What a dilation does to a line
A dilation moves every point along its own ray from the center. What happens to a straight line when all of its points move at once?
Where you are 10 in Similarity
Why this exists
A dilation is described one point at a time. The center O stays put, and every other point P slides along the ray from O through P until it is k times as far from O as it was. That tells you where any single point goes. It says nothing directly about a figure, and figures are made of lines and segments.
Dilate the three corners of a triangle and you get three new points. To join them with straight sides and call the result “the image of the triangle”, you need to know that the points of each side, all moving at once, land on the straight segment between the new corners and not on some curve. You would also like to know how each new side sits compared with the old one.
This lesson answers both questions, and the answer is short. A line through the center stays exactly where it is, and every other line is carried to a line parallel to it. That one fact is why a dilated figure has the same shape as the original, and it is the engine behind similar figures.
Definition
Fix a dilation with center O and scale factor k > 0. The image of a point P \neq O is the point P' on ray OP with OP' = k \cdot OP, and the image of O is O itself.
The image of a line l is the set of the images of all the points on l. It is written l' and read “l prime”. The image of a segment \overline{PQ} is defined the same way, as the set of the images of its points.
Two lines in a plane are parallel, written l \parallel m, when they never meet. Under this definition a line is not parallel to itself, so the results below keep “the same line” and “a parallel line” as two separate cases.
A line passes through the center when O is one of its points. Every line either does or does not, and the two kinds behave differently.
Intuition
Line m in Figure 1 passes through O. Every point of m moves along its own ray from O, and that ray lies along m itself. So P slides out to P' and Q slides out to Q', but no point leaves the line. The line as a whole is unchanged, even though every point on it except O has moved.
Line l misses O. Its points move out along different rays, fanning out from O, and each one ends up twice as far from O as it was. Points of l far from O travel further than points near O, by just the amount that keeps them in a straight row. The whole line ends up pushed away from O without being turned, like a ruler slid across a table, and a ruler slid without turning stays parallel to where it started.
With k less than 1 everything runs the other way: points slide in toward O, and l is pulled toward O, again without turning.
Results
Lines through the center. A dilation maps every line through its center onto itself: l' = l.
Lines that miss the center. If k \neq 1, a dilation maps every line that does not pass through its center to a different line parallel to it: l' \parallel l. The image lies on the same side of O as l, and its distance from O is k times the distance from O to l.
Segments. The image of \overline{PQ} is the segment \overline{P'Q'}. It lies on the same line as \overline{PQ} or on a line parallel to it, and it is k times as long:
In coordinates, with the center at the origin. The dilation sends (x, y) to (kx, ky), and lines go as follows:
The slope never changes; only the intercept is multiplied by k. A line through the origin has intercept 0, and k \cdot 0 = 0, which is the first result again.
With the center somewhere else. Draw the line through the center parallel to l, and measure the vertical gap from it up to l (a negative gap means l is below it; for vertical lines use the horizontal gap). The image l' has the same slope as l, and its gap from that line is k times as large.
Why it is true
A line through the center. Let l pass through O, and take any point P \neq O on l. Ray OP is part of l, and P' lies on ray OP, so P' is on l; and the image of O is O, which is on l. So every image lands on l. Every point X of l is also an image of something: if X = O it is the image of O, and otherwise the point P on ray OX with OP = \tfrac{1}{k}\,OX lies on l and has P' = X. So the image of l is all of l and nothing else.
A line that misses the center. Lay coordinate axes with the origin at O. This is allowed: the axes are a ruler laid on the page, and where it is laid changes no line and no distance. The dilation now sends (x, y) to (kx, ky), because (kx, ky) lies on the ray from the origin through (x, y), and its distance from the origin is \displaystyle \sqrt{k^2x^2 + k^2y^2} = k\sqrt{x^2 + y^2}, which is k times as far.
Suppose first that l is not vertical, with equation y = mx + b. A point of l is (t, mt + b) for some number t, and its image is
Write the image as (X, Y), so X = kt. Then Y = m(kt) + kb = mX + kb, so every image point lies on the line y = mx + kb. Every point of that line is an image, too: (X, mX + kb) is the image of \left(\tfrac{X}{k},\, m\tfrac{X}{k} + b\right), which lies on l. So the image of l is exactly the line y = mx + kb.
The two lines have the same slope m. Lines with the same slope are either the same line or parallel, and these two are the same only when kb = b. For k \neq 1 that means b = 0, which is exactly the case where l passes through the origin O. A vertical line x = c works the same way: its points (c, t) go to (kc, kt), which fill the vertical line x = kc, and that is the same line only when c = 0.
The distance from the center. Let F be the point of l nearest O, so \overline{OF} is perpendicular to l. Line OF passes through O, so the dilation keeps it in place, and F' lies on ray OF with OF' = k \cdot OF. Now F' is on l', and line OF, being perpendicular to l, is also perpendicular to the parallel line l'. So F' is the point of l' nearest O, and the distance from O to l' is k \cdot OF. Since F' is on ray OF, the image is on the same side of O as l.
The gap rule for any center. In axes with the origin at the center, the parallel through the center is y = mx and l is y = mx + b, where b is the vertical gap between them. The image is y = mx + kb, whose gap is kb. Sliding the axes back to where they were changes the equations but not the vertical gaps, so the rule holds in any axes.
Lengths. With P = (x_1, y_1) and Q = (x_2, y_2), the images are (kx_1, ky_1) and (kx_2, ky_2), and by the distance formula
where \displaystyle \sqrt{k^2} = k because k > 0. A point some fraction of the way from P to Q has coordinates built from those of P and Q by adding and scaling, and multiplying every coordinate by k keeps that fraction, so it goes to the point the same fraction of the way from P' to Q'. That is why the image of the segment is the whole segment \overline{P'Q'} and nothing more.
Worked examples
A line through the center
A dilation has its center at the origin and scale factor 3. Find the image of the line y = 2x, and the images of the points (1, 2) and (-2, -4) on it.
SolutionThe line y = 2x passes through the origin, since 0 = 2 \cdot 0. A line through the center is its own image, so the image is the line y = 2x again.
The points on it still move. With the center at the origin, the dilation multiplies each coordinate by 3:
Both images are on y = 2x, since 6 = 2 \cdot 3 and -12 = 2 \cdot (-6). The point (1, 2) is \displaystyle \sqrt{5} from the origin and its image is \displaystyle \sqrt{45} = 3\sqrt{5} from it, three times as far, as a dilation requires.
So “the line stays where it is” is a statement about the line, not about its points. Only the center itself stays put.
A line that misses the center
A dilation has its center at the origin and scale factor 2. Find the image of the line l with equation y = 2x + 3.
SolutionAt x = 0 the line has y = 3, not 0, so it misses the center and its image is a line parallel to it. Two points pin a line down, so take A = (1, 5) and B = (-2, -1), both on l because 2 \cdot 1 + 3 = 5 and 2 \cdot (-2) + 3 = -1. Doubling the coordinates gives
The slope of l', through A' and B', is
which is the slope of l, as the result promises. The intercept b comes from A': 10 = 2 \cdot 2 + b, so b = 6. The image is
Check. The shortcut says the intercept is multiplied by k, and 2 \cdot 3 = 6. The point B' is on the image too: 2(-4) + 6 = -2.
A fractional scale factor and a standard-form equation
A dilation has its center at the origin and scale factor \tfrac12. Find the image of the line 3x - 4y = 12.
SolutionA point (X, Y) is on the image exactly when it is the image of a point on the line. With k = \tfrac12, the point (X, Y) is the image of (2X, 2Y), because halving (2X, 2Y) gives (X, Y). So (X, Y) is on the image exactly when (2X, 2Y) is on the original line:
Dividing both sides by 2 in the last step does not change which points satisfy the equation. Back in the usual letters, the image is 3x - 4y = 6. The left side is unchanged and the right side was multiplied by \tfrac12, which is the rule ax + by = c \longmapsto ax + by = kc.
Check with intercepts. The original line crosses the axes at (4, 0) and (0, -3), since 3 \cdot 4 = 12 and -4 \cdot (-3) = 12. Their images are (2, 0) and \left(0, -\tfrac32\right), and both satisfy the new equation: 3 \cdot 2 = 6 and -4 \cdot \left(-\tfrac32\right) = 6. The image is nearer the origin, as it must be when k < 1.
A center that is not the origin
A dilation has center C = (1, 2) and scale factor 3. Find the image of the line y = x + 4.
SolutionFirst decide which kind of line this is. At x = 1 the line has y = 5, not 2, so it misses the center, and its image is a parallel line with the same slope, 1.
With the center away from the origin, multiplying coordinates by 3 is wrong. What gets tripled is each point’s offset from C: how far across and how far up it sits from C. For (0, 4) the offset is (0 - 1, 4 - 2) = (-1, 2), meaning 1 to the left of C and 2 above it; for (2, 6) it is (1, 4). Tripling an offset multiplies both of its numbers by 3, which keeps the point on the same ray from C but 3 times as far out. Adding the tripled offset to C, number by number, gives the image. The shorthand (1, 2) + 3(-1, 2) means exactly that: \big(1 + 3 \cdot (-1),\; 2 + 3 \cdot 2\big). So
The slope through the two images is \dfrac{14 - 8}{4 - (-2)} = 1, as expected, and from (4, 14) the intercept is 14 - 4 = 10. The image is y = x + 10.
The gap rule gives the same line faster. The line through C with slope 1 is y = x + 1, since 2 = 1 + 1. The original line sits 4 - 1 = 3 above it, so the image sits 3 \cdot 3 = 9 above it: y = x + 1 + 9, which is y = x + 10.
How far the image line sits
A line l is 6 cm from a point O. Find where its image lies under the dilation with center O and scale factor 2.5, and under the one with scale factor \tfrac13. In each case, how far apart are l and its image?
The line misses O, so each image is parallel to l, on the same side of O, and k times as far from O.
With k = 2.5, the image l_1 is 2.5 \cdot 6 = 15 cm from O. Both lines cross the same perpendicular from O, on the same side, so the gap between them is 15 - 6 = 9 cm, with l lying between O and l_1.
With k = \tfrac13, the image l_2 is \tfrac13 \cdot 6 = 2 cm from O. It lies between O and l, and 6 - 2 = 4 cm from l.
The dashed ray in Figure 3 shows why: one point A of l and its two images A_1 and A_2 lie on a single ray from O, at 2.5 times and \tfrac13 of the distance of A. In general a line at distance d from the center ends up at distance kd, which is (k - 1)d further out when k > 1 and (1 - k)d further in when k < 1.
Finding the scale factor
A dilation centered at the origin takes the line y = -x + 6 to the line y = -x + 2. Find the scale factor. Could a dilation centered at the origin take y = 2x + 1 to y = 3x + 3?
SolutionWith the center at the origin the slope stays and the intercept is multiplied by k, so
Check one point. (3, 3) is on the first line, since -3 + 6 = 3, and its image (1, 1) is on the second, since -1 + 2 = 1.
For the second pair the answer is no, and no calculation is needed. A dilation never changes a line’s slope, and the slopes 2 and 3 differ. The intercepts happen to be in the ratio 1 : 3, but that is not enough.
The image of a triangle
Triangle ABC has A = (1, 1), B = (4, 1) and C = (1, 3). Dilate it with center at the origin and scale factor 2, and compare the sides and angles of the image with those of ABC.
The corners go to A' = (2, 2), B' = (8, 2) and C' = (2, 6). No side of ABC lies on a line through the origin, so each side of the image is parallel to the side it came from and twice as long. The coordinates confirm it:
| Side and its image | Slope of each | Lengths |
|---|---|---|
| \overline{AB} and \overline{A'B'} | 0 | 3 and 6 |
| \overline{AC} and \overline{A'C'} | undefined | 2 and 4 |
| \overline{BC} and \overline{B'C'} | -\tfrac23 | \displaystyle \sqrt{13} and \displaystyle 2\sqrt{13} |
The last row needs the distance formula: \displaystyle BC = \sqrt{(1 - 4)^2 + (3 - 1)^2} = \sqrt{13} and \displaystyle B'C' = \sqrt{(2 - 8)^2 + (6 - 2)^2} = \sqrt{52} = 2\sqrt{13}.
Now the angles. At A the sides are horizontal and vertical, so \angle A is a right angle, and so is \angle A'. At B, one side points from B straight left and the other points up and to the left along slope -\tfrac23. At B' the sides are parallel to those, and because k is positive they point the same ways: B' is to the right of A' just as B is to the right of A. Two angles whose sides point in the same two directions have the same measure, so m\angle B' = m\angle B, and in the same way m\angle C' = m\angle C. Exercise 24 turns this into a proof with corresponding angles.
So the image has the same three angles as ABC and sides twice as long: the same shape, twice the size. Every dilation does this to every polygon, side by side.
Finding the center
A dilation with scale factor 3 has its center somewhere on the y-axis, and it takes the line y = 2x + 1 to the line y = 2x + 7. Find the center.
SolutionCall the center (0, c). The line through it parallel to the two given lines is y = 2x + c. The original line sits 1 - c above that line, measured vertically, and the image sits 7 - c above it. By the gap rule the image’s gap is 3 times the original’s:
The center is (0, -2).
Check. The point (0, 1) is on y = 2x + 1, and its offset from the center is (0, 3). Tripled, that is (0, 9), so its image is (0, -2 + 9) = (0, 7), which is on y = 2x + 7.
Here the result ran backwards: a fact about where lines go located a point that was never drawn.
Boundaries
The scale factor must not be 1. With k = 1 every point stays put, so every line is its own image, through the center or not. The parallel-line result needs k \neq 1.
The scale factor must not be 0. With k = 0 every point would go to O, and a whole line would collapse to a single point. That is why a dilation needs k \neq 0.
This lesson takes k > 0. Some books also allow a negative scale factor, which sends P to the point on the opposite side of O, |k| times as far. Lines through the center are still their own images and other lines still go to parallel lines, but now the image is on the opposite side of O, segments are |k| times as long, and each side of a figure points the opposite way.
The line stays, its points do not. A line through the center is its own image, but every point on it except the center moves along it. Knowing where a line goes says nothing about where a given point on it goes, beyond which line it lands on.
Directions survive, lengths do not. Every segment keeps its direction and is multiplied by k, so for k \neq 1 a dilation is never a rigid motion: it keeps the shape of a figure, not its size.
One line and its image do not pin down the dilation. Many centers and scale factors take a given line to a given parallel line (Exercise 20). And no dilation at all takes a line to a line that crosses it, since the image is always the same line or a parallel one.
In space. The same holds for lines in three dimensions: a line through the center is its own image and every other line goes to a parallel line. A plane through the center is likewise its own image, and every other plane goes to a parallel plane.
Where this goes next
Lines going to parallel lines is what the rest of similarity runs on. Applied to the sides of a triangle, it gives an image triangle with the same angles and every side k times as long, as in Example 7. Following a dilation with rigid motions gives the similarity transformations, and two figures are similar exactly when one can be carried onto the other this way. How the factor k then acts on perimeter and area is its own result.
Now put one corner of a triangle at the center. Dilating triangle OPQ from O stretches the sides \overline{OP} and \overline{OQ} along themselves by the factor k, and carries \overline{PQ} to a parallel segment \overline{P'Q'}. Read that picture the other way round and it says a line parallel to one side of a triangle cuts the other two sides in the same ratio, which is the triangle proportionality theorem. With k = \tfrac12 the segment \overline{P'Q'} joins the midpoints of \overline{OP} and \overline{OQ}, which is the midsegment, parallel to the third side and half as long.
Exercises
Through the center or not
- A dilation has its center at the origin and scale factor 4. Which of these lines are their own images: y = -3x, y = x + 1, x = 0, y = 5? Explain.
- A dilation has center (2, -1) and scale factor 3. Is the line y = 2x - 5 its own image? Explain.
- The point P = (3, -1) lies on the line y = -x + 2. Under the dilation with center at the origin and scale factor 2, find P'. Is P' on y = -x + 2? Which line is the image of y = -x + 2?
Images of lines, center at the origin
Each dilation has its center at the origin. Find the image of the line.
- y = 3x - 2, scale factor 4
- y = -\tfrac12 x + 6, scale factor \tfrac13
- x = 5, scale factor \tfrac52
- 2x + 5y = 10, scale factor 3
- x - 3y = 6, scale factor \tfrac12
Center somewhere else
Find the image of the line under the dilation with the given center and scale factor.
- Center (1, 1), scale factor 2, line y = x + 3
- Center (-2, 3), scale factor 3, line y = 5
- Center (4, 0), scale factor \tfrac12, line x = 10
- Center (0, 2), scale factor 4, line y = -2x + 5
Distances and lengths
- A line is 8 cm from a point O. Under the dilation with center O and scale factor 1.5, how far is its image from O, and how far from the original line?
- A line is 9 units from a point O. Under the dilation with center O and scale factor \tfrac23, how far is its image from O, and how far from the original line? Which of the two lines is nearer O?
- Let P = (1, 2) and Q = (4, 6). Under the dilation with center at the origin and scale factor 3, find P' and Q', the lengths PQ and P'Q', and the slopes of \overline{PQ} and \overline{P'Q'}.
- A dilation with center O takes a line 4 units from O to a line 10 units from O. Find the scale factor and the distance between the two lines.
Working backwards
- A dilation centered at the origin takes y = 5x - 6 to y = 5x - 2. Find the scale factor.
- A dilation centered at the origin with scale factor 2 takes a line l to the line y = -4x + 9. Find the equation of l.
- A dilation with scale factor 2 has its center at (0, c) on the y-axis, and it takes y = x + 1 to y = x + 5. Find c.
- Find two different dilations, each given by a center and a scale factor, that take the line y = 0 to the line y = 2.
Reasoning
- A dilation has scale factor k \neq 1, and the line l does not pass through its center. Explain why l and its image have no point in common.
- Line m passes through the center O of a dilation with scale factor 3, and P \neq O is a point on m. The image of m is m. Does that mean P' = P? Explain.
- Show that the dilation with center at the origin and scale factor k takes the line ax + by = c to the line ax + by = kc.
- Lines l and m cross at a point X. Show that under a dilation their images l' and m' cross at X', and that the angles at X' have the same measures as the angles at X.
- Explain why no dilation centered at the origin with a positive scale factor takes y = 2x + 4 to y = 2x - 4. Which scale factor would do it, if negative ones were allowed?
Answers
- y = -3x and x = 0. Each is satisfied by (0, 0), so each passes through the center and is its own image. The other two miss the origin and go to the parallel lines y = x + 4 and y = 20.
- Yes. At x = 2 the line has y = 2 \cdot 2 - 5 = -1, so it passes through the center, and a line through the center is its own image.
- P' = (6, -2). It is not on y = -x + 2, because -3 + 2 = -1 puts P on that line but -6 + 2 = -4, not -2. The image line keeps the slope and doubles the intercept, 2 \cdot 2 = 4, so it is y = -x + 4, and -6 + 4 = -2 confirms that P' is on it.
- y = 3x - 8. The slope stays and the intercept is multiplied by 4: 4 \cdot (-2) = -8.
- y = -\tfrac12 x + 2. The intercept becomes \tfrac13 \cdot 6 = 2.
- x = \tfrac{25}{2}. A vertical line x = c goes to x = kc, and \tfrac52 \cdot 5 = \tfrac{25}{2}.
- 2x + 5y = 30. The point (X, Y) is the image of \left(\tfrac{X}{3}, \tfrac{Y}{3}\right), so \tfrac{2X}{3} + \tfrac{5Y}{3} = 10, which is 2X + 5Y = 30.
- x - 3y = 3. The point (X, Y) is the image of (2X, 2Y), so 2X - 6Y = 6, and dividing by 2 gives X - 3Y = 3.
- y = x + 6. The parallel through the center is y = x, since 1 = 1. The line is 3 above it, so the image is 2 \cdot 3 = 6 above it. Check: (0, 3) goes to (1, 1) + 2(-1, 2) = (-1, 5), and -1 + 6 = 5.
- y = 9. The line is 5 - 3 = 2 above the horizontal line y = 3 through the center, so the image is 3 \cdot 2 = 6 above it, at y = 3 + 6 = 9.
- x = 7. The line is 10 - 4 = 6 to the right of the center, so the image is \tfrac12 \cdot 6 = 3 to the right of it, at x = 4 + 3 = 7.
- y = -2x + 14. The parallel through (0, 2) is y = -2x + 2. The line is 5 - 2 = 3 above it, so the image is 4 \cdot 3 = 12 above it: y = -2x + 2 + 12.
- 12 cm from O, and 4 cm from the original line. The image is 1.5 \cdot 8 = 12 cm from O on the same side, so it is 12 - 8 = 4 cm beyond the original line.
- 6 units from O, and 3 units from the original line; the image is nearer O. It is \tfrac23 \cdot 9 = 6 from O on the same side, and 9 - 6 = 3.
- P' = (3, 6) and Q' = (12, 18). \displaystyle PQ = \sqrt{3^2 + 4^2} = 5 and \displaystyle P'Q' = \sqrt{9^2 + 12^2} = 15, three times as long. Both slopes are \tfrac43: \tfrac{6 - 2}{4 - 1} and \tfrac{18 - 6}{12 - 3}.
- Scale factor \tfrac52, and the lines are 6 units apart. The distance is multiplied by k, so 4k = 10; the image is on the same side of O, so the gap is 10 - 4 = 6.
- k = \tfrac13. The slope matches, and the intercept is multiplied by k: -6k = -2.
- y = -4x + \tfrac92. The image keeps the slope -4 and has intercept 2b, so 2b = 9.
- c = -3. The parallel through the center is y = x + c. The gaps are 1 - c and 5 - c, and the image’s gap is twice the original’s: 5 - c = 2(1 - c), so c = -3. Check: (0, 1) goes to (0, -3) + 2(0, 4) = (0, 5), which is on y = x + 5.
- For example, center (0, -2) with scale factor 2, and center (0, -1) with scale factor 3. For the first, y = 0 is 2 above the center and the image is 2 \cdot 2 = 4 above it, at y = -2 + 4 = 2. For the second, the gaps are 1 and 3 \cdot 1 = 3, giving y = -1 + 3 = 2. The x-coordinate of the center does not matter, so there are many more.
- The image l' is parallel to l or equal to it. It is not equal, because l' is k times as far from the center as l, and k \neq 1. Two different parallel lines never meet, so l and l' share no point.
- No. P' is on ray OP with OP' = 3 \cdot OP, which is not OP, so P' \neq P. The line as a whole is unchanged, but every point on it except O slides along it.
- With the center at the origin, (X, Y) is the image of \left(\tfrac{X}{k}, \tfrac{Y}{k}\right), so (X, Y) is on the image exactly when a\tfrac{X}{k} + b\tfrac{Y}{k} = c. Multiplying both sides by k, which is not 0, gives aX + bY = kc, so the image is the line ax + by = kc.
- X is on both l and m, so X' is on both l' and m'. The image l' is l or parallel to it, and m' is m or parallel to it; since l and m cross, l' and m' are neither the same line nor parallel, so they cross, and only at X'. For the angles, suppose l' \neq l (if both lines pass through the center there is nothing to show, and otherwise rename them). Line m crosses l, so it also crosses the parallel line l', at a point Y, and the angles m makes with l at X equal the corresponding angles it makes with l' at Y. Then m and m' are parallel (or equal, in which case Y = X') and l' crosses both, so the angles at Y equal the corresponding angles at X'. So each angle at X' equals the matching one at X.
- With the center at the origin the intercept is multiplied by k, so 4k = -4 and k = -1, which is not positive. A positive k keeps the image on the same side of the center, and these two lines are on opposite sides of the origin. With k = -1, each point (x, y) goes to (-x, -y), and -y = 2(-x) + 4 is the line y = 2x - 4.