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5.4 — Parallelism Preserved and Protected
In earlier lessons you used rigid motions to reason about geometric figures. Today the focus is on one subtle question: when does a line segment point in the same direction as its transformed image? The answer depends on the transformation.
1. In class: A Solidify Understanding Task
Parallelism within a figure versus parallelism across two figures
A rigid motion moves a figure without changing lengths or angle measures. A translation slides, a rotation turns, and a reflection flips. Each maps a pair of parallel lines to another pair of parallel lines. For example, if the opposite sides of a parallelogram are parallel before a rigid motion, their images are still parallel.
But the statement “segment \(AB\) is parallel to its own image \(A'B'\)” is different. It compares the original figure with the image, not two lines within the same figure. It need not hold for a rotation or reflection.
Here, “parallel or on the same line” is the most precise wording: a transformed segment may overlap its original supporting line. Some definitions reserve parallel for distinct coplanar lines.
Translation: corresponding segments keep their direction
A translation moves every point by the same displacement. In coordinates, translating right \(p\) and up \(q\) gives \((x,y)\mapsto(x+p,y+q)\). Every segment keeps its length and direction.
Take \(A=(-3,1)\), \(B=(1,3)\). Translate by \((5,-2)\) to obtain \(A'=(2,-1)\), \(B'=(6,1)\). The directional change of \(AB\) is \((4,2)\). The directional change of \(A'B'\) is also \((4,2)\). Therefore their containing lines are parallel or coincide.
\[\boxed{\text{Translation: always parallel or collinear.}}\]Rotation: turning can change a segment’s direction
A rotation turns each segment through the same angle about a fixed center. A quarter-turn changes a horizontal segment into a vertical segment, so corresponding segments need not be parallel. A half-turn (180°) reverses every direction, which produces parallel (or collinear) segments. A full turn (360°) returns each segment to its original direction.
Rotate \(A=(-4,1)\) and \(B=(-1,3)\) through 180° about the origin. Then \(A'=(4,-1)\), \(B'=(1,-3)\). The first direction vector is \((3,2)\); the second is \((-3,-2)\). Since one vector is the negative of the other, the segments have the same line direction.
More precisely: for a nonzero segment in the plane, its rotated image is parallel or collinear with the original when the rotation angle is a multiple of 180°. Other rotation angles do not preserve the segment's direction.
\[\boxed{\text{Rotation: sometimes parallel or collinear.}}\]Reflection: two special directions remain parallel
A reflection flips a figure across a mirror line. Imagine reflecting across a horizontal mirror. A horizontal segment remains horizontal; a vertical segment remains vertical, although its direction reverses. A slanted segment usually changes to a different slope.
A segment in either of the following directions is parallel or collinear with its reflection: parallel to the mirror or perpendicular to the mirror. These directions are special because reflecting reverses one component of motion while leaving the other unchanged.
Reflect across the x-axis. Segment \(AB\) from \((1,2)\) to \((5,2)\) is horizontal and becomes \(A'B'\) from \((1,-2)\) to \((5,-2)\), also horizontal. Segment \(CD\) from \((3,1)\) to \((3,4)\) is vertical and becomes \(C'D'\) from \((3,-1)\) to \((3,-4)\), also vertical. A diagonal segment like \((1,1)\) to \((3,3)\) changes slope from 1 to −1, so it is not parallel to its image.
\[\boxed{\text{Reflection: sometimes parallel or collinear.}}\]2. Ready — Classify special quadrilaterals (problems 1–6)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Classify a quadrilateral using its stated properties and markings, not how closely the drawing resembles a familiar shape.
| Shape | Defining property |
|---|---|
| Parallelogram | Both pairs of opposite sides parallel |
| Rectangle | A parallelogram with four right angles |
| Rhombus | A parallelogram with four congruent sides |
| Square | A rectangle with four congruent sides; also a rhombus with right angles |
| Trapezoid | In this lesson, a quadrilateral with exactly one pair of parallel sides (exclusive definition) |
A figure can have multiple correct classifications. For instance, every square is a rectangle, rhombus, and parallelogram.
A quadrilateral has all four sides marked congruent and one right angle marked. It is a rhombus by equal sides. A rhombus with one right angle has four right angles, so it is a square. Thus it is also a rectangle and parallelogram.
Answer: square, rectangle, rhombus, parallelogram.
If a quadrilateral has two pairs of adjacent equal sides, but the two pairs have different lengths, we cannot conclude all four sides are equal or that opposite sides are parallel. The packet's Ready problem 5 shows this kind of kite, so none of the named classifications is forced.
Some textbooks define a trapezoid as having at least one pair of parallel sides, which would include parallelograms. Others require exactly one pair; under that convention, parallelograms are not trapezoids. The packet does not explicitly choose one. This supplement uses the exclusive definition, but your teacher's convention takes precedence. With the inclusive definition, also write “trapezoid” for parallelograms, rectangles, rhombi and squares.
1. Rectangle and parallelogram (adjacent sides have different tick marks). 2. Rhombus and parallelogram (all four sides marked equal; no right angle). 3. Trapezoid (one marked parallel pair). 4. Square, rectangle, rhombus and parallelogram. 5. None of these (kite). 6. Parallelogram (two opposite parallel pairs).
Check your work — Ready 1–6 answer key
Using the packet’s exclusive convention, a trapezoid has exactly one pair of parallel sides. Under the inclusive convention, add “trapezoid” to 1, 2, 4, and 6.
- Rectangle and parallelogram: both opposite-side pairs are congruent, and one angle is right.
- Rhombus and parallelogram: all four sides are marked congruent.
- Trapezoid: the marked parallel bases are JK and ML.
- Square, rectangle, rhombus, and parallelogram: all sides are congruent and one angle is right.
- None of the listed types; the markings describe a kite, with two unequal pairs of adjacent congruent sides.
- Parallelogram: both pairs of opposite sides are marked parallel.
3. Set — Read transformation diagrams and name parallel segments (problem 7)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Use notation such as \(\overline{FG}\parallel\overline{F'G'}\). Each endpoint in the pre-image must match the image with the same letter and a prime. The packet's drawings contain several crossing segments: crossings do not change which vertices correspond.
Translate \(\triangle PQR\) so its image is \(\triangle P'Q'R'\). The three corresponding side pairs are
\[\overline{PQ}\parallel\overline{P'Q'},\quad\overline{QR}\parallel\overline{Q'R'},\quad\overline{RP}\parallel\overline{R'P'}.\]If an original and image side lie on the same supporting line, say they are collinear instead of claiming two distinct parallel lines.
For a 180° rotation, each segment between labeled points remains parallel or collinear to the corresponding segment joining the primed points. If the diagram draws \(\overline{FH}\), \(\overline{HJ}\), and \(\overline{GJ}\), list respectively \(\overline{F'H'}\), \(\overline{H'J'}\), and \(\overline{G'J'}\) as their pairs. The fact that segments cross at other points is irrelevant.
In Set 7(c), locate the drawn segments whose direction matches that of the mirror. Each such segment is parallel to its own reflected counterpart. Name each pair using its endpoints and corresponding primed endpoints. If a segment is perpendicular to the mirror, it too has a parallel or collinear reflected counterpart, even though that broader observation is not the one emphasized in the packet statement.
Check your work — Set 7(a–c) answer key
- (a) \(\overline{FH}\parallel\overline{F\prime H\prime},\overline{HE}\parallel\overline{H\prime E\prime},\overline{EG}\parallel\overline{E\prime G\prime},\overline{GJ}\parallel\overline{G\prime J\prime},\overline{JF}\parallel\overline{J\prime F\prime}\). (b) The same five corresponding pairs are parallel after the half-turn (or lie on the same supporting line). (c) \(\overline{EH}\parallel\overline{E\prime H\prime}\); these sides are parallel to the mirror line.
4. Go — Triangle congruence and matching vertices (problems 8–11)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
First use the marks supplied in the drawing. Equal-length tick marks and equal-angle arcs are mathematical information, not decoration. Then write a congruence statement with matching letters in corresponding order. Only after that should you draw segments connecting corresponding vertices.
| Pattern | What must be marked | Caution |
|---|---|---|
| SSS | All three pairs of corresponding sides | Do not match vertices by picture orientation alone |
| SAS | Two sides and the angle between them | SSA is not a general triangle-congruence criterion |
| ASA | Two angles and the side between them | Match corresponding angle vertices first |
Suppose triangle \(ABC\) has one, two and three tick marks on sides \(AB, BC, CA\), and triangle \(DEF\) has corresponding marks on \(DE, EF, FD\). Then \(A\leftrightarrow D\), \(B\leftrightarrow E\), \(C\leftrightarrow F\), so \(\triangle ABC\cong\triangle DEF\) by SSS. The correspondence controls the order in the statement.
Suppose \(PQ\cong ST\), \(PR\cong SU\), and \(\angle QPR\cong\angle TSU\). The congruent angle is included between the congruent sides, so \(\triangle PQR\cong\triangle STU\) by SAS.
Suppose \(\angle M\cong\angle X\), \(\angle N\cong\angle Y\), and \(MN\cong XY\). The equal side lies between the two equal angles, so \(\triangle MNO\cong\triangle XYZ\) by ASA.
Connecting matching vertices: draw \(AA'\), \(BB'\), and \(CC'\). For a translation these connectors are parallel, congruent and point in the same direction; for a reflection they are perpendicular to one common mirror line and bisected by it; for a 180° rotation they pass through and are bisected by a common center. Triangle congruence alone does not force one of these connector patterns—the relative placement of the two drawn triangles matters.
Additional fully worked Go problems
In triangles \(JKL\) and \(RST\), side markings show \(JK=RS=6\), \(KL=ST=8\) and \(JL=RT=10\). State a congruence in matching order and justify it.
The sides at \(J\) have lengths 6 and 10, matching the sides at \(R\), so \(J\leftrightarrow R\). Similarly \(K\leftrightarrow S\) and \(L\leftrightarrow T\). Thus
\[\boxed{\triangle JKL\cong\triangle RST\text{ by SSS.}}\]Triangle \(PQR\) has \(PQ=7\), \(PR=9\), and \(\angle QPR=58^\circ\). Triangle \(XYZ\) has \(YX=7\), \(YZ=9\), and \(\angle XYZ=58^\circ\). Determine the correspondence, criterion, and the matching connector segments.
The equal included angles occur at \(P\) and \(Y\), so \(P\leftrightarrow Y\). The length-7 side pairs \(PQ\) and \(YX\), so \(Q\leftrightarrow X\); the remaining pair is \(R\leftrightarrow Z\). Therefore \(\triangle PQR\cong\triangle YXZ\) by SAS. Connect \(P\) to \(Y\), \(Q\) to \(X\), and \(R\) to \(Z\). Whether these connectors are parallel, concurrent or perpendicular to a common line depends on how the triangles are placed.
Two triangles have three corresponding equal sides but appear rotated in a drawing. Can you conclude that one is a 180° rotation of the other? No. SSS proves triangle congruence, but a 180° rotation is a more specific placement relationship. To establish it, the three corresponding-point connectors must have one common midpoint (the proposed center).
Check your work — Go 8–11 answer key
- \(\triangle CDB\cong\triangle EFG\), SSS. Plot/construction not shown. Connect C to E, D to F, B to G; for the translation suggested by the drawing, the joining segments should be parallel, congruent, and directed the same way. The markings prove congruence but do not specify an exact transformation.
- \(\triangle FEG\cong\triangle JLH\), SAS. Plot/construction not shown. Connect F to J, E to L, G to H; for the translation suggested by the drawing, the joining segments should be parallel, congruent, and directed the same way. The markings prove congruence but do not specify an exact transformation.
- \(\triangle FGH\cong\triangle MJK\), ASA. Plot/construction not shown. Connect F to M, G to J, H to K; for the reflection suggested by the drawing, the joining segments should be perpendicular to a common mirror line and bisected by it.
- \(\triangle FEG\cong\triangle JHL\), SAS. Plot/construction not shown. Connect F to J, E to H, G to L; a rotation consistent with the drawing would put each corresponding pair at equal distances from one center, turning each pair through the same angle. The perpendicular bisectors of the joining segments would pass through that center. The markings do not specify a center or rotation angle; a common midpoint is required only for a 180° rotation and is not established here.
5. Resources
Review Khan Academy: transformations and Khan Academy: triangle congruence if you need additional examples. The lesson's explanations and diagrams above are designed to be usable without consulting those websites.
6. Teacher's review (teachers and parents)
Section 5.4, Ready 1–6: unspecified trapezoid convention. The packet asks students to list all classes but does not say whether “trapezoid” means exactly one or at least one pair of parallel sides. This changes answers for most parallelograms, including rectangles, rhombi and squares. Specify the convention before assigning the classification questions.
Section 5.4, Go 8–11: connector question can be overgeneralized. Triangle congruence markings alone do not determine a unique motion or the pattern of segments joining corresponding vertices. The illustrated placements may suggest a transformation, but a student should justify any connector conclusion using the drawn placement or an explicitly named rigid motion, not merely SSS/SAS/ASA.
Section 5.4, Set 7(c): accurate but incomplete reflection case. A segment perpendicular to a reflection axis also has a parallel or collinear reflected image. The packet states only the parallel-to-axis sufficient condition, not an exhaustive characterization. Its claim remains true; make clear that it is a sufficient condition rather than the only case.
7. Packet coverage
| Section 5.4 component | Coverage |
|---|---|
| Main task: translations, rotations, reflections | In class |
| Ready 1–6: multiple quadrilateral classifications | Section 2 |
| Set 7(a)–(c): name parallel segments in transformation diagrams | Section 3 |
| Go 8–11: congruence, correspondence, joining vertices | Section 4 and three original worked problems |
Independent companion to Mathematics Vision Project, Secondary Mathematics II, Module 5: Geometric Figures (2017), Section 5.4. The lesson uses inline, mathematically checked SVG diagrams. The MathJax equation renderer requires an internet connection.
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