Math 8 · 8.G.A.4 · G.SRT.A.1

Dilations

How do you make a copy of a figure that is exactly the same shape but a different size, and how do you say where the copy goes?

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Scale factorMath 7 Similar figuresMath 8 The coordinate planeGrades 5 to 6

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Where you are 10 in Similarity

Geometry11

Dilations

Scale factor and perimeter, area and volume Similarity transformations ▾ Show all 10 (1 more)

Why this exists

A photocopier set to 150\%, a projector throwing a slide onto a wall, a plan redrawn at a larger scale: each makes a copy with the same shape and a different size. A scale factor says how much bigger the copy is, since every length is multiplied by one number. It does not say where the copy goes, or how to draw it. “Twice as big” fits a copy on the left of the page and a copy on the right equally well.

A dilation settles both questions with one rule. It names a single fixed point, the center, and moves every other point straight away from it or straight toward it, each one by the same factor. That decides where every point lands, and the copy that comes out has every length multiplied by the scale factor and every angle unchanged. A dilation is the move that makes “the same shape” something you can construct, and together with the rigid motions it is how similar figures are defined.

This lesson takes two days. Part 1 is the dilation itself: the center and the scale factor, what a dilation keeps and what it changes, and how to find the center of a dilation that has already been drawn. Part 2 puts dilations on the coordinate plane, first with the center at the origin and then with the center anywhere.

Definition

Choose a point O, the center, and a positive number k, the scale factor. The dilation with center O and scale factor k moves the points of the plane like this.

The center stays where it is.
Every other point P moves along the ray that starts at O and passes through P, to the point P' on that ray whose distance from O is k times the distance of P.

In symbols, P' is on ray OP and

OP' = k \cdot OP

The point P' is the image of P, read “P prime”, and P is the pre-image of P'. The image of a figure is the set of the images of all of its points. An arrow P \longmapsto P' is read “P goes to P'”.

When k > 1, every point moves away from O and the dilation is an enlargement. When 0 < k < 1, every point moves toward O and it is a reduction. When k = 1, nothing moves.

Figure 1Scale factor 2. Each point slides out along its own ray from O until it is twice as far from O.

The word ray matters. P' is on the same side of O as P: a dilation pushes points out or pulls them in, but never carries a point through the center to the other side.

Intuition

Think of every point as sitting on its own spoke from O. A dilation slides every point along its spoke, and on every spoke the distance from O is multiplied by the same number. With k = 3, a point 2 cm from O ends up 6 cm out, having moved 4 cm; a point 5 cm from O ends up 15 cm out, having moved 10 cm. Far points travel far and near points travel a little, in exact proportion to how far out they started. That proportion is what keeps the shape. Adding the same 2 cm to every distance instead would swell the parts of a figure near O more than the parts far away, and the shape would change.

The center decides where the copy lands, and nothing else. Put the center to the left of a triangle and an enlargement pushes the copy off to the right. Put it inside the triangle and the copy grows around it. Put it at a corner and that corner stays fixed while the copy grows out of it. Every one of these copies has the same size and the same shape; they sit in different places.

A lamp is a center in exactly this sense. Light travels in straight lines out from the lamp, so the shadow of a flat shape held parallel to a wall is a dilation of the shape, with the lamp as center. Move the shape toward the lamp and the scale factor grows.

Results

Part 1: what a dilation keeps and what it changes

Take a dilation with center O and scale factor k > 0.

The center is fixed. When k \neq 1 it is the only point that does not move.

Points stay on their rays. O, P and P' lie on one line, with P' on the same side of O as P. When k > 1, P lies between O and P'; when k < 1, P' lies between O and P. The distance a point moves is (k - 1) \cdot OP outward when k > 1, and (1 - k) \cdot OP inward when k < 1.

Every length is multiplied by k. For any two points P and Q,

P'Q' = k \cdot PQ

Every angle keeps its measure. Each line goes to itself or to a line parallel to it, which is the subject of what a dilation does to a line, and so the sides of each angle keep their directions.

The image is similar to the original. Matching angles are equal and every pair of matching lengths is in the ratio k.

Undoing and repeating. The dilation with the same center and scale factor \tfrac1k carries every image back to where it came from. A dilation with factor k_1 followed by one with the same center and factor k_2 is a single dilation with that center and factor k_1 k_2.

Finding the center and the factor. When k \neq 1, the line through any point and its image passes through the center, so two such lines that cross meet at the center. The scale factor is the ratio of any image length to the matching original length:

k = \frac{P'Q'}{PQ} = \frac{OP'}{OP}

Part 2: dilations on the coordinate plane

Center at the origin. The dilation with center (0, 0) and scale factor k multiplies both coordinates by k:

(x, y) \longmapsto (kx,\ ky)

Center anywhere. For the center (a, b), the offset of a point (x, y) is (x - a,\ y - b): how far across and how far up it sits from the center. The dilation multiplies the offset by k and adds it back on to the center:

\begin{aligned} x &\longmapsto a + k(x - a) \\ y &\longmapsto b + k(y - b) \end{aligned}

With a = b = 0 this is the first rule again.

Why it is true

The rule at the origin. Put the center at (0, 0) and take P = (x, y), not the origin. Walk from the origin toward P: after t times the distance to P you are at (tx, ty), so the points (tx, ty) with t \geq 0 make up ray OP. Since k > 0, the point (kx, ky) is on that ray. Its distance from the origin is

\sqrt{k^2x^2 + k^2y^2} = k\sqrt{x^2 + y^2} = k \cdot OP

using \displaystyle \sqrt{k^2} = k, which holds because k > 0. So (kx, ky) is on ray OP at k times the distance: it is P'.

The rule for any center. Measured from (a, b) as if it were the origin, the point (x, y) has coordinates (x - a,\ y - b), its offset. Measuring from a different point slides every coordinate by the same amounts and changes no distance and no ray, so the rule at the origin applies to offsets: the offset of P' is \big(k(x - a),\ k(y - b)\big). Adding the center back on gives the rule.

Lengths. Take P = (x_1, y_1) and Q = (x_2, y_2). In the rule for center (a, b), the first coordinates of P' and Q' differ by

\big(a + k(x_2 - a)\big) - \big(a + k(x_1 - a)\big) = k(x_2 - x_1)

and the second coordinates differ by k(y_2 - y_1). By the distance formula,

\begin{aligned} P'Q' &= \sqrt{k^2(x_2 - x_1)^2 + k^2(y_2 - y_1)^2} \\ &= k\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ &= k \cdot PQ \end{aligned}

Angles. Every side of a figure is carried to a segment on the same line or on a parallel line, pointing the same way because k > 0. Two angles whose sides point in the same two directions have the same measure, by corresponding angles along a line that crosses both pairs of sides.

Only one fixed point. If P \neq O then OP > 0, and OP' = k \cdot OP is a different distance when k \neq 1, so P' \neq P.

Undoing and repeating. Two dilations with center O keep every point on its own ray from O. The first takes P to distance k_1 \cdot OP, and the second multiplies that by k_2, giving k_1k_2 \cdot OP on the same ray. With k_2 = \tfrac1k after k_1 = k, the product is 1 and every point is back where it started.

Finding the center. For P \neq O and k \neq 1, the points O, P and P' are on one line and P' \neq P, so line PP' is that line and contains O. If a second point Q gives a different line QQ', the two lines share exactly one point, and the center is on both, so that point is the center.

Worked examples

Examples 1 to 5 go with Part 1, the dilation itself. Examples 6 to 9 go with Part 2, the coordinate plane.

Example 1

Where one point goes

The center of a dilation is O, and the point P is 9 cm from O. Where does P go when the scale factor is 2.5? When it is \tfrac23?

Solution

In both cases P' is on ray OP, and its distance from O is k times 9 cm.

With k = 2.5, OP' = 2.5 \cdot 9 = 22.5 cm. That is further out than P, so P' lies beyond P on the ray, 22.5 - 9 = 13.5 cm past it.

With k = \tfrac23, OP' = \tfrac23 \cdot 9 = 6 cm. That is nearer O than P, so P' lies between O and P, 9 - 6 = 3 cm in from P.

Check. A point moves (k - 1) \cdot OP outward in an enlargement: 1.5 \cdot 9 = 13.5. It moves (1 - k) \cdot OP inward in a reduction: \tfrac13 \cdot 9 = 3.

Example 2

The sides and angles of an image

A triangle has sides of 3 cm, 4 cm and 5 cm, and the angle between the two shorter sides is a right angle. The triangle is dilated with scale factor \tfrac25. Describe the image.

Solution

Every length is multiplied by \tfrac25:

\tfrac25 \cdot 3 = 1.2 \qquad \tfrac25 \cdot 4 = 1.6 \qquad \tfrac25 \cdot 5 = 2

Every angle keeps its measure, so the image is a right triangle with sides 1.2 cm, 1.6 cm and 2 cm, and its right angle is between the 1.2 cm and 1.6 cm sides.

Check. The image is a right triangle, so the Pythagorean theorem must hold for it: 1.2^2 + 1.6^2 = 1.44 + 2.56 = 4 = 2^2. It does. The perimeter went from 12 cm to 4.8 cm, and \tfrac25 \cdot 12 = 4.8, since a perimeter is a sum of lengths.

The center was never mentioned, and the answer did not need it. The center decides where the image sits, not its size or its shape.

Example 3

The center at a corner

Rectangle ABCD has AB = 6 and BC = 4. Dilate it with center A and scale factor \tfrac12.

Figure 2Center A, scale factor \tfrac12. The corner at the center stays put, and the image grows out of it.

Solution

The center is fixed, so A' = A.

B moves along ray AB to half its distance from A: AB' = \tfrac12 \cdot 6 = 3, so B' is the midpoint of \overline{AB}. In the same way AD' = \tfrac12 \cdot 4 = 2, so D' is the midpoint of \overline{AD}.

C moves along ray AC, the diagonal. The diagonal has length \displaystyle AC = \sqrt{6^2 + 4^2} = \sqrt{52} = 2\sqrt{13}, so \displaystyle AC' = \sqrt{13}, and C' is the midpoint of the diagonal.

The image AB'C'D' is a rectangle 3 by 2, tucked into the corner at A inside the original.

Check. Lay a grid with A at (0, 0), so B = (6, 0), C = (6, 4) and D = (0, 4). The midpoints are B' = (3, 0), C' = (3, 2) and D' = (0, 2), which are the corners of a 3 by 2 rectangle. Its diagonal is \displaystyle \sqrt{3^2 + 2^2} = \sqrt{13}, half of \displaystyle 2\sqrt{13}.

Example 4

Finding the center from a drawing

In Figure 3, triangle PQR has been dilated to triangle P'Q'R', and a ruler gives PQ = 4 cm and P'Q'= 6 cm. Find the center and the scale factor. If the center turns out to be 5 cm from P, how far is it from P'?

Figure 3The lines joining matching points all pass through the center.

Solution

The center, P and P' are on one line, so the center is on line PP'. For the same reason it is on line QQ'. Draw both lines and extend them back past the smaller triangle: they cross at one point, O, and that is the center. Line RR' also passes through O, which is a check on the drawing.

Every length is multiplied by k, so

k = \frac{P'Q'}{PQ} = \frac{6}{4} = 1.5

It is an enlargement, so P' is further from O than P is: OP' = 1.5 \cdot 5 = 7.5 cm, and PP' = 7.5 - 5 = 2.5 cm.

Check. The point moved (k - 1) \cdot OP = 0.5 \cdot 5 = 2.5 cm. That relation also runs backwards: knowing only k and the distance PP', the center is \dfrac{PP'}{k - 1} from P along line PP', on the side away from P'. Here \dfrac{2.5}{0.5} = 5.

Example 5

A shadow on a wall

A lamp shines on a wall 200 cm away. A flat cardboard bird 15 cm tall is held 40 cm from the lamp, parallel to the wall. How tall is its shadow? Where should the bird be held to throw a shadow 45 cm tall?

Solution

Light travels in straight lines from the lamp, so each point of the shadow is on the ray from the lamp through a point of the bird. The bird and the wall are parallel, so every one of those rays is cut by them in the same ratio, the ratio of their distances from the lamp. The shadow is the image of the bird under a dilation with center at the lamp and

k = \frac{200}{40} = 5

The shadow is 5 \cdot 15 = 75 cm tall.

A 45 cm shadow needs k = \tfrac{45}{15} = 3. The wall stays at 200 cm, so the bird must be at a distance d with \tfrac{200}{d} = 3, which gives d = \tfrac{200}{3} = 66\tfrac23 cm from the lamp.

Check. At 66\tfrac23 cm, the factor is 200 \div \tfrac{200}{3} = 3, and 3 \cdot 15 = 45. Moving the bird away from the lamp made the factor, and the shadow, smaller.

Example 6

Center at the origin

Triangle DEF has D = (-4, 2), E = (2, 4) and F = (6, -2). Dilate it with center at the origin and scale factor \tfrac12, and compare \overline{DE} with its image.

Solution

With the center at the origin, the dilation multiplies both coordinates by \tfrac12:

\begin{aligned} D = (-4, 2) &\longmapsto D' = (-2, 1) \\ E = (2, 4) &\longmapsto E' = (1, 2) \\ F = (6, -2) &\longmapsto F' = (3, -1) \end{aligned}

By the distance formula, \displaystyle DE = \sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10} and \displaystyle D'E' = \sqrt{3^2 + 1^2} = \sqrt{10}, half as long. The slopes are \tfrac26 = \tfrac13 and \tfrac13, so \overline{D'E'} is parallel to \overline{DE}.

Check. D' should be on the ray from the origin through D, halfway out. It is: (-2, 1) is exactly half of (-4, 2) in each coordinate, the midpoint of the segment from (0, 0) to D.

Example 7

A center that is not the origin

A dilation has center C = (2, 1) and scale factor 2. Find the image of triangle PQR with P = (3, 2), Q = (5, 2) and R = (4, 4).

Figure 4Center C = (2, 1), scale factor 2. Each point’s offset from C is doubled.

Solution

Work with offsets from the center. The offset of P is (3 - 2,\ 2 - 1) = (1, 1): one across and one up from C. In the same way Q has offset (3, 1) and R has offset (2, 3).

Doubling an offset keeps the point on the same ray from C and puts it twice as far out. The doubled offsets are (2, 2), (6, 2) and (4, 6), and adding each to C = (2, 1) gives

P' = (4, 3) \qquad Q' = (8, 3) \qquad R' = (6, 7)

Check. The rule gives the same for P: 2 + 2(3 - 2) = 4 and 1 + 2(2 - 1) = 3. And \overline{PQ} is horizontal with length 2, while \overline{P'Q'} is horizontal with length 4, twice as long.

Doubling the coordinates themselves would give (6, 4), (10, 4) and (8, 8). That is the dilation with center at the origin: a triangle of the same size and shape, sitting 2 to the right and 1 higher. Same factor, different center, different place.

Example 8

Finding the center and scale factor from coordinates

A dilation takes A = (1, 3) to A' = (5, 5) and B = (0, 0) to B' = (2, -4). Find its scale factor and its center.

Solution

The scale factor is the ratio of matching lengths. By the distance formula, \displaystyle AB = \sqrt{1^2 + 3^2} = \sqrt{10} and \displaystyle A'B' = \sqrt{3^2 + 9^2} = \sqrt{90} = 3\sqrt{10}, so

k = \frac{3\sqrt{10}}{\sqrt{10}} = 3

B is at the origin but it moved, so the origin is not the center. Call the center (a, b). The rule for A must give A'. The first coordinates give

\begin{aligned} a + 3(1 - a) &= 5 \\ 3 - 2a &= 5 \\ a &= -1 \end{aligned}

and the second coordinates give

\begin{aligned} b + 3(3 - b) &= 5 \\ 9 - 2b &= 5 \\ b &= 2 \end{aligned}

The center is (-1, 2).

Check with the other point. B = (0, 0) has offset (1, -2) from (-1, 2). Tripled that is (3, -6), and (-1 + 3,\ 2 - 6) = (2, -4), which is B'.

Check with lines. Line AA' has slope \tfrac{5 - 3}{5 - 1} = \tfrac12, so it is y = \tfrac12 x + \tfrac52. Line BB' is y = -2x. They meet where -2x = \tfrac12 x + \tfrac52, at x = -1 and y = 2, the same point.

Example 9

Undoing a dilation, and two in a row

(a) The dilation with center at the origin and scale factor 4 takes a point P to P' = (8, -12). Find P.

(b) Dilate with center at the origin and scale factor 2, then dilate the result with center (4, 0) and scale factor \tfrac12. Where does a point (x, y) end up? Is the combined move a dilation?

Solution

(a) The dilation with the same center and factor \tfrac14 undoes it, so it takes P' back to P. A quarter of 8 is 2 and a quarter of -12 is -3, so P = (2, -3). Check. 4 \cdot 2 = 8 and 4 \cdot (-3) = -12.

(b) The first dilation takes (x, y) to (2x, 2y). The second uses the rule with center (4, 0) and factor \tfrac12 on that point:

\begin{aligned} 2x &\longmapsto 4 + \tfrac12(2x - 4) = x + 2 \\ 2y &\longmapsto 0 + \tfrac12(2y - 0) = y \end{aligned}

So (x, y) \longmapsto (x + 2,\ y). Every point moves 2 to the right: this is a translation, not a dilation. No point stays fixed, and a dilation always has a fixed point, its center.

The factors multiply to 2 \cdot \tfrac12 = 1, so lengths come out unchanged, as a translation needs. With the same center twice, the two would have undone each other and nothing would have moved. With different centers, the second does not pull the points back to where they started.

Boundaries

The scale factor is positive here. A factor of 0 would send every point to the center and crush the whole plane to one point, so it is not allowed. Some books allow a negative factor: it sends P to the ray opposite ray OP, |k| times as far, which is the dilation with factor |k| followed by a half turn about O. With the center at the origin the rule (x, y) \longmapsto (kx, ky) still describes it.

A factor of 1 moves nothing. Every point is fixed, so there is no single center, and the method for finding the center fails because each line PP' shrinks to a point.

Finding the center needs two different lines. If P, Q and the center are on one line, then PP' and QQ' are that same line, and a third point off it is needed. If the lines joining matching points are parallel and never meet, no dilation produced the copy; a translation, for one, gives parallel joining lines.

The same factor in every direction. The rule (x, y) \longmapsto (x, 2y) doubles heights and leaves widths alone. It is not a dilation: it changes angles and the image is not similar to the original. That one-way move is a stretch, studied with graphs.

(kx, ky) is for the center at the origin only. With the center at (a, b), it is the offsets that are multiplied, not the coordinates.

A similar copy need not be a dilation image. A copy that is twice as big but turned, or flipped over, has sides that are not parallel to the original’s, so no dilation alone produces it. A dilation followed by a rigid motion does, which is a similarity transformation.

Repeating with different centers. The rule “multiply the factors” needs one shared center. Two dilations with different centers combine into a dilation with factor k_1k_2 about some third center, or, when k_1k_2 = 1, into a translation (Example 9).

In space. The definition works unchanged in three dimensions: every point slides along its ray from the center to k times the distance, and every length is multiplied by k.

Where this goes next

A dilation moves points; the next question is what it does to whole lines, and the answer is what a dilation does to a line: a line through the center stays where it is and every other line goes to a parallel line. That fact is what makes the angles of an image equal to the original’s.

Following a dilation with rotations, reflections and translations gives the similarity transformations, and two figures are similar exactly when one of these carries one onto the other. Once lengths scale by k, perimeters scale by k too, and areas by k^2, which is its own result.

Put the center at a corner of a triangle, as in Example 3, and the two sides from that corner are stretched along themselves while the third side moves to a parallel segment. Read the other way, that picture says a line parallel to one side of a triangle cuts the other two sides in the same ratio: the triangle proportionality theorem.

Two right triangles with the same acute angle can be placed with that angle at a shared corner, and then one is a dilation of the other from that corner. So the ratio of two of their sides depends on the angle alone, which is where sine, cosine and tangent come from.

The offset rule of Part 2 multiplies an offset, a pair of numbers, by k. That is scaling a vector, and a dilation is the picture of that algebra.

Exercises

Part 1: one point at a time

A dilation has center O and scale factor 3. A point P is 5 cm from O. Find OP' and PP'.
A dilation has center O and scale factor \tfrac14, and OP = 12 cm. Find OP' and PP', and say which of P and P' is nearer O.
A dilation with center O takes a point 6 cm from O to a point 4.5 cm from O. Find the scale factor. Is it an enlargement or a reduction?
P' is on ray OP beyond P, with OP = 4 cm and PP' = 6 cm. Find the scale factor.
P' is on ray OP between O and P, with OP = 10 cm and PP' = 6 cm. Find the scale factor.

Part 1: images of figures

A dilation with scale factor 2.5 is applied to a segment 7 cm long and to an angle of 40^\circ. Describe the two images.
A triangle has sides of 6 cm, 9 cm and 12 cm. Under a dilation, the longest side of its image is 8 cm. Find the scale factor and the other two sides of the image.
A square with side 5 cm is dilated with scale factor 3, centered at the point where its diagonals cross. Find the side and the perimeter of the image, and say where the image’s diagonals cross.
One rectangle is 4 cm by 6 cm and another is 6 cm by 10 cm. Can a dilation take the first onto the second?
A lamp is 50 cm from a flat cut-out 12 cm tall, and 300 cm from a wall parallel to the cut-out. How tall is the shadow?
Triangle ABC is dilated with center A and scale factor 3. AB = 2 cm and AC = 5 cm. Find AB', AC' and BB', and say where A' is.

Part 1: centers and repeats

A dilation with center O and scale factor 4 is followed by one with center O and scale factor \tfrac38. Which single dilation has the same effect, and which dilation undoes it?
A dilation with scale factor 3 moves a point P a distance PP' = 8 cm. How far is P from the center?
A reduction with scale factor \tfrac14 moves a point P a distance PP' = 9 cm. Find OP and OP'.

Part 1: reasoning

Explain why, when k \neq 1, the center is the only point that a dilation leaves where it is.
A dilation takes P to P' and Q to Q', where k \neq 1. Explain why lines PP' and QQ' both pass through the center. When do these two lines fail to pin the center down?
Explain why every dilation with scale factor 2 gives an image of the same size and shape, whatever the center. What does the center decide?

Part 2: center at the origin

Each dilation in Exercises 18 to 21 has its center at the origin.

Find the image of the triangle with vertices (3, -1), (0, 4) and (-2, -2) under scale factor 2.
Find the image of (9, -6) under scale factor \tfrac23.
A dilation takes (-8, 12) to (-2, 3). Find the scale factor.
Which of these rules are dilations with center at the origin, and with what scale factor? (x, y) \longmapsto (3x, 3y); (x, y) \longmapsto (3x, y); (x, y) \longmapsto (x + 3, y + 3); (x, y) \longmapsto \left(\tfrac{x}{2}, \tfrac{y}{2}\right).

Part 2: center anywhere

A dilation has center (1, -2) and scale factor 2. Find the image of (4, 2), and the distances of the point and its image from the center.
A dilation has center (4, 4) and scale factor \tfrac12. Find the image of the square with vertices (0, 0), (8, 0), (8, 8) and (0, 8).
A dilation has center (-3, 1) and scale factor 3. Find the image of the triangle with vertices (-1, 1), (-3, 2) and (-2, 3).
Write the rule for the dilation with center (2, -3) and scale factor 4 in the form (x, y) \longmapsto (px + q,\ ry + s).

Part 2: working backwards

A dilation takes A = (2, 1) to A' = (5, 7) and B = (4, 1) to B' = (11, 7). Find its scale factor and its center.
A dilation with scale factor \tfrac14 takes (6, 8) to (3, 5). Find its center.
Dilate with center at the origin and scale factor 2, then with center (6, 0) and scale factor 3. Find the rule for the combined move, and show that it is a single dilation by finding its center and scale factor.
Dilate with center at the origin and scale factor \tfrac13, then with center (0, 6) and scale factor 3. Find the rule for the combined move. Is it a dilation?

Part 2: reasoning

Show that the dilation with center (a, b) and scale factor k, using its coordinate rule, leaves (a, b) fixed, and, when k \neq 1, leaves no other point fixed.
Show, using coordinates, that a dilation with any center multiplies the length of every segment by its scale factor.
A dilation takes (0, 0) to itself and (2, 0) to (6, 0). Explain why its center must be the origin, and find its scale factor.
Explain why a triangle and its image under a dilation are similar.

Answers

OP' = 15 cm and PP' = 10 cm. OP' = 3 \cdot 5 = 15, and P' is beyond P on the ray, so PP' = 15 - 5 = 10.
OP' = 3 cm and PP' = 9 cm, and P' is nearer O. OP' = \tfrac14 \cdot 12 = 3; a reduction pulls P in toward O, so PP' = 12 - 3 = 9.
k = \tfrac34, a reduction. k = \dfrac{OP'}{OP} = \dfrac{4.5}{6} = 0.75, which is less than 1.
k = \tfrac52. P' is beyond P, so OP' = 4 + 6 = 10 and k = \tfrac{10}{4}.
k = \tfrac25. P' is between O and P, so OP' = 10 - 6 = 4 and k = \tfrac{4}{10}.
A segment 17.5 cm long and an angle of 40^\circ. Lengths are multiplied by the factor, 2.5 \cdot 7 = 17.5, and angles keep their measure.
k = \tfrac23, and the other sides are 4 cm and 6 cm. Multiplying by a positive number keeps the order of sizes, so the longest side goes to the longest: k = \tfrac{8}{12} = \tfrac23. Then \tfrac23 \cdot 6 = 4 and \tfrac23 \cdot 9 = 6.
Side 15 cm, perimeter 60 cm, and the diagonals still cross at the same point. 3 \cdot 5 = 15 and 4 \cdot 15 = 60, which is 3 times the original 20. The center does not move, and each diagonal lies on a line through the center, so each is stretched along itself and the image’s diagonals cross at the center.
No. A dilation multiplies every length by one factor. The short sides would need k = \tfrac64 = \tfrac32 and the long sides k = \tfrac{10}{6} = \tfrac53, and these differ.
72 cm. The shadow is a dilation of the cut-out with center at the lamp and k = \tfrac{300}{50} = 6, and 6 \cdot 12 = 72.
AB' = 6 cm, AC' = 15 cm, BB' = 4 cm, and A' = A. The center is fixed. B' is on ray AB with AB' = 3 \cdot 2 = 6, so BB' = 6 - 2 = 4; and AC' = 3 \cdot 5 = 15.
The dilation with center O and scale factor \tfrac32, undone by the one with center O and scale factor \tfrac23. With one center the factors multiply, 4 \cdot \tfrac38 = \tfrac32, and \tfrac32 \cdot \tfrac23 = 1.
4 cm, and then OP' = 12 cm. An enlargement moves a point (k - 1) \cdot OP = 2 \cdot OP, so 2 \cdot OP = 8.
OP = 12 cm and OP' = 3 cm. A reduction moves a point in by (1 - k) \cdot OP, so \tfrac34 \cdot OP = 9, and OP' = \tfrac14 \cdot 12 = 3.
Any point P \neq O has OP > 0, and its image has OP' = k \cdot OP, which differs from OP when k \neq 1. A point at a different distance from O is a different point, so P' \neq P. The center is the image of itself by definition, so it is the only fixed point.
P' is on ray OP, so O, P and P' are on one line, and P' \neq P by Exercise 15, so line PP' is that line and contains O. The same holds for QQ'. If the two lines are different they meet in exactly one point, which must be O. They fail when P, Q and O are on one line, since then PP' and QQ' are the same line; a third point off that line settles it.
Every dilation with factor 2 doubles every length and keeps every angle, and neither fact depends on where the center is, so all the images have the same size and shape. The center decides where the image sits, because points move along rays from the center; Example 7 shows two centers giving the same triangle in two places.
(6, -2), (0, 8) and (-4, -4). Both coordinates of each vertex are doubled.
(6, -4). \tfrac23 \cdot 9 = 6 and \tfrac23 \cdot (-6) = -4.
k = \tfrac14. From -8k = -2 and 12k = 3; both give \tfrac14, as they must for a dilation.
(x, y) \longmapsto (3x, 3y) with factor 3, and (x, y) \longmapsto \left(\tfrac{x}{2}, \tfrac{y}{2}\right) with factor \tfrac12. The rule (3x, y) sends (1, 1) to (3, 1), which is not on the ray from the origin through (1, 1), so it is a sideways stretch. The rule (x + 3, y + 3) moves the origin, which a dilation centered there must keep; it is a translation.
(7, 6); the point is 5 from the center and its image 10. The offset is (4 - 1,\ 2 - (-2)) = (3, 4), doubled (6, 8), so the image is (1 + 6,\ -2 + 8). The distances are \displaystyle \sqrt{3^2 + 4^2} = 5 and \displaystyle \sqrt{6^2 + 8^2} = 10.
(2, 2), (6, 2), (6, 6) and (2, 6). The offsets from (4, 4) are (-4, -4), (4, -4), (4, 4) and (-4, 4); halved and added back to (4, 4) they give these points. The center is the center of the square, so the image is a square of side 4 around the same middle.
(3, 1), (-3, 4) and (0, 7). The offsets from (-3, 1) are (2, 0), (0, 1) and (1, 2); tripled they are (6, 0), (0, 3) and (3, 6), added back to (-3, 1).
(x, y) \longmapsto (4x - 6,\ 4y + 9). 2 + 4(x - 2) = 4x - 6 and -3 + 4(y + 3) = 4y + 9. Check: (2, -3) goes to (8 - 6,\ -12 + 9) = (2, -3), so the center stays put.
k = 3 and the center is \left(\tfrac12, -2\right). AB = 2 and A'B' = 6, so k = 3. Then a + 3(2 - a) = 5 gives 6 - 2a = 5 and a = \tfrac12, and b + 3(1 - b) = 7 gives 3 - 2b = 7 and b = -2. Check with B: \tfrac12 + 3\left(4 - \tfrac12\right) = 11 and -2 + 3(1 + 2) = 7.
(2, 4). From a + \tfrac14(6 - a) = 3, multiplying by 4, 3a + 6 = 12 and a = 2; from b + \tfrac14(8 - b) = 5, 3b + 8 = 20 and b = 4. Check: the offset of (6, 8) is (4, 4), a quarter of that is (1, 1), and (2 + 1,\ 4 + 1) = (3, 5).
(x, y) \longmapsto (6x - 12,\ 6y), the dilation with center \left(\tfrac{12}{5}, 0\right) and scale factor 6. The first step gives (2x, 2y); then 6 + 3(2x - 6) = 6x - 12 and 0 + 3(2y - 0) = 6y. A dilation with factor 6 and center (a, b) has rule (6x - 5a,\ 6y - 5b), which matches when 5a = 12 and b = 0.
(x, y) \longmapsto (x,\ y - 12), a translation 12 down, not a dilation. The first step gives \left(\tfrac{x}{3}, \tfrac{y}{3}\right); then 0 + 3\left(\tfrac{x}{3} - 0\right) = x and 6 + 3\left(\tfrac{y}{3} - 6\right) = y - 12. No point stays fixed, although \tfrac13 \cdot 3 = 1 keeps every length.
At (a, b) the offsets are 0, so the rule gives a + k \cdot 0 = a and b + k \cdot 0 = b. If (x, y) is fixed, then a + k(x - a) = x, so (k - 1)(x - a) = 0; since k \neq 1, x = a, and in the same way y = b.
Take P = (x_1, y_1) and Q = (x_2, y_2). In the rule for center (a, b) the a terms cancel, so the first coordinates of P' and Q' differ by k(x_2 - x_1), and the second by k(y_2 - y_1). Write u = x_2 - x_1 and v = y_2 - y_1. The distance formula gives \displaystyle P'Q' = \sqrt{k^2u^2 + k^2v^2}, which is \displaystyle k\sqrt{u^2 + v^2} = k \cdot PQ, using k > 0.
k = 3. The point (2, 0) moved, so k \neq 1, and then the center is the only fixed point (Exercise 15). The origin is fixed, so it is the center. Then the rule is (x, y) \longmapsto (kx, ky), and 2k = 6 gives k = 3.
The dilation multiplies every side of the triangle by the same factor k and keeps every angle, so matching angles are equal and matching sides are in one ratio, which is what similar means. A dilation is also one of the moves allowed in the transformation definition of similarity, so the image is similar by that definition too.