Verify experimentally that a dilation fixes lines through its center, takes other lines to parallel lines, and scales every segment by the scale factor
Several sessions: ruler-and-protractor experiments or coordinate work, a proof discussed as a class, modeled examples, independent practice and a written argument.
math-2017:G.SRT.1 Verify experimentally the properties of dilations given by a center and a scale factor:
math-2017:G.SRT.1a A dilation takes a line not passing through the center of the dilation to a parallel line and leaves a line passing through the center unchanged.
math-2017:G.SRT.1b The dilation of a line segment is longer or shorter in the ratio given by the scale factor.
Provisional targets. Clause-by-clause educational review is still pending.
📘Section 1 · LearnLearn the idea
Triangle ABC with A(1, 1), B(4, 1), C(2, 3) is dilated from the origin O with scale factor 2 to A′(2, 2), B′(8, 2), C′(4, 6); the dashed rays from O pass through each vertex and its image. AB = 3 becomes A′B′ = 6, AC = √5 becomes 2√5 and BC = 2√2 becomes 4√2, all in the ratio 2. Line AB (y = 1) goes to the parallel line y = 2, while the line through O and C (y = 1.5x) is unchanged. Axis ticks are one unit apart.
Learning goal: Verify experimentally the properties of dilations given by a center and a scale factor: a dilation takes a line not passing through the center of the dilation to a parallel line and leaves a line passing through the center unchanged, and the dilation of a line segment is longer or shorter in the ratio given by the scale factor.
The definition, with its conditions. A dilation is specified by two things: a center, which is a point O, and a scale factor, which is a positive number k. The dilation sends each point P other than O to the point P′ on the ray from O through P whose distance from O is k times the distance OP. In symbols, OP′ = k · OP, with P′ on ray OP. The center is its own image, because k · 0 = 0. If k is greater than 1 every other point moves away from the center and the figure is enlarged; if k is between 0 and 1 every other point moves toward the center and the figure is reduced; if k = 1 nothing moves. A dilation with k ≠ 1 is not a rigid motion, since it changes lengths, but this lesson shows that it changes them in a very disciplined way. When the center is the origin of a coordinate grid the rule is (x, y) → (kx, ky). When the center is some other point (h, j), the rule is (x, y) → (h + k(x − h), j + k(y − j)): measure the displacement from the center, multiply it by k, and add it back to the center.
What "verify experimentally" asks for. Draw a figure, choose a center and a scale factor, construct the image with a ruler (measure OP, multiply by k, mark P′ on the ray) or compute it on a grid, and then measure. A ruler and protractor cannot prove a property for every dilation, but they let you see it hold in several cases and, just as important, see how it fails when a condition is dropped. The coordinate version lets you check the same properties with exact arithmetic.
Property (a), first half: a line through the center is unchanged. Take the center O at the origin and k = 2, and the line y = 1.5x, which passes through O. The point C(2, 3) lies on it, since 3 = 1.5 · 2. Its image is C′(4, 6), and 6 = 1.5 · 4, so C′ lies on the same line. Every point of the line behaves the same way: a point (t, 1.5t) goes to (2t, 3t) = (2t, 1.5 · 2t), still on the line. The reason is built into the definition: the image of P lies on ray OP, and when the line already passes through O, that ray is part of the line. Individual points slide along the line, but the line as a set of points is left unchanged.
Property (a), second half: a line not through the center goes to a parallel line. Now take triangle ABC with A(1, 1), B(4, 1), C(2, 3) and the same dilation, center O, k = 2. The images are A′(2, 2), B′(8, 2), C′(4, 6). The line AB is y = 1; it does not pass through the origin. Its image is the line through A′ and B′, which is y = 2. The two lines are parallel and distinct. Line BC has slope (3 − 1)/(2 − 4) = −1 and equation y = −x + 5; it does not pass through O since 0 ≠ 5. The image line through B′(8, 2) and C′(4, 6) has slope (6 − 2)/(4 − 8) = −1 and equation y = −x + 10. Same slope, different intercept: parallel. On paper, lay a set square along AB and slide it to A′B′; the edge stays in the same direction. The condition matters: the line OC through the center did not move to a parallel line, it stayed where it was.
Property (b): a segment is longer or shorter in the ratio given by the scale factor. Segment AB runs from (1, 1) to (4, 1); its length is 3. Segment A′B′ runs from (2, 2) to (8, 2); its length is 6, which is 2 · 3. Segment AC runs 1 right and 2 up, so its length is √(1² + 2²) = √5 ≈ 2.236; A′C′ runs 2 right and 4 up, length √(4 + 16) = √20 = 2√5 ≈ 4.472, again exactly twice. Segment BC has run 2 and rise 2, length √8 = 2√2 ≈ 2.828, and B′C′ has run 4 and rise 4, length √32 = 4√2 ≈ 5.657. Every ratio image ÷ pre-image is 2, the scale factor. With k = 1/2 the image would be shorter, in the ratio 1/2. The reason: the run and the rise of any segment are both multiplied by k, so the length, which is built from the run and rise, is multiplied by k as well.
A center that is not the origin. Take center P(1, 2) and k = 3. The point Q(3, 3) is 2 right and 1 up from P, so its image is 3 · 2 = 6 right and 3 · 1 = 3 up from P: Q′ = (7, 5). Check the distances: PQ = √(2² + 1²) = √5 and PQ′ = √(6² + 3²) = √45 = 3√5, three times as far, and Q′ lies on ray PQ because its displacement from P is a positive multiple of Q's. Students who apply (x, y) → (3x, 3y) here get (9, 9), a point that is not on ray PQ at all; the rule (kx, ky) belongs only to dilations centered at the origin.
The misconception. Many students expect a dilation to change the direction of lines the way a rotation does, or to change angles the way a stretch in one direction would. It does neither. Because every line goes to a parallel line or to itself, every angle between two lines is carried to an angle between two parallel lines, which has the same measure. Length is the only thing a dilation changes, and it changes every length by the same factor. That is why a dilated figure looks like a photograph of the original: same shape, different size.
The experiment on paper. Draw a point O near a corner of a sheet. Draw a triangle a few centimeters away and a line through O crossing the triangle. With a ruler, measure the distance from O to each vertex, double it, and mark the image vertex on the same ray. Join the image vertices. Record, for each side, the length before and after, and the ratio; record whether the image of the line through O is the same line; and check with a set square that each side of the image is parallel to the corresponding side of the pre-image. Expect each image length to be within a millimeter or two of twice the original length, so each ratio comes out close to 2.0 (a 1 mm error in a 10 cm image side moves the ratio by only 0.02 when the original side is 5 cm). A ratio far from 2, such as 1.7, does not show that the property failed, because the property is proved exactly; it signals a construction or measuring error to look for, such as a distance from O doubled wrongly, an image point marked off its ray, or a side measured carelessly.
✏️Section 2 · ModelWork through the examples
Worked, center not at the origin: Dilate the segment from (3, 4) to (5, 2) with center (2, 1) and k = 2. Displacements from the center: (3, 4) is 1 right and 3 up, so its image is 2 right and 6 up, at (4, 7). (5, 2) is 3 right and 1 up, so its image is 6 right and 2 up, at (8, 3). Original length: run 2, rise −2, √8 = 2√2 ≈ 2.83. Image length: run 4, rise −4, √32 = 4√2 ≈ 5.66, twice as long. Slopes: original (2 − 4)/(5 − 3) = −1, image (3 − 7)/(8 − 4) = −1, so the image segment lies on a line parallel to the original.
Worked, a line through the center: Center at the origin, k = 3, and the line y = 2x. The point (1, 2) on the line goes to (3, 6), and 6 = 2 · 3, so the image point is still on y = 2x. Any point (t, 2t) goes to (3t, 6t), which also satisfies y = 2x. The line is left unchanged, even though every point except the origin has moved.
Worked reduction: Center at the origin and k = 1/3. The segment from (3, 9) to (9, 3) has run 6 and rise −6, length √72 = 6√2 ≈ 8.49. Its image runs from (1, 3) to (3, 1), run 2 and rise −2, length √8 = 2√2 ≈ 2.83. The ratio 2√2 ÷ 6√2 = 1/3 is the scale factor; the image is shorter in exactly that ratio. Both segments have slope −1, so they are parallel.
Worked non-example of the rule: A student dilates Q(3, 3) from center P(1, 2) with k = 3 and writes (9, 9). Check: (9, 9) is 8 right and 7 up from P, but Q is 2 right and 1 up, and 8/2 = 4 while 7/1 = 7, so (9, 9) is not even on ray PQ. The correct image is P plus three times the displacement: (1 + 6, 2 + 3) = (7, 5), and PQ′ = √45 = 3√5 is three times PQ = √5.
🧩Section 3 · DoPractice, demonstrate, and explain
Recording task: On grid paper draw triangle ABC with A(1, 1), B(4, 1), C(2, 3), and dilate it from the origin with k = 2 to A′(2, 2), B′(8, 2), C′(4, 6). Draw the rays from the origin through each vertex and confirm each image vertex is on its ray. Record a table with a row for each side: pre-image endpoints, image endpoints, pre-image length, image length, ratio (AB: 3 and 6, ratio 2; AC: √5 and 2√5, ratio 2; BC: 2√2 and 4√2, ratio 2). Add a second table for the lines: AB (y = 1) and its image (y = 2), BC (y = −x + 5) and its image (y = −x + 10), and the line OC (y = 1.5x) and its image, noting which are parallel and which is unchanged.
Independent performance: Choose a center that is not the origin, a scale factor between 0 and 1, and a quadrilateral with vertices on grid points that does not contain the center. Compute every image vertex with the rule (x, y) → (h + k(x − h), m + k(y − m)). Then, with a ruler, construct the same image on paper by measuring along the rays, and compare. Success criteria: every computed image vertex agrees with the ruler construction to within a millimeter; every side of the image has length equal to k times its pre-image length, shown by run-and-rise calculation; every side of the image is parallel to its pre-image side, shown by equal slopes; one line drawn through the center is shown to be unchanged by checking two of its points; and one sentence explains why the rule (kx, ky) would have given the wrong answer for your center.
✅Section 4 · Check8-question lesson practice test
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
Problem 1
Dilate the segment from (2, 6) to (8, 2) from the origin with k = 1/2. Give the image endpoints, both lengths, and the ratio of the lengths.
Show worked solution
The rule is (x, y) → (x/2, y/2), so the image runs from (1, 3) to (4, 1). The original has run 6 and rise −4, length √(36 + 16) = √52 = 2√13 ≈ 7.21. The image has run 3 and rise −2, length √(9 + 4) = √13 ≈ 3.61. The ratio image ÷ pre-image is √13 ÷ 2√13 = 1/2, the scale factor: the image segment is shorter in exactly that ratio.
Problem 2
The line y = 3x − 6 is dilated from the origin with k = 2. Find the image line and explain how you know the two lines are parallel rather than identical.
Show worked solution
A point on the line, (2, 0), goes to (4, 0); another, (3, 3), goes to (6, 6). The image line through (4, 0) and (6, 6) has slope 6/2 = 3 and equation y = 3x − 12. Both lines have slope 3, so they are parallel. They are not the same line because the original does not pass through the center: the origin gives 0 ≠ −6. A line is left unchanged by a dilation with k ≠ 1 only when it passes through the center.
Problem 3
The line y = −2x passes through the origin. Show that a dilation from the origin with any positive k leaves it unchanged, even though, for k ≠ 1, it moves the point (1, −2).
Show worked solution
A point on the line has the form (t, −2t). The dilation sends it to (kt, −2kt), and −2kt = −2(kt), so the image is on the same line. The point (1, −2) goes to (k, −2k), a different point when k ≠ 1, but it is still on y = −2x. Every point slides along the line and every point of the line is reached, so the line as a whole is unchanged. This is the first half of property (a).
Problem 4
Dilate P(5, 1) from center C(2, 3) with k = 3, and show that the image lies on ray CP three times as far from C as P does.
Show worked solution
P is 3 right and 2 down from C. Multiply the displacement by 3: 9 right and 6 down, so P′ = (2 + 9, 3 − 6) = (11, −3). CP = √(3² + 2²) = √13 and CP′ = √(9² + 6²) = √117 = 3√13, three times as far. P′ is on ray CP because its displacement (9, −6) is a positive multiple of (3, −2).
Problem 5
A student measures a side of a triangle as 4.0 cm, dilates the triangle with k = 2.5 by measuring along rays from a center, and measures the image side as 9.8 cm. The class agreed before measuring that a ruler-and-pencil construction can be off by up to 3 mm in any length. Does this contradict property (b)?
Show worked solution
No. Property (b) predicts exactly 2.5 × 4.0 = 10.0 cm, and 9.8 cm differs by 2 mm, inside the agreed 3 mm tolerance, so the measurement agrees with the prediction. A tolerance has to be set before the comparison is judged; without one, a 2 mm gap could be neither accepted nor rejected. The property itself is exact: the coordinate version gives exactly 2.5 times the length, because run and rise are both multiplied by 2.5. A much larger discrepancy, such as 7 cm, would point to a mistake in the construction or the measuring, not to a failure of the property.
Problem 6
Triangle with vertices (0, 0), (4, 0), (0, 3) is dilated from the origin with k = 3. Give the image, the three side lengths before and after, and the angle at the origin before and after.
Show worked solution
The image is (0, 0), (12, 0), (0, 9). The sides 4, 3 and 5 (the hypotenuse, √(16 + 9)) become 12, 9 and 15 (√(144 + 81) = √225), each three times as long. The angle at the origin is a right angle before and after, because the two legs still lie along the axes. A dilation changes lengths by the scale factor and leaves every angle unchanged.
Problem 7
Why does a dilation with center O and scale factor k ≠ 1 send a line not through O to a parallel line? Give a reason using the coordinate rule with O at the origin.
Show worked solution
With O at the origin the rule is (x, y) → (kx, ky). Two points (x₁, y₁) and (x₂, y₂) on the line have images whose run is k(x₂ − x₁) and rise is k(y₂ − y₁). The slope of the image, k(y₂ − y₁) ÷ k(x₂ − x₁), equals the original slope (and a vertical line, whose run is 0, goes to a vertical line), so the image line is parallel to or the same as the original. Because k ≠ 1, it cannot be the same line unless the original passes through O (with k = 1 the dilation moves nothing and every line is its own image): if a point P of the line and its image P′ (a different point when k ≠ 1) were both on the line, the line would contain two points of line OP and so would be line OP itself, which passes through O.
Problem 8
The vertex Q of a figure is 6 units from the center of a dilation and its image Q′ is 4 units from the center. What is the scale factor, and is the image larger or smaller than the original?
Show worked solution
The scale factor is 4/6 = 2/3, because OQ′ = k · OQ gives k = OQ′/OQ. Since k is between 0 and 1 the dilation is a reduction: every segment of the image is shorter than its pre-image in the ratio 2/3, and Q′ lies on ray OQ between O and Q.