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What rigid motions preserve

A figure has been slid, flipped or turned. Which of its measurements can you still trust without measuring again, and which might have changed?

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Why this exists

Moving a figure is useful only if you know what survives the move. A tiler who has measured one tile wants every copy laid by sliding, flipping or turning to have the same edges and corners, so the pattern closes up with no gaps. The lessons on translations, reflections and rotations showed where each motion sends a point. This lesson asks what all three keep.

Definition

A rigid motion is a translation, a reflection, a rotation, or a sequence of these. If a motion sends the point A to A', then A' is the image of A, written A \mapsto A'. A property is preserved when every figure and its image share it. Rigid motions preserve:

  • lengths: a segment goes to a segment of the same length, A'B' = AB;
  • lines: a line goes to a line, and a ray to a ray;
  • angle measures: m\angle A'B'C' = m\angle ABC;
  • parallel lines: if \ell \parallel m, then \ell' \parallel m';
  • betweenness: if Q lies on \overline{PR}, then Q' lies on \overline{P'R'}.

What can change is where the figure sits, the direction each side points, and its orientation: whether going round triangle ABC in the order A, B, C runs counterclockwise or clockwise. To test it, walk from A to B; if C is on your left, the order is counterclockwise.

Intuition

-6 -4 -2 2 4 6 -6 -4 -2 2 4 6 A B C A' B' C' A' B' C' A' B' C'
Figure 1Triangle ABC and three images of it. A quarter turn about the origin (top left), a slide (bottom left) and a reflection in the x-axis (bottom right). Only the reflected copy has its labels running clockwise.

Trace a figure on tracing paper. A translation slides the paper, a rotation turns it, and a reflection turns it face down. The paper never stretches or creases, so lengths, corners, parallel edges and midpoints arrive as they left. What can change is where it lands, which way it points, and which face is up. In Figure 1 the turned and slid copies still run counterclockwise; the reflected copy, the tracing seen from the back, runs clockwise.

Results

Lengths are kept, so the image of a polygon has the same sides, perimeter and area.

Lines, betweenness, angles and parallels are kept, so a midpoint goes to the midpoint of the image, points on a line stay on a line in the same order, and perpendicular lines stay perpendicular.

What can change. Position. Direction: a translation keeps every slope, a rotation or reflection can change it. Orientation: translations and rotations keep it and a reflection reverses it, so a sequence keeps it exactly when it uses an even number of reflections.

A dilation is not rigid. A dilation enlarges or shrinks from a center. It keeps angles, parallel lines and betweenness, but multiplies every length by its scale factor, so unless that factor is 1 it is not a rigid motion.

Why it is true

On a grid, a segment has a run (change in x) and a rise (change in y). Its length is \displaystyle \sqrt{\text{run}^2 + \text{rise}^2} by the Pythagorean theorem (this is the distance formula), and its slope is \text{rise} \div \text{run}.

Translations. (x, y) \mapsto (x + a,\ y + b) adds the same a to both ends’ x-coordinates and the same b to both y-coordinates, so run and rise are unchanged: same length, same slope.

Reflections in the axes. (x, y) \mapsto (-x,\ y) changes the sign of the run and keeps the rise, and (x, y) \mapsto (x,\ -y) does the opposite. Since (-r)^2 = r^2, the length is unchanged.

Quarter and half turns. (x, y) \mapsto (-y,\ x) makes the new run the opposite of the old rise and the new rise the old run, so \text{run}^2 + \text{rise}^2 is unchanged. A half turn about any point (a, b), (x, y) \mapsto (2a - x,\ 2b - y), changes the sign of both, so it keeps every length and every slope. Reflections in other lines and turns through other angles keep lengths too; rigid motions defined precisely shows why.

Betweenness. Q lies on \overline{PR} exactly when PQ + QR = PR; a point off the segment is a detour, with PQ + QR > PR. All three lengths are kept, so Q' lies on \overline{P'R'}. So segments go to segments, and lines to lines.

Angles. Put A and C on the two rays of \angle ABC. Triangle A'B'C' has the same three sides as triangle ABC, and three sides fix a triangle’s angles, as three stiff sticks joined at their ends cannot flex. That fact, the side-side-side test, is taken as known here. So m\angle A'B'C' = m\angle ABC.

Parallels. Each motion can be undone by one of the same kind. If \ell' and m' met at X', undoing the motion would carry X' to a point on both \ell and m. Parallel lines share no point, so \ell' and m' cannot meet.

Worked examples

Example 1

A slide keeps lengths and parallel sides

Parallelogram JKLM has J(-4, 1), K(-1, 1), L(0, 3) and M(-3, 3). Translate it by (x, y) \mapsto (x + 5,\ y - 4) and show that side lengths and parallel sides are kept.

Solution

Adding 5 to each x and subtracting 4 from each y:

J'(1, -3), \quad K'(4, -3), \quad L'(5, -1), \quad M'(2, -1)
-6 -4 -2 2 4 6 -4 -2 2 4 J K L M J' K' L' M'
Figure 2Example 1. The slide moves every corner 5 right and 4 down, so every side keeps its run, its rise and its slope.

The horizontal sides all have length 3: JK = -1 - (-4) = 3 and J'K' = 4 - 1 = 3, and likewise ML = M'L' = 3. Each slanted side, before and after, has run 1 and rise 2 (from K' to L', run 5 - 4 = 1 and rise -1 - (-3) = 2), so each has length \displaystyle \sqrt{1^2 + 2^2} = \sqrt{5}.

The slopes are 0 and 2 in both figures. Equal slopes mean parallel lines, so J'K'L'M' is again a parallelogram, and every side points exactly the way it did.

Example 2

A flip keeps a right angle and reverses the order

Triangle DEF has D(1, 1), E(5, 3) and F(0, 3). Reflect it in the x-axis, (x, y) \mapsto (x,\ -y). Use slopes to show the angle at D is right before and after, and compare lengths and orientation.

Solution

The images are D'(1, -1), E'(5, -3) and F'(0, -3). The slopes at D and at D' are

\begin{aligned} DE: \tfrac{3 - 1}{5 - 1} &= \tfrac{1}{2}, & DF: \tfrac{3 - 1}{0 - 1} &= -2 \\ D'E': \tfrac{-3 - (-1)}{5 - 1} &= -\tfrac{1}{2}, & D'F': \tfrac{-3 - (-1)}{0 - 1} &= 2 \end{aligned}

Each pair multiplies to -1, so both angles are right angles. The slopes changed; the angle did not. The sides are \displaystyle \sqrt{20}, \displaystyle \sqrt{5} and 5 in both triangles (run 4 and rise \pm 2, run -1 and rise \pm 2, and a horizontal 5), and 20 + 5 = 25 = 5^2.

Walking from D up to E, the point F is above line DE, on your left: counterclockwise. Walking from D' down to E', the point F' is below, on your right: clockwise. The reflection kept every length and the right angle and reversed the orientation.

Example 3

A quarter turn keeps points in line and in order

P(2, -1), Q(4, 0), R(6, 1) and S(8, 2) lie on one line in that order, with Q the midpoint of \overline{PR}. Turn them 90^\circ counterclockwise about the origin, (x, y) \mapsto (-y,\ x). Show the images are on one line in the same order with Q' the midpoint of \overline{P'R'}, and compare slopes.

Solution

The images are P'(1, 2), Q'(0, 4), R'(-1, 6) and S'(-2, 8).

-4 -2 2 4 6 8 -2 2 4 6 8 P Q R S P' Q' R' S' O
Figure 3Example 3. The four points stay on one line, in the same order and the same distance apart, but the line now points a different way.

Each step from P' to Q' to R' to S' has run -1 and rise 2. Equal steps with equal slope -2 put the images on one line, each a step beyond the last, so the order is kept. Each step has length \displaystyle \sqrt{(-1)^2 + 2^2} = \sqrt{5}, the same as the original steps of run 2 and rise 1.

The midpoint of \overline{P'R'} is

\left(\frac{1 + (-1)}{2},\ \frac{2 + 6}{2}\right) = (0, 4)

which is Q'. But the slope went from \tfrac{1}{2} to -2, and \tfrac{1}{2} \times (-2) = -1: the turn kept the line straight and its points in order, and pointed it at right angles to before.

Example 4

Two floor tiles that make a strip

Floor tiles are shaped like trapezoid ABCD with \overline{AB} \parallel \overline{DC}, AB = 50 cm, DC = 30 cm, AD = BC = 20 cm, m\angle A = m\angle B = 60^\circ and m\angle C = m\angle D = 120^\circ. Wynn lays a second tile by giving a copy a half turn about N, the midpoint of \overline{BC}. Find the second tile’s sides and angles, and show the two tiles make a parallelogram 80 cm long.

Solution
60^\circ
120^\circ
120^\circ
60^\circ
 A B C D A' D'
Figure 4Example 4. The shaded tile is the first one turned half a turn about the midpoint of \overline{BC}. At B and at C the two corners add to 180^\circ, so the edges run straight on.

A half turn about N sends each point to the far side of N, the same distance away, so B \mapsto C and C \mapsto B. Name the second tile A'B'C'D', with B' at C and C' at B. A half turn is rigid, so the second tile has sides 50, 20, 30 and 20 cm, and angles m\angle A' = m\angle B' = 60^\circ and m\angle C' = m\angle D' = 120^\circ.

At B, the first tile’s 60^\circ and the second tile’s \angle C' of 120^\circ sit on either side of the shared edge, and

60^\circ + 120^\circ = 180^\circ

so \overline{BA} and \overline{BD'} form one straight line. The bottom edge runs AB + BD' = 50 + 30 = 80 cm. At C the corners are 120^\circ and 60^\circ, so the top edge is straight too: DC + CA' = 30 + 50 = 80 cm.

The outline AD'A'D has top and bottom on the parallel lines DC and AB, and its right end \overline{D'A'} is the half-turn image of its left end \overline{DA}, so the two have the same slope. It is a parallelogram with sides 80 cm and 20 cm.

Example 5

A dilation keeps angles but not lengths

Triangle GHI has G(1, 1), H(3, 1) and I(1, 2). Apply the dilation from the origin with scale factor 3, (x, y) \mapsto (3x,\ 3y). Which preserved properties survive?

Solution

The images are G'(3, 3), H'(9, 3) and I'(3, 6).

2 4 6 8 10 2 4 6 G H I G' H' I' O
Figure 5Example 5. The dilation by 3 from O keeps every angle and every side’s direction, but every length is three times as long.

GH = 2 but G'H' = 6, and GI = 1 but G'I' = 3. \overline{HI} has run -2 and rise 1, length \displaystyle \sqrt{5}, while \overline{H'I'} has run -6 and rise 3, length \displaystyle \sqrt{45} = 3\sqrt{5}. Every length is tripled, so this is not a rigid motion.

Yet \overline{GH} and \overline{G'H'} are both horizontal, \overline{GI} and \overline{G'I'} both vertical, and the slanted sides both have slope -\tfrac{1}{2}. Each side is parallel to its image, so every corner is made by the same pair of directions and every angle is kept. The midpoint (2, 1) of \overline{GH} goes to (6, 3), the midpoint of \overline{G'H'}. Only length fails: same shape, not same size, which is similarity.

Example 6

Which motion could it be?

Triangle KLM has K(1, 1), L(4, 1) and M(1, 3). Ulric lists four triangles, each giving the images of K, L and M in that order. For each, decide whether a single translation, reflection or rotation could have produced it, and which.

  • (a) (5, 0), (8, 0), (5, 2)
  • (b) (-1, 1), (-1, 4), (-3, 1)
  • (c) (1, -1), (4, -1), (1, -3)
  • (d) (-4, 0), (-1, 0), (-4, 3)
Solution

In KLM, \overline{KL} points right with length 3, \overline{KM} points up with length 2, and M is on the left of the walk from K to L: counterclockwise. Every rigid motion keeps lengths, only a translation keeps every direction, and only a reflection reverses orientation.

(d) From (-4, 0) to (-4, 3) is 3, but KM = 2, so no rigid motion produced (d).

(a) The sides point right 3 and up 2, as before, so directions are kept: a translation, (x, y) \mapsto (x + 4,\ y - 1), which does send K, L, M to the listed points.

(b) The sides go up 3 and left 2: lengths kept, directions not, so not a translation. Walking up from (-1, 1) to (-1, 4), the point (-3, 1) is on your left: still counterclockwise, so not a reflection. It is a rotation: (x, y) \mapsto (-y,\ x) sends K, L, M to exactly these points.

(c) Walking right from (1, -1) to (4, -1), the point (1, -3) is on your right: clockwise. Only a reflection reverses orientation, so this must be a reflection, here (x, y) \mapsto (x,\ -y).

Boundaries

Only distances within the moving figure. A point’s distance from something that stays put can change: Example 1 moves J(-4, 1), at distance \displaystyle \sqrt{17} from the origin, to J'(1, -3), at distance \displaystyle \sqrt{10}. Coordinates and slopes can change too.

Orientation is kept only without flips. A triangle and its mirror image have the same sides and angles, yet no slide or turn alone carries one onto the other with each corner going to its partner.

One kept property does not bring the others. Example 5 keeps every angle and triples every length. The stretch (x, y) \mapsto (x,\ 4y) keeps parallel lines parallel but sends the perpendicular lines y = x and y = -x to y = 4x and y = -4x, whose slopes multiply to -16.

Where this goes next

These properties are what make congruence work: when rigid motions carry one figure onto another, matching sides and angles are at once equal. Rigid motions defined precisely proves the list for every reflection and rotation, and adding dilations gives similarity. The angle facts of parallel lines and transversals are rigid motions at work: sliding one crossing along the transversal onto the other carries each angle onto its partner.

Exercises

What a rigid motion keeps

  1. A reflection carries \triangle UVW onto \triangle U'V'W'. UV = 9 cm, VW = 7 cm and m\angle V = 48^\circ. Find U'V', V'W' and m\angle V'.
  2. Q lies on \overline{PR} with PQ = 4 and QR = 6. A rotation carries P, Q, R to P', Q', R'. Find P'R' and say where Q' lies.
  3. Lines \ell and m are parallel. A 90^\circ rotation carries them to \ell' and m'. Is \ell' \parallel m'? Is \ell' \parallel \ell?
  4. True or false: (a) a translation keeps every slope; (b) a reflection keeps orientation; (c) a rigid motion keeps area; (d) a half turn keeps every slope.

Checking on coordinates

  1. Translate P(-3, 2) and Q(1, 5) by (x, y) \mapsto (x + 2,\ y - 6) and show P'Q' = PQ.
  2. Reflect triangle T(2, 0), U(6, 0), V(6, 3) in the y-axis, (x, y) \mapsto (-x,\ y). Find the images, the side lengths of both triangles, and each orientation.
  3. Rotate the segment from A(1, 3) to B(5, 5) by 90^\circ clockwise about the origin, (x, y) \mapsto (y,\ -x). Compare the lengths and slopes of \overline{AB} and \overline{A'B'}.
  4. The parallel lines y = 2x + 1 and y = 2x - 4 pass through (0, 1) and (1, 3), and through (0, -4) and (2, 0). Reflect the four points in the x-axis, show the image lines are parallel, and give their equations.
  5. M(1, 3) is the midpoint of \overline{CD}, with C(-2, 1) and D(4, 5). Translate all three by (x, y) \mapsto (x - 3,\ y + 2) and show M' is the midpoint of \overline{C'D'}.
  6. Use slopes to show E(1, 2), F(3, 1), G(4, 3) make a right angle at F. Give them a half turn about the origin, (x, y) \mapsto (-x,\ -y), and show \angle F' is still right.
  7. Reflect (0, 1), (2, 2) and (6, 4) in the x-axis. Show the images are on one line, with the image of (2, 2) between the others.

Dilations for contrast

  1. Apply (x, y) \mapsto \left(\tfrac{1}{2}x,\ \tfrac{1}{2}y\right) to J(2, 4), K(6, 4), L(6, 0). Which of the lengths, the right angle at K, and the slope of \overline{JL} are kept?
  2. A 5 by 3 rectangle is dilated with scale factor 2. Which of its side lengths, angles, parallel sides, perimeter and area are kept?

Floor tiles

  1. A tile is a parallelogram with sides 12 cm and 8 cm and one angle of 75^\circ. Tamsin-Jo lays the next tile as its reflection in a 12 cm side. Find the new tile’s sides and angles.
  2. Xenia’s tiles are 20 cm by 10 cm and Yoel’s are 30 cm by 15 cm. Could a rigid motion carry one of Xenia’s onto one of Yoel’s? Could a dilation, followed by a rigid motion if needed, and with what scale factor?

Reasoning

  1. Explain why a translation keeps every slope but a quarter turn changes the slope of every slanted segment.
  2. Vesna says the triangles TUV and T'U'V' of Exercise 6 could be matched, T to T', U to U' and V to V', by slides and turns alone. Explain why she is wrong.
  3. Explain why a rigid motion carries the midpoint of a segment to the midpoint of its image.
  4. Explain why two parallel lines cannot meet after a rotation.
  5. A rule keeps every angle measure and keeps parallel lines parallel. Must it be a rigid motion?
Answers
  1. U'V' = 9 cm, V'W' = 7 cm and m\angle V' = 48^\circ: a reflection keeps lengths and angles.
  2. P'R' = 10, and Q' lies on \overline{P'R'}, 4 from P' and 6 from R', since betweenness and lengths are kept.
  3. Yes, \ell' \parallel m': parallel lines go to parallel lines. No: a quarter turn turns every line through 90^\circ, so \ell' \perp \ell.
  4. (a) True: run and rise are unchanged. (b) False: it reverses orientation. (c) True: the image is the same figure in a new place. (d) True: run and rise both change sign, and \tfrac{-\text{rise}}{-\text{run}} = \tfrac{\text{rise}}{\text{run}}.
  5. P'(-1, -4) and Q'(3, -1). Both segments have run 4 and rise 3, so \displaystyle PQ = P'Q' = \sqrt{4^2 + 3^2} = 5.
  6. T'(-2, 0), U'(-6, 0) and V'(-6, 3). Both triangles have sides 4, 3 and 5. Walking from T to U, V is on your left: counterclockwise. Walking from T' to U' (leftwards), V' is on your right: clockwise.
  7. A'(3, -1) and B'(5, -5). Runs and rises 4, 2 and 2, -4 give both lengths \displaystyle \sqrt{20} = 2\sqrt{5}. The slopes are \tfrac{1}{2} and -2: length kept, slope not.
  8. Images (0, -1), (1, -3) and (0, 4), (2, 0). Both slopes are -2 and the lines cross the y-axis at different points, so they are parallel: y = -2x - 1 and y = -2x + 4.
  9. C'(-5, 3), D'(1, 7) and M'(-2, 5); the midpoint of \overline{C'D'} is \left(\tfrac{-5 + 1}{2},\ \tfrac{3 + 7}{2}\right) = (-2, 5), which is M'.
  10. Slopes EF = -\tfrac{1}{2} and FG = 2 multiply to -1. The images E'(-1, -2), F'(-3, -1) and G'(-4, -3) give slopes \tfrac{1}{-2} = -\tfrac{1}{2} and \tfrac{-2}{-1} = 2, again multiplying to -1, so \angle F' is right.
  11. (0, -1), (2, -2) and (6, -4). Both steps have slope -\tfrac{1}{2}, so the points are on one line, and the step lengths \displaystyle \sqrt{5} and \displaystyle 2\sqrt{5} add to \displaystyle \sqrt{45} = 3\sqrt{5}, the whole distance, so (2, -2) is between.
  12. J'(1, 2), K'(3, 2) and L'(3, 0). Lengths are halved (JK = 4, J'K' = 2; \displaystyle JL = 4\sqrt{2}, \displaystyle J'L' = 2\sqrt{2}). The right angle is kept, and both \overline{JL} and \overline{J'L'} have slope -1.
  13. The image is 10 by 6. Kept: the right angles and parallel sides. Not kept: side lengths, perimeter (16 becomes 32) and area (15 becomes 60).
  14. Sides 12, 8, 12 and 8 cm; angles 75^\circ, 105^\circ, 75^\circ and 105^\circ, since neighboring angles of a parallelogram add to 180^\circ and 180 - 75 = 105.
  15. No rigid motion, since 20 \neq 30 and lengths are kept. Yes, a dilation with scale factor \tfrac{3}{2}, since 20 \times \tfrac{3}{2} = 30 and 10 \times \tfrac{3}{2} = 15, then a slide or turn to set it in place.
  16. A translation leaves run and rise unchanged. A quarter turn makes the new run -\text{rise} and the new rise \text{run}, so slope m becomes -\tfrac{1}{m}, which equals m only if m^2 = -1. No number does that.
  17. Slides and turns keep orientation. T, U, V run counterclockwise and T', U', V' clockwise, so only a sequence with a flip can match them corner to corner.
  18. The midpoint M of \overline{AB} lies between A and B with AM = MB. Betweenness is kept, so M' is on \overline{A'B'}, and lengths are kept, so A'M' = M'B'.
  19. If the images met at X', turning back would carry X' to a point on both original lines. The originals share no point, so the images cannot meet.
  20. No. The dilation of Example 5 keeps every angle and every pair of parallel lines but triples every length. Keeping lengths is the test.
kg1128 · updated 2026-10-09 · CC-BY-SA-4.0