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6.1 — Photocopy Faux Pas
Imagine copying a small drawing at 200% magnification. The drawing looks twice as tall and twice as wide, but why might it move off the edge of the page? And is it “twice as big” or “four times as big”? Both statements can describe different measurements. A dilation explains exactly what happens.
1. In class: A Develop Understanding Task
Understanding a dilation
A dilation is a transformation that changes the size of a figure while keeping its shape. The original figure is the pre-image; the resized figure is the image. Each point of the pre-image is paired with a corresponding point of the image.
A dilation is determined by two pieces of information: a fixed center of dilation (a reference point) and a positive scale factor \(k\) (the multiplier for distances from that center).
If \(O\) is the center and \(P\) is any point, its image \(P'\) is on ray \(OP\), with distance \(OP'=k\cdot OP\). The center itself stays fixed. Thus, a dilation changes not only side lengths but also how far each point lies from the chosen center.
Point \(A\) is 3 cm to the right of center \(O\). Point \(B\) is 5 cm above \(O\). Dilate by \(k=2\).
\(OA'=2(3)=6\) cm and \(OB'=2(5)=10\) cm. Put \(A'\) on the ray from \(O\) through \(A\), 6 cm from \(O\). Put \(B'\) on the ray through \(B\), 10 cm from \(O\). The directions from the center remain the same; distances double.
Why the center controls placement
A scale factor specifies the size of the image, but it does not by itself specify where the image will be. Moving the center changes where the enlarged drawing lands. This is why a photocopier can enlarge an image accurately and still place it incorrectly on a sheet.
A 4-by-6 cm sketch is enlarged to 200%. Its copy measures \(8\) by \(12\) cm. If enlargement is taken about the sketch's upper-left corner, that corner stays fixed and the copy grows rightward and downward. If enlargement is instead centered about the sketch's center, the copy expands equally to either side of that center.
Takeaway: To center the final copy on the paper, align the effective dilation center with the intended center of the final image, accounting for how the copying equipment anchors and positions the image.
Length factors are different from area factors
The scale factor \(k\) multiplies every corresponding length by \(k\). For a rectangle with width \(w\) and height \(h\), an enlargement has width \(kw\), height \(kh\), and area
\[A_{\text{new}}=(kw)(kh)=k^2wh=k^2 A_{\text{original}}.\]
Thus the area factor is \(k^2\). This follows because both dimensions are scaled, not because the copier applies the length factor twice to each single side.
| Copy setting | Length scale factor \(k\) | Area multiplier \(k^2\) |
|---|---|---|
| 50% | \(\tfrac12\) | \(\tfrac14\) |
| 200% | \(2\) | \(4\) |
| 300% | \(3\) | \(9\) |
An original picture is 5 inches wide and 7 inches high. A 200% copy is \(2(5)=10\) inches wide and \(2(7)=14\) inches high. The original area is \(5\cdot7=35\) square inches; the new area is \(10\cdot14=140\) square inches.
\(140\div35=4\). So the picture is twice as wide and tall, but four times the area. If one quarter of the complete enlarged image has the same area as the original, the area statement is consistent with 200% enlargement.
Constructing a dilation with rays
You can create or complete an enlarged tracing using a straightedge, ruler, or a rubber-band stretcher that keeps each original point, its image, and the fixed center aligned.
Suppose a point on a drawing lies 2.5 cm from the center and the photocopier uses a 200% scale. The corresponding point lies \(2(2.5)=5\) cm from the same center along the same ray. Repeat for points around the outline. Curved parts require more sample points to preserve the outline accurately.
The packet depicts a sketch and a cropped enlargement. To reconstruct the remaining image, use the center implied by the aligned original and enlarged features, then extend each ray at twice the original distance. The copy's off-center position is a consequence of the placement/anchor used for the enlargement. Problem 3 asks for a placement that centers the resulting image, not merely for positioning the original sketch at the visual center of the starting paper.
2. Ready — Find scale factors and missing lengths (problems 1–6)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
For corresponding sides, the scale factor is the ratio image length divided by pre-image length:
\[k=\frac{\text{image side length}}{\text{pre-image side length}}.\]
Then apply \(\text{image length}=k(\text{pre-image length})\). A missing pre-image length is found by dividing the image length by \(k\).
A small triangle has a side of 4 cm and its image has the corresponding side of 10 cm.
\(k=10/4=2.5\). A second pre-image side of 6 cm becomes \(2.5(6)=15\) cm. All corresponding sides must have the same ratio if the figures are related by this dilation.
A reduced image has a side length of 3 cm. The reduction scale factor is \(\tfrac38\). What was the original length?
Let \(L\) be the original length. Then \(3=\tfrac38 L\), so \(L=3\div\tfrac38=3\cdot\tfrac83=8\) cm. Do not multiply by \(\tfrac38\) again when finding the original.
A smaller right triangle has vertical leg 4 units. A similar larger right triangle has corresponding vertical leg 10 units. Find the full horizontal leg if the small horizontal leg is 6 units.
Scale factor \(k=10/4=2.5\). Full horizontal leg \(=2.5(6)=15\) units. It does not matter that the small triangle lies inside the large triangle: identify the corresponding sides first.
Check your work — Ready 1–6 answer key
- \(k=2\).
- \(k=3\); both vertical sides are \(3\), and both slanted sides are \(4.5\).
- \(k=\frac15\); missing pre-image top side \(12.5\), missing image left side \(0.4\).
- \(k=\frac32\); other outer side \(6\), other inner side \(2.25\).
- \(k=5\); other straight side \(10\), arc length \(17.5\).
- \(k=3\). The small triangle has sides \(3,4,5\); the large triangle has sides \(9,12,15\).
3. Set — Identify the centers and tools of real-world dilations (problems 7–12)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
In a physical situation, the “center” can be an actual geometric point or an idealized viewpoint or optical center. The “tool” is the device or method that creates the observed change in apparent or reproduced size.
| Situation from the packet | Useful way to identify the center | Tool / mechanism |
|---|---|---|
| 7. Photographer steps back to fit family | Camera's effective viewpoint / optical center | Camera and change of camera position |
| 8. Projected movie image fills screen | Projector's optical center, in a simplified ray model | Projector zoom optics |
| 9. Thumb comparisons to distant objects | Observer's eye | Line of sight and thumb as visual reference |
| 10. Animator draws receding telephone poles | Idealized vanishing point / viewpoint in a perspective model | Drawing or animation software |
| 11. Three-rubber-band tracing project | Fixed anchor of the stretcher | Rubber-band dilation tool |
| 12. Copier set at 300% | Machine's enlargement reference point / anchor | Photocopier |
Camera zoom, walking backward, looking at a waterfall, and perspective drawing involve projection and viewpoint. For this assignment, identify the intended viewing or projection center and the tool that changes apparent scale. In general, scenes at different depths do not all transform by one exact plane-wide dilation with a single scale factor; the packet is using these situations as accessible models.
A machine is set to 300%. Its length scale factor is \(k=300/100=3\); a 2-inch segment becomes 6 inches long and a 5-square-inch patch becomes \(3^2(5)=45\) square inches. The tool is the copier; the effective center depends on the copier's scan and placement conventions.
Check your work — Set 7–12 answer key
- Center: the camera lens (optical center); tool: the camera and its changing distance from the family.
- Center: the projector lens; tool: the projector zoom lens.
- Center: Melanie’s eye; tool: her thumb held at a fixed distance from her eye, comparing apparent heights.
- Center: the scene’s viewpoint (the vanishing point in the picture); tool: the computer’s perspective/animation software.
- Center: the anchored end of the rubber-band device; tool: the rubber-band stretcher.
- Center: the copier’s fixed reference corner, typically the top-left corner; tool: the copier enlargement setting, with scale factor \(3\).
4. Go — Recognize linear, exponential, and quadratic functions (problems 13–18)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
This is a review of earlier function families, not a new property of dilations. You must classify information given by a table, equation, or graph and justify the classification.
| Function family | Typical equation | How to recognize it |
|---|---|---|
| Linear | \(y=mx+b\) | Straight-line graph; equal changes in \(x\) give equal first differences in \(y\). |
| Exponential | \(y=ab^x\), \(b>0\), \(b\ne1\) | Multiplicative pattern for equal \(x\)-steps; characteristic increasing or decreasing curved graph. |
| Quadratic | \(y=ax^2+bx+c\), \(a\ne0\) | Parabola-shaped graph; equal \(x\)-steps give constant second differences of \(y\). |
Important: The first- and second-difference tests assume equally spaced \(x\)-values. If the input gaps differ, compute slopes \(\Delta y/\Delta x\) for a linear table, or analyze the quadratic/exponential rule directly. A small finite table can be consistent with a function family without uniquely proving its global rule.
Classify the table.
| \(x\) | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| \(y\) | 3 | 6 | 11 | 18 |
First differences: \(6-3=3\), \(11-6=5\), \(18-11=7\). Second differences: \(5-3=2\) and \(7-5=2\). The constant second difference indicates a quadratic pattern.
Consider \((x,y)=(1,4),(4,13),(10,31),(12,37)\). The input gaps are 3, 6, and 2, so the raw output differences \(9,18,6\) should not be tested for equality. Instead compute slopes:
\(9/3=3\), \(18/6=3\), \(6/2=3\). Equal rates of change suggest the linear rule \(y=3x+1\), which matches all the pairs.
Determine the type of function represented by \(x=0,1,2,3\) and \(y=5,10,20,40\). Justify your answer.
Consecutive output ratios are \(10/5=2\), \(20/10=2\), and \(40/20=2\). Since each increase of 1 in \(x\) doubles \(y\), the table represents an exponential pattern, modeled by \(y=5(2^x)\).
A table shows \((x,y)=(-2,13),(0,5),(2,5),(4,13)\). Classify the pattern and find one quadratic rule consistent with it.
The \(x\)-steps are all 2; the first differences are \(-8,0,8\), and the second differences are \(8,8\). Thus the table follows a quadratic pattern. The values are symmetric around \(x=1\), so try \(y=a(x-1)^2+c\). From \(x=0\), \(a+c=5\); from \(x=2\), the same equation holds. From \(x=4\), \(9a+c=13\). Subtracting gives \(8a=8\), hence \(a=1\) and \(c=4\). One matching rule is \(\boxed{y=(x-1)^2+4}\).
Classify \(f(x)=6x-4\), \(g(x)=2(3^x)\), and \(h(x)=-2x^2+5x+1\). Then describe their graph shapes.
\(f\) is linear because its highest power of \(x\) is 1; its graph is a straight line. \(g\) is exponential because the input is an exponent; its graph grows multiplicatively. \(h\) is quadratic because it has a nonzero \(x^2\) term; its graph is a downward-opening parabola.
The table's \(x\)-values are 2, 7, 14, and 25, while \(y\)-values are −5, 5, 19, and 41. The corresponding changes are \(\Delta x=5,7,11\) and \(\Delta y=10,14,22\). Each rate \(\Delta y/\Delta x\) equals 2, so the entries fit the linear rule \(y=2x-9\). Comparing only the raw \(y\)-differences would give the wrong impression.
Check your work — Go 13–18 answer key
- Quadratic: first differences are \(5,7,9\), so second differences are constantly \(2\); the data fit \(y=x^2+3\).
- Exponential: the curve approaches zero on the left and grows by a constant multiplicative factor over equal horizontal steps.
- Quadratic: \(y=3x^2+3x\) has a nonzero squared term.
- Linear: \(y=7x-10\) has constant slope \(7\).
- Quadratic: the graph is a parabola.
- Linear: each ratio \(\frac{\Delta y}{\Delta x}=2\), even though the input steps differ; \(y=2x-9\).
5. Additional learning resources
The following links are supplementary; they are not required to use this self-contained lesson.
| Resource | What it teaches / why use it | Scope / What to skip |
|---|---|---|
| Math Is Fun — Resizing | Written and interactive explanation of enlargements, centers and ratios. Especially useful for the main task and Ready. | Close match. Focus on construction and resizing examples. |
| Khan Academy — Performing dilations | Video showing how to scale points about a center that need not be the origin. | Some extra material. Coordinate calculations extend beyond the geometric drawing task; focus on center and multiplier. |
| Khan Academy — Dilate triangles (practice) | Check placement and scale-factor reasoning with short interactive questions. | Close match. Focus on geometric scaling; do not jump ahead to similarity proofs. |
| Khan Academy — Dilations intro | Brief conceptual video contrasting size-changing dilations with rigid motions. | Close match. An optional quick review before the main lesson. |
A suitably matched Professor Leonard lecture specific to these introductory geometric dilations was not verified, so none is assigned.
6. Quick self-check
1. A drawing is enlarged by 150%. What is its length multiplier? Its area multiplier?
2. An image side is 18 cm long and the original corresponding side was 12 cm. Find \(k\).
3. Does a 200% enlargement guarantee centering? Explain.
4. What kind of function has first differences \(4,6,8,10\) for consecutive whole-number inputs?
Check answers
1. \(k=1.5\), area multiplier \(2.25\). 2. \(18/12=1.5\). 3. No; placement depends on the center and alignment. 4. A quadratic pattern; second differences are all 2.
Teacher's review of the packet
Section 6.1, Set 7–10 — physical modeling ambiguity. The exercises describe changes of perspective, zoom, and distance as dilations without defining an exact mathematical model or the intended meaning of “center.” Depending on whether one means optical center, viewpoint, vanishing point, or center of a planar enlargement, answers can differ. A clearer prompt would specify that students should use an idealized ray model and identify the viewing/projection center.
Section 6.1, main task 3 — copier reference/placement is implicit. The prompt asks where the original should have been placed for the image to be centered, but no precise geometric mapping from copier glass to output page is stated in words. Students can reason from the packet illustration; an unambiguous numerical construction would need the dilation center or copier anchor explicitly marked.
Section 6.1, Go 18 — nonuniform inputs. This is mathematically solvable, but may catch students who apply the standard first-difference test without checking input spacing. The task could specify that students should compare rates \(\Delta y/\Delta x\) when \(x\)-steps are unequal. This is important instructional guidance, rather than an error in the data.
7. Coverage of the original packet
| Packet item | Mathematics and skills | Where taught here |
|---|---|---|
| Main task 1–3 | Dilation center, placement, ray construction, completing the enlarged image | In class; packet clarification |
| Main task 4–5 | Length versus area scaling; essential dilation parameters | In class |
| Ready 1–6 | Scale factor from corresponding sides, missing measures, enlargements and reductions | Section 2, examples 5–7 |
| Set 7–12 | Interpret dilation centers and tools in real contexts, including 300% | Section 3, context table and example 8 |
| Go 13–18 | Linear/exponential/quadratic equations, graphs, tables, rates of change | Section 4, examples 9–10 and problems A–C |
Source: Mathematics Vision Project, Secondary Math II, Module 6 — Similarity & Right Triangle Trigonometry (2017), section 6.1, packet pages 1–5 (PDF pages 4–8). Supplement independently authored; no affiliation or endorsement.
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