Define rotations, reflections and translations precisely in terms of angles, circles, perpendicular lines, parallel lines and line segments

Learning goal: Develop precise definitions of rotation, reflection and translation using only angles, circles, perpendicular lines, parallel lines and line segments, and use each definition to locate the image of a point and to check a claimed image.

From experience to definition. Sliding, flipping and turning a transparency give a feel for rigid motions, but a feel is not a definition. A definition must say, for every point P in the plane, exactly where its image P′ is, in terms already defined. The three definitions below do that, and each one uses the vocabulary of the earlier lessons: perpendicular lines and segments for reflections, circles and angles for rotations, parallel lines and segments for translations.

Reflection across a line. The reflection across line m is the transformation that sends each point P not on m to the point P′ such that m is the perpendicular bisector of segment PP′, and sends each point on m to itself. Unpacking "perpendicular bisector": the segment PP′ is perpendicular to m, and m passes through the midpoint of PP′, so P and P′ are the same distance from m on opposite sides. This is a complete instruction. To locate P′, drop the perpendicular from P to m, find the foot F, and continue the same distance PF beyond F. To check a claimed image, verify both conditions: the midpoint of PP′ lies on m, and PP′ is perpendicular to m. Checking only one is a common error: the point directly across but at the wrong distance has the perpendicular property but not the midpoint property, and a point at the right distance on the other side but slid along the line has the midpoint property but not the perpendicular one (reflecting P(2, 5) across y = 1, the point (4, −3) gives midpoint (3, 1) on the line, yet the segment is not vertical).

A worked reflection. Reflect P(4, 1) across the line y = x. Claimed image: P′(1, 4). Midpoint of PP′: ((4 + 1) ÷ 2, (1 + 4) ÷ 2) = (2.5, 2.5), and 2.5 = 2.5, so the midpoint lies on y = x. Slope of PP′: (4 − 1) ÷ (1 − 4) = 3 ÷ (−3) = −1, and the line y = x has slope 1; the product is −1, so PP′ is perpendicular to the line. Both conditions hold and y = x is the perpendicular bisector of PP′, so (1, 4) is the image. A second example without a grid rule: reflect P(2, 5) across the horizontal line y = 1. The perpendicular from P to the line is vertical, meeting it at F(2, 1); PF = 4, so continue 4 more units to P′(2, −3). The midpoint of PP′ is (2, 1), on the line, as required.

Rotation about a point. The rotation about center C through angle θ (counterclockwise unless stated otherwise) sends C to itself and sends every other point P to the point P′ that satisfies two conditions: P′ lies on the circle with center C and radius CP, so CP′ = CP; and the angle PCP′, measured counterclockwise from ray CP to ray CP′, equals θ. The circle condition fixes the distance; the angle condition fixes the position on the circle. Both are needed: the circle alone leaves infinitely many candidates, and the angle alone would allow any point on ray CP′. A rotation through 180° is a half turn: P′ is the point on the far side of C with CP′ = CP, so C is the midpoint of PP′.

A worked rotation. Rotate P(3, 1) about the origin through 90° counterclockwise. Claimed image: P′(−1, 3). Circle condition: OP = √(9 + 1) = √10 and OP′ = √(1 + 9) = √10, equal. Angle condition: the vectors (3, 1) and (−1, 3) have dot product 3 × (−1) + 1 × 3 = −3 + 3 = 0, so the angle between them is 90° (the dot product of displacements (a, b) and (c, d) is ac + bd, and it is 0 exactly when they are perpendicular: for non-vertical displacements, ac + bd = 0 is the slope test (b/a)(d/c) = −1 rearranged, and the form ac + bd = 0 also covers a vertical displacement); and (−1, 3) is reached from (3, 1) by turning counterclockwise (the x-coordinate becomes the y-coordinate, matching the rule (x, y) → (−y, x)). Both conditions hold. To rotate about a center other than the origin, say P(5, 2) about C(2, 1) through 90° counterclockwise, work with the displacement from C: P − C = (3, 1), which rotates to (−1, 3), and P′ = C + (−1, 3) = (1, 4). Check: CP = √(9 + 1) = √10 and CP′ = √(1 + 9) = √10; dot product of (3, 1) and (−1, 3) is 0.

Translation along a segment. The translation by the directed segment from A to B sends every point P to the point P′ such that segment PP′ is parallel to segment AB (or lies along line AB, when P does), has the same length as AB and points in the same direction (from P toward P′ as from A toward B). Equivalently, when P is not on line AB, the four points A, B, P′, P form a parallelogram ABP′P. Parallel lines carry the direction, the segment carries the distance, and together they say "move the same amount the same way". If P lies on line AB the parallelogram collapses, but the instruction is the same: move along the line by distance AB in the direction from A to B. On a grid the translation by the segment from A(1, 2) to B(4, 3) is the rule (x, y) → (x + 3, y + 1), and P(0, 0) goes to P′(3, 1); PP′ has run 3 and rise 1, exactly the run and rise of AB, so it is parallel, equal in length and in the same direction.

Why these are the right definitions. Each definition produces exactly one image for each point, so each transformation really is a function of the plane. Each also preserves distance, though proving that is the work of a later lesson; here the goal is to see that the descriptions are precise enough to be proved about. An instruction such as "turn it a bit" or "flip it over" cannot be proved about because it does not say where any point goes.

The misconception. Students often define a reflection as "the same picture on the other side" and a rotation as "turning". Neither locates a single point. The test of a definition is whether two students working independently from it will place P′ in the same spot; the perpendicular bisector condition and the circle-plus-angle condition pass that test, and the informal phrases do not.

Worked, reflection across a vertical line: Reflect Q(7, 3) across the line x = 2. The perpendicular from Q to the line is horizontal, meeting it at F(2, 3); QF = 5. Continue 5 units past F to Q′(−3, 3). Check: midpoint of QQ′ is (2, 3), on the line; QQ′ is horizontal, perpendicular to the vertical line. Both conditions hold, so x = 2 is the perpendicular bisector of QQ′.

Worked, rotation of 180°: Rotate R(6, −2) about C(1, 1) through 180°. For a half turn C is the midpoint of RR′, so R′ = (2 × 1 − 6, 2 × 1 − (−2)) = (−4, 4). Check the circle: CR = √(25 + 9) = √34 and CR′ = √(25 + 9) = √34. Check the angle: the displacements (5, −3) and (−5, 3) point in exactly opposite directions, which is a 180° turn.

Worked, translation by a segment: Translate the triangle with vertices (0, 0), (2, 0), (0, 3) by the segment from A(1, 1) to B(4, −1). The run is 3 and the rise is −2, so the rule is (x, y) → (x + 3, y − 2), giving (3, −2), (5, −2), (3, 1). For the vertex (0, 3) and its image (3, 1), the segment joining them has run 3 and rise −2, the same as AB, so it is parallel to AB, has the same length √13 and the same direction; the points A, B, (3, 1), (0, 3) form a parallelogram.

Worked, rejecting a wrong image: A student claims the reflection of P(1, 5) across the line y = 2 is (1, 0). Midpoint of P and (1, 0): (1, 2.5), which is not on y = 2, so the line is not the perpendicular bisector and the claim is wrong. The correct image is 3 units below the line, at (1, −1): midpoint (1, 2) is on the line and the segment is vertical, perpendicular to the horizontal line.

Recording task: On grid paper draw the line y = x and the points P(4, 1), Q(−2, 3) and R(5, 5). Using only the definition, construct the reflection of each point: drop a perpendicular, mark the foot, continue the same distance. Record a table with the point, its image, the midpoint of the segment joining them and the slope of that segment; expected images P′(1, 4), Q′(3, −2), R′(5, 5), with R on the line and therefore fixed. Then draw a circle of radius 3 about C(2, 2), mark P(5, 2) on it and mark the images of P under rotations about C of 90°, 180° and 270° counterclockwise, which are (2, 5), (−1, 2) and (2, −1), checking each with the circle and angle conditions.

Independent performance: Write the three definitions in your own words, then exchange a set of five points and instructions with a partner (for example "reflect (3, −4) across y = −1", "rotate (2, 6) about (2, 1) through 90° clockwise", "translate (−1, 0) by the segment from (0, 0) to (4, 5)") and locate each image using only the definitions, without memorized coordinate rules. Success criteria: every image is correct; each reflection is checked by showing the midpoint is on the line and the segment is perpendicular to it; each rotation is checked by showing equal distances to the center and the correct angle (by dot product or by a protractor); each translation is checked by comparing run and rise with the defining segment; and a closing paragraph explains why "flip it over the line" is not a definition and what the perpendicular bisector condition adds.