Describe transformations as functions of the plane and compare those that preserve distance and angle with those that do not

Learning goal: Represent transformations in the plane with transparencies, graph paper and geometry software; describe a transformation as a function whose inputs are points of the plane and whose outputs are other points; and compare transformations that preserve distance and angle measure, such as a translation, with transformations that do not, such as a horizontal stretch.

A transformation is a function. In algebra a function takes a number in and gives a number out. A transformation of the plane does the same thing with points: every point of the plane is an input, and the transformation assigns to it exactly one output point, its image. We write T(P) = P′ and read it "T takes P to P′". Because the whole plane is the domain, a transformation moves everything, not just the vertices of the figure you happen to be looking at; the figure's image is the set of images of all its points. On a coordinate grid a transformation is written as a rule on coordinates, for example T(x, y) = (x + 3, y − 1). That rule is a function in exactly the algebra sense: the input is one point, an ordered pair (x, y), and the output is one point, an ordered pair, with exactly one output for each input.

Representing a transformation. Three representations are used in this course. A transparency laid over a drawing and then slid, turned or flipped shows a rigid transformation physically; the transparency cannot stretch, so what you see is guaranteed to preserve distance. Graph paper lets you apply a coordinate rule point by point and check each image by arithmetic. Dynamic geometry software lets you define a transformation once and then drag the original figure, watching the image move, which is the fastest way to test whether a property such as "the image is always the same size" really holds for every input. Each representation answers a different question; use the transparency to see, the grid to verify, and the software to explore.

Preserving distance and angle. A transformation preserves distance when, for every pair of points P and Q, the distance between the images P′ and Q′ equals the distance between P and Q. It preserves angle measure when every angle and its image have the same measure. Translations, rotations and reflections preserve both; they are called rigid motions, or isometries, and a figure and its image under a rigid motion are the same size and shape. Other transformations are just as much functions but fail one or both properties. A dilation by factor 2 keeps every angle but doubles every distance. A horizontal stretch by factor 2 changes both distances and angles. A shear, which slides each horizontal line sideways by an amount proportional to its height, keeps horizontal lengths but changes almost every other length and angle.

A worked comparison: translation versus horizontal stretch. Take triangle ABC with A(1, 1), B(3, 1) and C(1, 4). Its sides are AB = 2 (horizontal), AC = 3 (vertical) and BC, which has run 2 and rise 3, so BC = √(2² + 3²) = √13, about 3.61. The angle at C is the angle between the vertical side CA and the slanted side CB; its tangent is 2 ÷ 3, so it is about 33.7°.

Apply the translation T(x, y) = (x + 3, y − 1): A′(4, 0), B′(6, 0), C′(4, 3). A′B′ is horizontal of length 6 − 4 = 2; A′C′ is vertical of length 3; B′C′ has run 2 and rise 3, length √13. Every distance survived, so every angle did too, and the angle at C′ is still about 33.7°. The whole triangle slid 3 right and 1 down without changing shape.

Now apply the horizontal stretch S(x, y) = (2x, y) to the same triangle: A′(2, 1), B′(6, 1), C′(2, 4). A′B′ is horizontal of length 6 − 2 = 4, double the original. A′C′ is vertical of length 3, unchanged, because a horizontal stretch does not alter vertical distances. B′C′ has run 4 and rise 3, so B′C′ = √(16 + 9) = 5, compared with √13 ≈ 3.61 before. The angle at C′ now has tangent 4 ÷ 3, so it is about 53.1°, not 33.7°. S is a perfectly good function of the plane, but it does not preserve distance (AB went from 2 to 4) and does not preserve angle (C went from 33.7° to 53.1°). One counterexample of each kind is enough: a property that must hold for every pair of points fails as soon as one pair breaks it.

What "preserve" requires. Notice that the stretch left AC unchanged. A student who checked only that side would wrongly conclude that S preserves distance. Preservation is a claim about all pairs of points, so a single unchanged length proves nothing, while a single changed length disproves the claim. When testing a transformation, deliberately choose a segment that is not aligned with the direction the rule leaves alone.

The misconception. Many students believe that any transformation described by a neat rule is a rigid motion, or that "transformation" and "rigid motion" mean the same thing. They do not. Rigid motions are the special transformations that preserve distance and angle; dilations, stretches and shears are transformations that are not rigid. The distinction matters because the definition of congruence, in a later lesson, rests on rigid motions specifically. Calling a stretch a congruence would make a 2-by-3 rectangle "congruent" to a 4-by-3 rectangle, which defeats the purpose of the word.

Recognizing a rigid motion from its rule. A rule of the form (x, y) → (x + a, y + b) is a translation. The rules (x, y) → (−x, y), (x, −y), (y, x) are reflections across the y-axis, the x-axis and the line y = x. The rules (x, y) → (−y, x) and (y, −x) are rotations of 90° counterclockwise and clockwise about the origin, and (−x, −y) is a rotation of 180°. Rules that multiply one or both coordinates by a number other than 1 or −1, such as (2x, y) or (3x, 3y), or that add one coordinate to the other, such as (x + y, y), are not rigid motions. A rule (x, y) → (ax, by) is rigid exactly when a and b are each 1 or −1, which gives the identity, the two axis reflections and the 180° rotation: the points (0, 0) and (1, 0) are 1 apart, and their images are |a| apart. Do not turn this into a rule about coefficients: the rotation (x, y) → (0.8x − 0.6y, 0.6x + 0.8y), through about 36.9° about the origin, has no coefficient equal to ±1 and is still rigid (it sends (5, 0) to (4, 3), and both are 5 from the origin). The honest test remains the definition. One changed distance proves a rule is not rigid; to show a rule is rigid, compute the distance between two arbitrary points (x₁, y₁) and (x₂, y₂) and between their images, and show the two are equal for every choice.

Worked, a function on all points: Under T(x, y) = (x − 2, y + 5), find the images of (0, 0), (7, −3) and (2, −5), and the input whose image is (4, 4). T(0, 0) = (−2, 5); T(7, −3) = (5, 2); T(2, −5) = (0, 0). The origin is an output as well as an input, which shows T is a rule on the whole plane and not a rule that "moves the figure away from the origin". For the last part solve x − 2 = 4 and y + 5 = 4: the input is (6, −1). Every output has exactly one input, so T is one-to-one.

Worked, testing a dilation: D(x, y) = (3x, 3y) applied to P(1, 0), Q(0, 2) and the origin O. Images: P′(3, 0), Q′(0, 6), O′(0, 0). Distance PQ = √(1 + 4) = √5 ≈ 2.24; distance P′Q′ = √(9 + 36) = √45 ≈ 6.71, three times as long. The angle at O between OP and OQ is 90° (one ray along the x-axis, one along the y-axis), and the angle at O′ is also 90°. This one angle is an illustration, not a proof (that a dilation preserves every angle is proved in the similarity unit), but the one changed distance is proof that D does not preserve distance: it is a transformation but not a rigid motion.

Worked, testing a shear: H(x, y) = (x + y, y) applied to the square with vertices (0, 0), (2, 0), (2, 2), (0, 2). Images: (0, 0), (2, 0), (4, 2), (2, 2). The bottom side still runs from (0, 0) to (2, 0), length 2, and the top side runs from (2, 2) to (4, 2), length 2. But the left side now runs from (0, 0) to (2, 2), length √8 ≈ 2.83 instead of 2, and the angle at the origin, between the ray toward (2, 0) and the ray toward (2, 2), is 45° instead of 90°. Horizontal lengths survive; the shape is a parallelogram, not a square.

Worked, confirming a rotation is rigid: R(x, y) = (−y, x) applied to P(3, 1) and Q(5, 4). Images: P′(−1, 3), Q′(−4, 5). PQ has run 2 and rise 3, length √13. P′Q′ has run −3 and rise 2, length √(9 + 4) = √13. The distance is preserved. Checking the angle at the origin with the slope test: OP has slope 1/3 and OP′ has slope 3 ÷ (−1) = −3, and (1/3)(−3) = −1, so OP and OP′ are perpendicular. The same holds for every point (x, y) with x and y both nonzero: OP has slope y/x, OP′ to (−y, x) has slope x/(−y), and their product is −1. If x or y is 0, one segment is horizontal and the other vertical, still perpendicular. Both segments have length √(x² + y²), so every point other than the origin turns exactly 90° about the origin, as a rotation must.

Recording task: On one sheet of graph paper draw triangle ABC with A(1, 1), B(3, 1), C(1, 4). Apply the translation T(x, y) = (x + 3, y − 1) and the horizontal stretch S(x, y) = (2x, y), drawing both images in different colors. Record a table with rows AB, AC, BC and angle C, and columns "original", "after T", "after S", filling in AB = 2, 2, 4; AC = 3, 3, 3; BC = √13, √13, 5; angle C ≈ 33.7°, 33.7°, 53.1°. Then lay a transparency over the original, and record which of the two images you can reach by sliding the transparency and which you cannot, with a sentence explaining why the transparency cannot produce the other one.

Independent performance: Invent three transformation rules of your own: one that is a rigid motion, one that preserves angles but not distances, and one that preserves neither. Apply each rule to a scalene triangle of your choice with grid-point vertices, and for each rule compute two side lengths and one angle before and after. Success criteria: every rule is written as a function of (x, y); every image coordinate is correct when checked against the rule; the side you use to test distance is not one the rule leaves alone; each rule is classified correctly with the computed evidence cited; and a closing paragraph explains, in function language, why a transformation is defined on every point of the plane and not only on the vertices you drew.