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Mathematics LibreTexts

5.4: Reflection across a line

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Assume the point P and the line (AB) are given. To find the reflection P of P across (AB), one drops a perpendicular from P onto (AB), and continues it to the same distance on the other side.

According to Theorem 5.3.1, P is uniquely determined by P.
Note that P=P if and only if P(AB).

Proposition 5.4.1

Assume P is a reflection of the point P across (AB). Then AP=AP and if AP, then BAPBAP.

Proof

Note that if P(AB), then P=P. By Corollary 2.4.1, BAP=0 or π. Hence the statement follows.

截屏2021-02-04 上午9.36.19.png

If P(AB), then PP. By the construction of P, the line (AB) is a perpendicular bisector of [PP]. Therefore, according to Theorem 5.3.1, AP=AP and BP=BP. In particular, ABPABP. Therefore, BAP=±BAP.

Since PP and AP=AP, we get that BAPBAP. That is, we are let with the case

BAP=BAP.

Corollary 5.4.1

The reflection across a line is a motion of the plane. Moreover, if PQR is the reflection of PQR, then

QPRQPR.

Proof

Note that the composition of two reflections across the same line is the identity map. In particular, any reflection is a bijection.

截屏2021-02-04 上午9.42.22.png

Fix a line (AB) and two points P and Q; denote their reflections across (AB) by P and Q. Let us show that

PQ=PQ;

that is, the reflection is distance-preserving, Without loss of generality, we may assume that the points P and Q are distinct from A and B. By Proposition 5.4.1, we get that

BAPBAP,AP=AP, BAQBAQ,AQ=AQ.

It follows that

PAQPAQ.

By SAS, PAQPAQ and 5.4.1 follows. Moreover, we also get that

APQ±APQ.

From 5.4.2 and the theorem on the signs of angles of triangles (Theorem 3.3.1) we get

APQAPQ.

Repeating the same argument for a pair of points P and R, we get that

APRAPR.

Subtracting 5.4.4 from 5.4.3, we get that

QPRQPR.

Exercise 5.4.1

Show that any motion of the plane can be presented as a composition of at most three reflections across lines.

Hint

Choose an arbitrary nondegenerate triangle ABC. Suppose that ˆAˆBˆC denotes its image after the motion.

If AˆA, apply the reflection across the perpendicular bisector of [AˆA]. This reflection sends A to ˆA. Let B and C denote the reflections of B and C respectively.

If BˆB, apply the reflection across the perpendicular bisector of [BˆB]. This reflection sends B to ˆB. Note that ˆAˆB=ˆAB; that is, ˆA lies on the perpendicular bisector. Therefore, ˆA reflects to itself. Suppose that C denotes the reflection of C.

Finally, if CˆC, apply the reflection across (ˆAˆB). Note that ˆAˆC=ˆAC and ˆBˆC=ˆBC; that is, (AB) is the perpendicular bisector of [CˆC]. Therefore, this reflection sends C to ˆC.

Apply Exercise 4.4.3 to show that the composition of the constructed reflections coincides with the given motion.

The motions of plan can be divided into two types, direct and indirect. The motion f is direct if

QPR=QPR

for any PQR and P=f(P), Q=f(Q) and R=f(R); if instead we have

QPRQPR

for any PQR, then the motion f is called indirect.

Indeed, by Corollary 5.4.1, any reflection is an indirect motion. Applying the exercise above, any motion is a composition of reflections. If the number of reflections is odd then the composition indirect; if the number is even, then the motion is direct.

Exercise 5.4.2

Let X and Y be the reflections of P across the lines (AB) and (BC) respectively. Show that

XBY2ABC.

Answer

Note that XBA=ABP,PBC=CBY. Therefore,

XBYXBP+PBY2(ABP+PBC)2ABC.


This page titled 5.4: Reflection across a line is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Anton Petrunin via source content that was edited to the style and standards of the LibreTexts platform.

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