection changes height while the parallel slide changes horizontal position. These independent changes explain why the two motions commute.

A nonzero glide has no fixed point: every point changes its position along the mirror. A slide across the mirror does not meet the parallel condition, so the commuting claim above does not apply to it. Example 1 used such a perpendicular slide and obtained different orders.

Congruence means an exact rigid match

Two figures are congruent when a sequence of translations, reflections, and rotations carries one whole figure exactly onto the other. For example, a triangle reflected and then slid still matches its starting cutout exactly. The sequence changes placement but preserves every distance and every angle measure at each step.

The correspondence records which points match. If $A,B,C$ go to $D,E,F$ respectively, write $\triangle ABC\cong\triangle DEF$. This order pairs $AB$ with $DE$ and the angle at $B$ with the angle at $E$. Choosing a different letter order can make a different, incorrect claim.

A nonrigid mapping fails to preserve some distance or angle. Consider the horizontal stretch $(x,y)\mapsto(2x,y)$. It sends $(0,0)$ and $(2,0)$, distance $2$ apart, to $(0,0)$ and $(4,0)$, distance $4$ apart. Therefore it is globally nonrigid.

A vertical segment from $(0,0)$ to $(0,3)$ does keep its length under this stretch. One preserved segment does not make the plane mapping rigid; rigidity requires preservation for every pair of points and every angle. A stretch therefore cannot be included as a rigid step in the congruence definition.

Common mistakes