HSG.CO.A.4: Defining Rotations, Reflections, and Translations
In plain English: HSG.CO.A.4 is the Common Core geometry standard that asks students to write precise definitions of rotations, reflections and translations using angles, circles, perpendicular lines, parallel lines and line segments. For example, a reflection across a line makes that line the perpendicular bisector of every segment from a point to its image. It is usually taught early in high school Geometry.
Develop definitions of rotations, reflections, and translations in terms of angles, circles, perpendicular lines, parallel lines, and line segments.
Common Core State Standards for Mathematics · Domain: Congruence (CO) · Cluster: Experiment with transformations in the plane Also written as HSG-CO.A.4 or G-CO.4 · Official standard
Students move from informal words such as slide, turn and flip to precise definitions of the three rigid transformations. Each definition is built from terms students already know from HSG.CO.A.1: a rotation uses a circle and an angle, a reflection uses a perpendicular bisector, and a translation uses parallel, congruent, same-direction line segments. The goal is a definition precise enough that two people who follow it always get the same image point.
Students test draft definitions against counterexamples, repair them, and then use the finished definitions on the coordinate plane: they check whether a given point is really the image, and they work backward from a point and its image to the center, the line of reflection or the translation.
Learning Objectives
By the end of this lesson, students will be able to:
Define a rotation about a point O through an angle θ using a circle centered at O and the angle ∠POP′
Define a reflection across a line ℓ as the transformation that makes ℓ the perpendicular bisector of each segment PP′, with points on ℓ fixed
Define a translation along a directed segment AB using segments PP′ that are parallel to AB, congruent to AB and point the same way
Find a counterexample to an incomplete definition and repair the definition
Use a definition to decide whether a point is the image of another point, and to find the line of reflection from a point and its image
Prior Knowledge Required
Students should already be comfortable with:
Precise definitions of angle, circle, perpendicular line, parallel line and line segment HSG.CO.A.1
Transformations as functions that take points to points HSG.CO.A.2
Informal properties of rotations, reflections and translations 8.G.A.1
Midpoint and slope formulas on the coordinate plane, and the fact that perpendicular slopes multiply to -1
Give each pair a sheet of tracing paper, a small scalene triangle on grid paper, and three images of it: one slid, one turned and one flipped. Then post the prompt:
Warm-Up Prompt
"Your partner cannot see the paper. Write instructions that tell them exactly where vertex P goes for each of the three images. Your instructions must work for any point, not only P."
Collect a few sets of instructions and read them aloud. Typical first drafts say "turn it a quarter turn" (turn around what point?), "flip it over the line" (how far on the other side, and in which direction?) and "slide it up and right" (how far?). Tell students that the lesson is about writing instructions that leave no choices open. That is what a definition does.
Direct Instruction20 minutes
Build each definition from the HSG.CO.A.1 vocabulary, using Diagram 1. For each one, name the tool that carries the idea: the circle keeps the distance to the center, the angle fixes how far to turn, the perpendicular bisector fixes both the direction and the distance of a flip, and parallel segments of equal length fix a slide.
Rotation about point O through angle θ (with a stated direction, usually counterclockwise): O is its own image. Every other point P goes to the point P′ on the circle centered at O through P for which m∠POP′ = θ, measured in the stated direction. So OP′ = OP.
Reflection across line ℓ: every point on ℓ is its own image. Every point P not on ℓ goes to the point P′ for which ℓ is the perpendicular bisector of segment PP′.
Translation along directed segment AB: every point P goes to the point P′ for which segment PP′ is congruent to AB, parallel to AB (or on the same line as AB) and points in the same direction as A to B. When P is not on line AB, ABP′P is a parallelogram.
Test for precision: for every input point, the definition must give exactly one output point. If you can draw two different points that both fit, the definition is missing a condition.
Work through the examples below. For each one, check the answer against the definition, not against a memorized coordinate rule.
Reflection: check with the perpendicular bisector
Reflect P(1, 4) across the line ℓ: y = x - 1. Find P′ and show that ℓ is the perpendicular bisector of PP′.
Equation: P′ = (5, 0). The midpoint of PP′ is (3, 2), which is on ℓ because 2 = 3 - 1, and the slope of PP′ is -1, perpendicular to the slope 1 of ℓ
Rotation: circle and angle
Rotate P(5, 2) by 90° counterclockwise about O(1, 1). Use the circle and the angle to check the image.
Equation: P′ = (0, 5). OP = OP′ = √17, and the slopes of OP (1/4) and OP′ (-4) multiply to -1, so m∠POP′ = 90°
Translation: parallel, congruent segments
The translation along the directed segment from D(-2, 1) to E(3, 3) maps P(-1, -3) to P′. Find P′ and show that DEP′P is a parallelogram.
Equation: P′ = (4, -1). The diagonals DP′ and EP have the same midpoint (1, 0), so DEP′P is a parallelogram and PP′ ∥ DE with PP′ = DE = √29
Working backward: find the line of reflection
A reflection maps A(2, 1) to A′(4, 5). Find the line of reflection.
Equation: ℓ is the perpendicular bisector of AA′: it passes through M(3, 3) with slope -1/2, so ℓ: y = -x/2 + 9/2
Close the direct instruction by comparing fixed points. A rotation through an angle that is not a multiple of 360° fixes only its center. A reflection fixes every point of its line. A translation along a segment of nonzero length fixes no points. Students use these facts later to recognize a transformation from its effect.
Guided Practice15 minutes
Pairs receive four draft definitions. For each draft, they draw a counterexample (two different points that both satisfy the draft, or an image the draft allows that is clearly wrong) and then repair the draft:
"A rotation about O maps P to a point P′ with OP′ = OP." (No angle: every point on the circle qualifies.)
"A reflection across ℓ maps P to a point P′ on the other side of ℓ at the same distance from ℓ." (No perpendicular: P′ can slide along a line parallel to ℓ.)
"A translation moves every point the same distance." (No direction: two points could move the same distance in different directions.)
"A reflection across ℓ maps P to a point P′ so that PP′ is perpendicular to ℓ." (No bisector: P′ can be any point on the perpendicular.)
Then pairs use the repaired reflection definition on one coordinate case: is B′(-1, 6) the image of B(3, 2) under the reflection across y = x + 3? (Yes: the midpoint (1, 4) lies on the line, and the slope of BB′ is -1, perpendicular to slope 1.) Listen for pairs who check only one of the two conditions.
Independent Practice10-15 minutes
Students work alone on three problems and justify each answer with the matching definition:
Rotate P(5, 1) by 90° clockwise about C(2, -1). (P′(4, -4): CP = CP′ = √13 and the slopes 2/3 and -3/2 multiply to -1.)
Translate K(2, -2) along the directed segment from M(1, 1) to N(-3, 4). (K′(-2, 1): KK′ and MN both move 4 left and 3 up, so they are parallel, congruent and point the same way.)
A reflection maps R(-3, 2) to R′(1, -2). Find the line of reflection. (Midpoint (-1, 0), slope of RR′ = -1, so ℓ: y = x + 1.)
Closure5 minutes
Exit ticket: (1) Write the definition of a reflection across line ℓ in one sentence that uses the words "perpendicular bisector". (2) A rotation about O maps P to P′. Name two facts about OP, OP′ and ∠POP′ that must be true. (3) Which of the three transformations can have no fixed points at all? (A translation along a segment of nonzero length.)
Differentiation Strategies
For Struggling Students
Give a definition frame with blanks: "A reflection across ℓ maps each point P not on ℓ to the point P′ such that ℓ is the ____ of segment PP′."
Let students fold tracing paper along ℓ and poke through P to see P′ before checking the midpoint and the right angle with an index card corner
Start the coordinate checks with a horizontal or vertical line of reflection, where the perpendicular direction is easy to see
For Advanced Students
Ask students to explain why a rotation of 180° about O can be defined without an angle direction, and why O is then the midpoint of every segment PP′
Ask for the image of a general point (x, y) under the reflection across y = x + k, and a proof from the definition that the formula works
Ask students to show that the translation along AB is the same as the translation along any other segment CD with ABDC a parallelogram
Assessment Guidance
What to Look For
Strong definitions name the fixed points and give conditions that single out one image point: a rotation needs the center, the circle (equal distance) and the angle with its direction; a reflection needs both "perpendicular" and "bisector"; a translation needs parallel, congruent and same direction. When a student checks an image, look for both conditions, not one. A student who verifies only that the midpoint lies on ℓ, or only that the slopes are perpendicular, has not yet used the whole definition.
02
Classroom Activities
3 Activities
1
Draft, Break, Repair
20 minGroups of 3
Groups write a definition for one transformation, trade it with another group, and try to break it with a counterexample drawn on tracing paper. The goal is a definition that survives every attack.
Procedure
Assign each group one transformation: rotation, reflection or translation. Groups write a definition using at least two of these terms: angle, circle, perpendicular line, parallel line, line segment
Groups trade definitions. The receiving group has 5 minutes to draw a point P and two different candidate images that both fit the definition, or an image that fits but is plainly wrong
The original group repairs its definition and names the term that fixed the problem
Repeat with a second trade so every definition is attacked twice
Discussion Questions
Which word in your final definition removed the most wrong images?
Why does a rotation definition need a direction (clockwise or counterclockwise) unless the angle is 180°?
Why must a reflection definition say what happens to points on the line itself?
Modification for Distance Learning
Groups post definitions on a shared slide. The attacking group adds a screenshot from dynamic geometry software showing two points that both satisfy the definition.
2
Construct from the Definition
20 minPairs
Pairs locate images using only the tool named in each definition, then check with tracing paper. This connects each word of the definition to a physical construction step.
Stations
Rotation: given O, P and θ = 60°, draw the circle centered at O through P with a compass, then use a protractor to mark P′ on the circle with m∠POP′ = 60° counterclockwise
Reflection: given ℓ and P, draw the line through P perpendicular to ℓ with an index card corner, mark where it meets ℓ as M, and use the compass to mark P′ with MP′ = MP
Translation: given directed segment AB and point P, draw the line through P parallel to AB (slide a ruler along an index card), then mark P′ with PP′ = AB in the direction from A to B
Check
Trace the figure and the point, then turn, flip or slide the tracing paper to see whether P lands on the constructed P′
Measure OP and OP′, MP and MP′, or PP′ and AB, and record whether they match to the nearest millimeter
Challenge Variation
Give the pair only P and P′ and ask them to construct every possible center of a 90° rotation that maps P to P′ (there are two, one for each direction), and the one line of reflection.
3
Evidence Cards
15 minPairs
Each pair sorts 8 cards. Every card gives measured facts about points and their images. Pairs decide which transformation the facts prove, or whether the facts are not enough, and cite the definition.
The 8 Cards
Card 1: OP = OP′ = 5 cm and m∠POP′ = 70° counterclockwise (rotation about O)
Card 2: line m meets PP′ at its midpoint and at a right angle (reflection across m)
Card 3: PP′ ∥ QQ′ ∥ AB, each 6 cm long, all pointing the same way (translation along AB)
Card 4: PP′ ⊥ m, but m does not pass through the midpoint of PP′ (not a reflection across m)
Card 5: OP = OP′ = 4 cm, no angle given (not enough information)
Card 6: O is the midpoint of PP′ and of QQ′ (rotation of 180° about O)
Card 7: PP′ and QQ′ are parallel and 3 cm long, but point in opposite directions (not a translation)
Card 8: PP′ and QQ′ are both perpendicular to line n, and both have their midpoints on n (reflection across n)
Discussion Questions
Which cards needed two conditions before you could decide?
What one extra fact would make Card 5 a rotation?
Cards 1 and 5 both say OP = OP′. Why does only one of them identify the transformation?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: The Three Definitions
Rotation: P′ lies on the circle centered at O through P, and m∠POP′ = θ. Reflection: ℓ is the perpendicular bisector of PP′ (right angle at M, and MP = MP′). Translation: PP′ and QQ′ are parallel to AB, congruent to AB and point the same way, so ABP′P is a parallelogram.
Diagram 2: Finding the Line of Reflection
Drawn to scale. A reflection maps A(2, 1) to A′(4, 5). By the definition, the line of reflection is the perpendicular bisector of AA′: through the midpoint M(3, 3) with slope -1/2. The point (5, 2) lies on ℓ, so the reflection leaves it where it is.
04
Homework Assignment
~30 min
HSG.CO.A.4 Homework: Defining Rotations, Reflections, and Translations
Directions: Write every definition in complete sentences. For coordinate problems, show the midpoint, slope or distance calculations that prove your answer satisfies the definition, not only a coordinate rule.
Part 1: Writing and Testing Definitions (Problems 1-3)
Write a precise definition of (a) a rotation about point O through angle θ, (b) a reflection across line ℓ and (c) a translation along directed segment AB. Underline the word or words in each definition that come from this list: angle, circle, perpendicular line, parallel line, line segment.
A student defines a reflection this way: "A reflection across line ℓ maps each point P to a point P′ on the other side of ℓ that is the same distance from ℓ." Let ℓ be the x-axis, P = (1, 3) and P′ = (4, -3). Show that P and P′ fit the student's definition but that ℓ is not the perpendicular bisector of PP′. Then rewrite the definition so this cannot happen.
Point A(5, 0) is rotated 90° counterclockwise about the origin O. (a) On which circle must A′ lie? Give its center and radius. (b) Find A′. (c) Find the image B′ of B(0, -3) under the same rotation, and verify that OB′ = OB and that ∠BOB′ is a right angle.
Part 2: Using the Definitions on the Coordinate Plane (Problems 4-6)
The translation along the directed segment from S(-1, 2) to T(4, 0) maps U(-3, -1) to U′. Find U′. Then use slopes and lengths to show that STU′U is a parallelogram.
A reflection maps G(-2, -1) to G′(4, 1). (a) Find an equation of the line of reflection. (b) Where does the reflection send H(0, 3)? Explain using the definition.
A rotation about C(1, 2) maps J(4, 2) to J′(1, 5). (a) Find the angle and direction of the rotation, and show that CJ = CJ′. (b) Find the image of K(-1, 2) under the same rotation.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Precision of Definitions
Each definition names fixed points and gives conditions that determine one image
Definitions mostly correct, one condition missing
Definitions vague or incorrect
Use of Geometric Terms
Circle, angle, perpendicular bisector and parallel segments used correctly
Terms used with one misuse
Terms missing
Coordinate Verification
Midpoints, slopes and distances shown and correct
Correct answers with incomplete verification
No verification or incorrect
Counterexample and Repair
Counterexample explained and definition fixed
Counterexample shown but repair incomplete
Missing
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Which statement is a complete definition of the reflection across line ℓ?
Answer: B
The perpendicular bisector condition fixes both the direction (perpendicular) and the distance (bisector), so each P has exactly one image, and the definition also says what happens to points on ℓ. Choice A allows P′ to slide parallel to ℓ. Choice D allows P′ to be any point on the perpendicular. Choice C describes a slide along ℓ, not a flip.
Question 2 of 20 · Multiple Choice
A rotation about O through 50° counterclockwise maps P to P′. Which pair of conditions defines P′?
Answer: D
P′ must be on the circle centered at O through P (so OP′ = OP) and must make a 50° counterclockwise angle with OP. Choice B leaves the distance open: every point on the ray at 50° qualifies. Choice A puts P′ on the same line as P, which only happens for 0° or 180°.
Question 3 of 20 · Multiple Choice
The translation along directed segment AB maps P to P′. Which condition does the definition require?
Answer: A
A translation moves every point the same distance in the same direction, which is exactly the statement that PP′ matches the directed segment AB. Choice D would make APBP′ a parallelogram with diagonals AB and PP′, so P′ would be the image of P under the 180° rotation about the midpoint of AB, not under a translation. Choice C describes a rotation about A.
Question 4 of 20 · Multiple Choice
Which points are fixed (mapped to themselves) by the reflection across line ℓ?
Answer: C
The definition states that each point on ℓ is its own image. A point not on ℓ moves to the other side of ℓ, so it is not fixed. Choice A describes a translation along a nonzero segment. Choice B confuses the reflection with a rotation, which has one fixed point.
Question 5 of 20 · Multiple Choice
What is the image of (3, -2) under the reflection across the line y = x + 2?
Answer: D
For P(3, -2) and P′(-4, 5), the midpoint is (-1/2, 3/2), which is on y = x + 2, and the slope of PP′ is 7/(-7) = -1, perpendicular to slope 1. Choice A reflects across y = x and ignores the +2. Choice B has a midpoint (-1/2, -1/2) that is not on the line.
Question 6 of 20 · Multiple Choice
What is the image of (4, -1) under the rotation of 90° counterclockwise about the origin?
Answer: B
The image (1, 4) is on the same circle (both points are √17 from O), and the slopes of the two radii, -1/4 and 4, multiply to -1, so the angle is 90°. Going from (4, -1) to (1, 4) is counterclockwise. Choice A is the 90° clockwise image. Choice C is the 180° image. Choice D is √17 from O but its radius has slope -4, which is not perpendicular to slope -1/4.
Question 7 of 20 · Multiple Choice
The translation along the directed segment from (2, 5) to (-1, 7) maps the origin to which point?
Answer: C
The segment from (2, 5) to (-1, 7) moves 3 left and 2 up, so the origin moves to (-3, 2). Choice A moves in the opposite direction, which reverses the direction of the segment. Choice B is the endpoint of the segment itself.
Question 8 of 20 · Multiple Choice
A reflection maps P(1, 1) to P′(5, 3). What is the line of reflection?
Answer: A
The line of reflection is the perpendicular bisector of PP′: it passes through the midpoint (3, 2) and has slope -2 because PP′ has slope 1/2. So y - 2 = -2(x - 3), which is y = -2x + 8. Choice B is the line through P and P′. Choice C passes through the midpoint but is not perpendicular. Choice D is perpendicular but misses the midpoint.
Question 9 of 20 · Multiple Choice
A rotation about O maps P to P′, and OP = 6. Which statement must be true?
Answer: B
The definition of a rotation keeps the distance to the center: OP′ = OP = 6, so P′ is on the circle centered at O with radius 6. Choice A is true only for a 60° rotation. Choices C and D are true only for a 180° rotation.
Question 10 of 20 · Multiple Choice
A translation maps P to P′ and Q to Q′, and P and Q are not on a line parallel to the translation. What kind of quadrilateral is PP′Q′Q?
Answer: C
By the definition, PP′ and QQ′ are both congruent and parallel to the same directed segment and point the same way. A quadrilateral with one pair of sides that are both parallel and congruent is a parallelogram. Choice B misses that PQ and P′Q′ are also parallel.
Question 11 of 20 · Multiple Choice
Which rigid transformation has exactly one fixed point?
Answer: D
A rotation through an angle that is not a multiple of 360° fixes only its center O: every other point moves along its circle. Choice A fixes a whole line of points. Choice B fixes no points.
Question 12 of 20 · Multiple Choice
A student writes: "A rotation of 70° about O maps P to the point P′ with OP′ = OP and m∠POP′ = 70°." Which condition is missing?
Answer: A
The circle centered at O through P contains two points that make a 70° angle with OP, one on each side of ray OP, so P would have two images. Naming the direction picks exactly one of them. Choice C would force the angle to be 90°. Choice D is wrong because a definition must give exactly one image for each point.
Question 13 of 20 · Multiple Choice
A rotation about C(2, 3) maps A(6, 3) to A′(2, 7). What is the rotation?
Answer: B
CA = CA′ = 4, and CA points right while CA′ points straight up, so m∠ACA′ = 90°. Turning from the right to up is counterclockwise. Choice A would send A to (2, -1). Choice C would send A to (-2, 3).
Question 14 of 20 · Multiple Choice
Is Q′(-1, 4) the image of Q(3, 0) under the reflection across the line y = x + 1?
Answer: C
Both conditions of the definition hold: 2 = 1 + 1, so the midpoint is on the line, and the slopes -1 and 1 multiply to -1. Choices A and B each claim a condition fails, but both hold. Choice D confuses reflections with rotations, which need an angle.
Question 15 of 20 · Short Answer
Write a definition of a translation along directed segment AB that uses the terms "parallel" and "line segment".
Sample answer: the translation along AB maps each point P to the point P′ such that segment PP′ is parallel to AB (or on line AB), has the same length as AB, and points in the same direction as from A to B. A complete answer includes all three conditions: parallel, same length and same direction.
Question 16 of 20 · Short Answer
A translation maps D(-2, 4) to D′(1, -1). Find the image of E(5, 0), and use the definition to explain why DD′ and EE′ are parallel and congruent.
E′ = (8, -5). The translation moves every point 3 right and 5 down. DD′ and EE′ both have slope -5/3 and length √34, so they are parallel and congruent, and both point in the same direction.
Question 17 of 20 · Short Answer
Find the image of N(4, 0) under the rotation of 180° about M(1, -2). Explain why M is the midpoint of NN′.
N′ = (-2, -4). A 180° rotation puts N′ on the circle centered at M through N, directly opposite N, so N, M and N′ are on one line and MN = MN′ = √13. That makes M the midpoint: ((4 + (-2))/2, (0 + (-4))/2) = (1, -2).
Question 18 of 20 · Short Answer
A student says a reflection across ℓ maps P to "a point P′ with PP′ perpendicular to ℓ". Let ℓ be the y-axis. Give a point P and a point P′ that fit the student's statement but are not a reflection pair, and fix the statement.
Sample answer: P(2, 1) and P′(-5, 1). PP′ is horizontal, so it is perpendicular to the y-axis, but its midpoint (-3/2, 1) is not on the y-axis. Fix: ℓ must be the perpendicular bisector of PP′, so the midpoint must also lie on ℓ. The true image of (2, 1) is (-2, 1).
Question 19 of 20 · Short Answer
Find the image of (-2, -5) under the rotation of 90° counterclockwise about the origin. Verify your answer using the circle and the angle.
(5, -2). Both points are √29 from the origin, so the image is on the right circle. The radii have slopes 5/2 and -2/5, which multiply to -1, so the angle is 90°, and moving from (-2, -5) to (5, -2) turns counterclockwise.
Question 20 of 20 · Short Answer
Find the image of (3, 4) under the reflection across the line y = -1, and explain how the definition gives the answer.
(3, -6). The segment from P to P′ must be perpendicular to the horizontal line y = -1, so it is vertical and x stays 3. The line must bisect the segment: P is 5 units above y = -1, so P′ is 5 units below, at y = -6. The midpoint (3, -1) is on the line.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.CO.A.4 mean?
It means students write their own precise definitions of rotations, reflections and translations using basic geometric objects. A rotation is described with a circle and an angle, a reflection with a perpendicular bisector, and a translation with parallel line segments of equal length and direction. The standard is about the definitions themselves, not only about moving shapes.
Is HSG.CO.A.4 taught in Geometry or Algebra?
It is a high school Geometry standard. It is usually taught in the first unit on transformations, right after the definitions of angle, circle, perpendicular and parallel lines in HSG.CO.A.1, and before congruence in HSG.CO.B.6.
How is this different from what students learned about transformations in 8th grade?
In grade 8 (8.G.A.1 to 8.G.A.3), students observe what transformations do and use coordinate rules. In HSG.CO.A.4 they explain what a transformation is, precisely enough that the definition works for any point and any line, center or segment, including ones that are not on the axes. Those definitions are what later congruence proofs rely on.
Do students still need coordinate rules such as (x, y) → (-y, x)?
They are useful shortcuts, but they are not definitions. The rule (x, y) → (-y, x) works only for a 90° counterclockwise rotation about the origin. The definition (same distance from the center, given angle) works for every center. Ask students to check a coordinate answer with the definition: equal distances and a right angle for a 90° rotation.
Why does the definition of a reflection use the perpendicular bisector?
Because it is the shortest way to say two things at once. "Perpendicular" fixes the direction from P to its image, and "bisector" fixes the distance, so exactly one point qualifies. Definitions that say only "same distance from the line" or only "perpendicular to the line" allow many wrong images.
Is a 180° rotation the same as a reflection?
No. A 180° rotation about O fixes only O and makes O the midpoint of every segment PP′. A reflection across a line fixes a whole line and reverses orientation: a figure labeled clockwise comes back labeled counterclockwise. A 180° rotation keeps the orientation. The 180° rotation is sometimes called a point reflection, which causes the confusion.
What is a directed line segment in the definition of a translation?
It is a segment with a starting point and an ending point, such as from A to B. It carries a length and a direction. The translation along AB moves every point that same length in that same direction, which is why the segments from points to their images are all parallel to AB and congruent to it.
What mistakes do students make with these definitions?
Leaving out one condition is a frequent error: a rotation without the angle direction, a reflection without "bisector", a translation without "same direction". Another is forgetting to say what happens to the center or to points on the line of reflection. When checking an image on the coordinate plane, many students test only one condition, such as the midpoint, and stop.
How is HSG.CO.A.4 assessed?
Typical tasks ask students to choose or write a correct definition, to find the error in an incomplete one, to decide whether a given point is the image of another under a stated transformation, and to find the center or line of a transformation from a point and its image. Written justification that cites the definition carries much of the credit.
Why do these definitions matter for later geometry?
Congruence in high school geometry is defined through rigid motions (HSG.CO.B.6), and the triangle congruence criteria are explained from the properties of these definitions (HSG.CO.B.7, HSG.CO.B.8). Proofs about lines and angles (HSG.CO.C.9) also use them: for example, a point on the perpendicular bisector of a segment is equidistant from its endpoints because the reflection across that bisector swaps the endpoints.
07
Related Standards
6 standards
These standards connect to HSG.CO.A.4: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSG.CO.A.1Prerequisite
Know precise definitions of angle, circle, perpendicular and parallel lines, segments
