Transformations · applications

Applications: Transformations

10 question types · Secondary 3 · each worked step by step with a figure that follows the steps

O-Level · GCSE Higher

01

A Sofa Moved Twice on a Floor Plan, and the Vector That Brings It Back

methodAdd Each Vector to Every Corner in Turn, Add the Two Vectors for the Single Move, and Change Both Signs for the Way Back

On a floor plan marked in meters, a sofa is the rectangle with corners A(1, 1), B(4, 1), C(4, 2) and D(1, 2). It is slid across the floor by the vector 52, and then, after a rug has been laid, by the vector −23. (a) Find the coordinates of the corners of the sofa after both moves. (b) Find the single vector that would have made both moves at once, and the vector that would slide the sofa straight back to where it started.

xy4726ABCDA'move 1:52adds 5 to x and 2 to yA'(6, 3), B'(9, 3), C'(9, 4), D'(6, 4)
The first move adds 5 to each x-coordinate and 2 to each y-coordinate: A(1, 1) goes to A'(6, 3).
The first move adds 5 to every x-coordinate and 2 to every y-coordinate. So A(1, 1) goes to (1 + 5, 1 + 2) = (6, 3), B(4, 1) goes to (9, 3), C(4, 2) goes to (9, 4) and D(1, 2) goes to (6, 4).
step 1 of 5

A translation slides every point of a shape by the same column vector: the top number is added to the x-coordinate and the bottom number to the y-coordinate. The shape keeps its size, its shape and the way it faces. Two translations, one after the other, make a single translation whose vector is the sum of the two, and the vector that undoes a translation has both of its numbers with the opposite sign.

  1. The first move adds 5 to every x-coordinate and 2 to every y-coordinate. So A(1, 1) goes to (1 + 5, 1 + 2) = (6, 3), B(4, 1) goes to (9, 3), C(4, 2) goes to (9, 4) and D(1, 2) goes to (6, 4).
  2. The second move subtracts 2 from every x-coordinate and adds 3 to every y-coordinate. So (6, 3) goes to (6 − 2, 3 + 3) = (4, 6), (9, 3) goes to (7, 6), (9, 4) goes to (7, 7) and (6, 4) goes to (4, 7).
  3. (a) After both moves the corners are at (4, 6), (7, 6), (7, 7) and (4, 7). The sofa is still 3 m long and 1 m deep, because a translation keeps every length.
  4. One move that does the work of both is found by adding the two vectors: 52 + −23 = 5 − 22 + 3 = 35. Check: A(1, 1) moved by this vector lands at (4, 6), the corner found in two moves.
  5. (b) The single vector is 35. To slide the sofa straight back, every point must move 3 m in the negative x-direction and 5 m in the negative y-direction, so the vector back is −3−5. Check: C''(7, 7) moved by it lands at (4, 2), where C started.

answer(a) (4, 6), (7, 6), (7, 7) and (4, 7); (b) the single vector is 35, and the vector back is −3−5

techniqueTranslation of Shapes

examsO-Level · GCSE Higher

Common pitfalls

  • Subtracting the vector instead of adding it, which sends A to (1 − 5, 1 − 2) = (−4, −1), outside the room. The top number of the vector is how far the sofa moves in the positive x-direction, so it is added to each x-coordinate.
  • Giving the way back as the second vector with its signs changed, 2−3. That undoes only the second move and leaves the sofa where the first move put it; the way back must undo the whole of 35.
02

A Hiking Map on a Tablet: the Slide That Puts the Campsite in the Middle of the Window

methodFind the Vector from the Center of the Window to the Campsite, Then Add It to Every Corner

A hiking app shows a rectangular window of a map. On the map's grid, marked in kilometers, the window has corners (0, 0), (6, 0), (6, 4) and (0, 4). A hiker taps a campsite at (10, 7), and the app slides the window across the map, without turning it or changing its size, until the campsite is at the center of the window. (a) Find the vector by which the window is translated. (b) Find the corners of the window after the slide, and decide whether a waterfall at (14, 6) can be seen in the window.

xy71359(3, 2)campsitewaterfallcenter: x = (0 + 6)/2 = 3, y = (0 + 4)/2 = 2
The center of the window is halfway along each side, at (3, 2).
The center of the window is halfway along each side: x = 0 + 62 = 3 and y = 0 + 42 = 2. So the center starts at (3, 2).
step 1 of 5

Sliding the window without turning it or changing its size is a translation, so every point of the window, its center included, moves by the same vector. The vector from a point to its image is the image's coordinates minus the point's coordinates, written as a column.

  1. The center of the window is halfway along each side: x = 0 + 62 = 3 and y = 0 + 42 = 2. So the center starts at (3, 2).
  2. The center must end at the campsite, (10, 7). The vector from (3, 2) to (10, 7) is 10 − 37 − 2 = 75.
  3. (a) The window is translated by 75, which is 7 km in the x-direction and 5 km in the y-direction.
  4. Every corner moves by the same vector: (0, 0) goes to (7, 5), (6, 0) goes to (13, 5), (6, 4) goes to (13, 9) and (0, 4) goes to (7, 9). Check: the center of the new window is (7 + 132, 5 + 92) = (10, 7), which is the campsite.
  5. (b) The new window has corners (7, 5), (13, 5), (13, 9) and (7, 9), so it shows x from 7 to 13 and y from 5 to 9. The waterfall's y-coordinate, 6, is in that range, but its x-coordinate, 14, is 1 km beyond the right-hand edge at x = 13. The waterfall cannot be seen in the window.

answer(a) 75; (b) the corners are (7, 5), (13, 5), (13, 9) and (7, 9), and the waterfall at (14, 6) cannot be seen: it is 1 km beyond the right-hand edge

techniqueTranslation of Shapes

examsO-Level · GCSE Higher

Common pitfalls

  • Moving the corner (0, 0) onto the campsite and translating by 107. That puts the campsite at the bottom left corner of the window, not at its center.
  • Subtracting the wrong way round and giving −7−5. The vector from the center to the campsite is the campsite's coordinates minus the center's; the reverse vector moves the window away from the campsite.
03

A Shadow Puppet Between a Lamp and a Screen: the Size of the Shadow and Where to Hold the Puppet

methodDivide the Lamp's Distance to the Screen by Its Distance to the Puppet for the Scale Factor, Then Carry Each End of the Puppet Along Its Ray from the Lamp

A side view of a shadow puppet show is drawn on a grid in which one unit is 10 cm. A small lamp is at L(0, 3), and a flat puppet is held upright between P(2, 2) and Q(2, 4), facing a screen along the line x = 6. The shadow on the screen is an enlargement of the puppet with center L. (a) Find the scale factor, and the coordinates of the ends of the shadow. (b) The puppeteer wants a shadow 40 cm tall, and moves the puppet in the x-direction only, keeping its ends at heights y = 2 and y = 4. At what x-coordinate should the puppet be held, and where are the ends of the shadow then?

xy233screen x = 6PQLlamp to screen 6, lamp to puppet 2k = 6/2 = 3
The lamp is 6 units from the screen and 2 units from the puppet, so k = 62 = 3.
Measured across the grid, the lamp is 6 − 0 = 6 units from the screen and 2 − 0 = 2 units from the puppet. So the scale factor is k = 62 = 3.
step 1 of 5

Light travels in straight lines from the lamp, so each point of the shadow lies on the ray from the lamp through the matching point of the puppet. That makes the shadow an enlargement with center L: the image of a point X lies on the ray LX, k times as far from L as X is. Because the puppet and the screen are both upright, k is the lamp's distance to the screen divided by its distance to the puppet.

  1. Measured across the grid, the lamp is 6 − 0 = 6 units from the screen and 2 − 0 = 2 units from the puppet. So the scale factor is k = 62 = 3.
  2. From L(0, 3) to P(2, 2) is 2−1, and three times that is 6−3, so the image of P is at (0 + 6, 3 − 3) = (6, 0). In the same way, from L to Q(2, 4) is 21, so the image of Q is at (0 + 6, 3 + 3) = (6, 6).
  3. (a) The scale factor is 3, and the shadow runs from (6, 0) to (6, 6). Check: the shadow is 6 units, or 60 cm, tall, which is 3 times the puppet's 2 units.
  4. The puppet is 2 units, or 20 cm, tall, so a shadow 40 cm tall needs k = 4020 = 2. If the puppet is held at x = d, the lamp is d units from it, so k = 6d. Then 6d = 2 gives d = 3.
  5. (b) The puppet should be held at x = 3, halfway between the lamp and the screen. From L to (3, 2) is 3−1, and twice that reaches (6, 1); from L to (3, 4) is 31, and twice that reaches (6, 5). The shadow runs from (6, 1) to (6, 5), which is 4 units, or 40 cm, tall.

answer(a) the scale factor is 3, and the shadow runs from (6, 0) to (6, 6); (b) at x = 3, with the shadow from (6, 1) to (6, 5)

techniqueEnlargement with a Center

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing the distances the other way round, 26 = 13, which would make the shadow smaller than the puppet. The screen is farther from the lamp than the puppet is, so the shadow is larger and k is more than 1.
  • Measuring from the screen instead of from the lamp, which gives 64 = 1.5. Every distance in an enlargement is measured from its center, and the center here is the lamp.
04

A Photo Enlarged in an Editor: the Point It Grows From and Where a Sticker Lands

methodJoin Two Corners to Their Images and Extend the Lines Until They Meet, Then Multiply the Vector from That Point by the Scale Factor

In a photo editor, the grid is marked in centimeters. A photo PQRS has corners P(3, 2), Q(5, 2), R(5, 3) and S(3, 3). After it is enlarged, the photo P'Q'R'S' has corners P'(7, 4), Q'(13, 4), R'(13, 7) and S'(7, 7). (a) Find the center of the enlargement and the scale factor. (b) A sticker on the small photo has one corner at (4, 2.5). Where is that corner on the enlarged photo?

xy1713147PQRSP'Q'R'S'S(3, 3) to S'(7, 7): up 4 for 4 acrossgradient 1: y = x
The line through S(3, 3) and S'(7, 7) has gradient 1: it is y = x.
The line through S(3, 3) and S'(7, 7) rises 4 for 4 across, so its gradient is 1, and since it passes through (3, 3) its equation is y = x.
step 1 of 5

In an enlargement, every point and its image lie on a straight line through the center, so the lines joining corners to their images all pass through the center. The scale factor is the length of a side of the image divided by the matching side of the object. The image of any point is then found by multiplying its vector from the center by the scale factor.

  1. The line through S(3, 3) and S'(7, 7) rises 4 for 4 across, so its gradient is 1, and since it passes through (3, 3) its equation is y = x.
  2. The line through P(3, 2) and P'(7, 4) rises 2 for 4 across, so its gradient is 12 and its equation is y − 2 = 12(x − 3). Where it meets y = x: x − 2 = 12(x − 3), so 2x − 4 = x − 3, which gives x = 1 and then y = 1.
  3. The side PQ is 5 − 3 = 2 cm long and the side P'Q' is 13 − 7 = 6 cm long, so the scale factor is 62 = 3.
  4. (a) The center of the enlargement is (1, 1) and the scale factor is 3. Check: from (1, 1) to P(3, 2) is 21, and three times that, 63, reaches (7, 4), which is P'.
  5. (b) From the center (1, 1) to the sticker's corner (4, 2.5) is 31.5. Three times that is 94.5, so on the enlarged photo the corner is at (1 + 9, 1 + 4.5) = (10, 5.5). Check: it lies inside P'Q'R'S', between x = 7 and x = 13 and between y = 4 and y = 7, as a point of the photo must.

answer(a) the center is (1, 1) and the scale factor is 3; (b) (10, 5.5)

techniqueEnlargement with a Center

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Multiplying the sticker's coordinates by 3 to get (12, 7.5), which is not even on the enlarged photo. That enlarges from the origin, but this photo grew from (1, 1), so it is the vector from (1, 1) that is multiplied by 3.
  • Taking the scale factor as the difference of the lengths, 6 − 2 = 4. An enlargement multiplies lengths, so the scale factor is the ratio of the lengths, 62 = 3.
05

A Camera Obscura: the Image of a Tree on the Back Wall

methodTake the Pinhole as the Center, Compare the Distances on the Two Sides of It for the Scale Factor, and Multiply the Vector from the Pinhole to Each End of the Tree

A camera obscura is a dark room with a small hole in one wall; light through the hole forms an image of the scene outside on the opposite wall. A side view is drawn on a grid in meters, with the pinhole at O(0, 0), 2 m above the flat ground, which is the line y = −2. A tree 6 m tall stands 12 m in front of the pinhole, from its base B(−12, −2) to its top T(−12, 4). The back wall of the room is the line x = 3. The image of the tree on the back wall is an enlargement of the tree with center O. (a) Find the scale factor of the enlargement. (b) Find the coordinates of the images of the top and of the base of the tree, and the height of the image.

xy24B(−12, −2)T(−12, 4)Otree: 12 m in front of O, wall: 3 m behind O3/12 = 1/4
The tree is 12 m in front of the pinhole and the back wall 3 m behind it: the image is 312 = 14 as far from O.
The tree is 12 m in front of the pinhole and the back wall is 3 m behind it. So the image is 312 = 14 as far from O as the tree is, on the opposite side.
step 1 of 5

Light from each point of the tree passes straight through the pinhole, so a point and its image lie on one line through O, on opposite sides of it. That is an enlargement with center O and a negative scale factor. The image of a point X is at the end of k times the vector OX, and the minus sign in k sends it to the other side of the center.

  1. The tree is 12 m in front of the pinhole and the back wall is 3 m behind it. So the image is 312 = 14 as far from O as the tree is, on the opposite side.
  2. (a) The scale factor is k = −14 = −0.25. Its size, 14, says the image is a quarter of the size of the tree, and its minus sign says the image is on the other side of the pinhole.
  3. The top of the tree is T(−12, 4), so OT = −124. Multiplying by −14 gives 3−1, so the image of the top is at (3, −1), on the back wall.
  4. The base of the tree is B(−12, −2). Multiplying −12−2 by −14 gives 30.5, so the image of the base is at (3, 0.5).
  5. (b) The image of the top is at (3, −1) and the image of the base is at (3, 0.5), so the image is 0.5 − (−1) = 1.5 m tall, with the top of the tree at the bottom. Check: 1.5 m is 14 of the tree's 6 m, and since the floor is at y = −2, the image runs from 1 m to 2.5 m above the floor.

answer(a) k = −14 = −0.25; (b) the top at (3, −1) and the base at (3, 0.5), and the image is 1.5 m tall

techniqueEnlargement with a Center

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Giving the scale factor as 14, with no sign. A positive scale factor would put the image on the same side of the pinhole as the tree, with the top at (−3, 1), which is outside the room.
  • Using 123 = 4 as the scale factor, which would make the image 24 m tall. The scale factor is the image's distance from the center divided by the object's distance, 3 divided by 12, not the other way round.
06

Two Mirrors of a Kaleidoscope at 45 Degrees, and the Image of an Image

methodReflect in One Mirror and Then in the Other, Compare the Final Image with the Bead Corner by Corner, and Repeat in the Other Order

In a kaleidoscope, two long mirrors meet along a line at an angle of 45°. Looking down the tube, a grid in millimeters puts the mirrors along the positive x-axis and along the line y = x, meeting at O(0, 0). A triangular bead between them has corners A(4, 1), B(6, 1) and C(5, 2). (a) The bead is reflected in the mirror along the x-axis, and that image is reflected in the line y = x. Find the coordinates of the corners of the final image A''B''C''. (b) Describe fully the single transformation that maps the bead onto A''B''C''. Is the result the same if the bead is reflected in y = x first?

xy−226−6−26mirror 1mirror 2OABCin the x-axis: (x, y) → (x, −y)(4, 1) → (4, −1), (6, 1) → (6, −1), (5, 2) → (5, −2)
Reflecting in the x-axis changes the sign of each y-coordinate.
Reflecting in the x-axis changes the sign of each y-coordinate: A(4, 1) goes to (4, −1), B(6, 1) goes to (6, −1) and C(5, 2) goes to (5, −2).
step 1 of 6

A reflection in the x-axis changes the sign of the y-coordinate, so (x, y) goes to (x, −y). A reflection in the line y = x swaps the two coordinates, so (x, y) goes to (y, x). To describe a single transformation fully, name it and give what fixes it: for a rotation, its center, its angle and its direction.

  1. Reflecting in the x-axis changes the sign of each y-coordinate: A(4, 1) goes to (4, −1), B(6, 1) goes to (6, −1) and C(5, 2) goes to (5, −2).
  2. Reflecting in y = x swaps the coordinates: (4, −1) goes to (−1, 4), (6, −1) goes to (−1, 6) and (5, −2) goes to (−2, 5).
  3. (a) The final image has corners A''(−1, 4), B''(−1, 6) and C''(−2, 5).
  4. Compare A with A'' from O. OA = √42 + 12 = √17 and OA'' = √(−1)2 + 42 = √17, so they are equally far from O. The gradient of OA is 14 and the gradient of OA'' is 4−1 = −4, and 14 × (−4) = −1, so OA and OA'' are at right angles. The same holds for B and C, and each image is a quarter turn counterclockwise from its corner.
  5. In the other order, y = x first sends A(4, 1) to (1, 4), and then the x-axis sends it to (1, −4). In the same way B ends at (1, −6) and C at (2, −5). Each of these is a quarter turn clockwise about O from its corner.
  6. (b) The single transformation is a rotation of 90° counterclockwise about O(0, 0), which is twice the 45° between the mirrors. Reflecting in y = x first gives a rotation of 90° clockwise about O instead, so the result is not the same: the order of the two reflections matters.

answer(a) A''(−1, 4), B''(−1, 6) and C''(−2, 5); (b) a rotation of 90° counterclockwise about O(0, 0); reflecting in y = x first gives a rotation of 90° clockwise about O, so the order matters

techniqueCombined Transformations · Reflection · Rotation

examsGCSE Higher

Common pitfalls

  • Describing the result as a reflection, because it was made from reflections. The corners A, B, C run counterclockwise round the bead, and so do A'', B'', C''; one reflection turns a shape over and reverses that order, and a second reflection turns it back.
  • Giving the rotation without its center or direction. A quarter turn clockwise about O puts A at (1, −4), and a quarter turn about another point puts it somewhere else again, so the center, the angle and the direction are all part of the answer.
07

A Mat Between the Parallel Mirrors of a Dance Studio, and Its Image in the Far Mirror

methodReflect Each Corner in the Near Mirror and Then in the Far One, and Follow a General Point (x, y) Through Both Mirrors

A dance studio has mirrors on two opposite walls, which lie along the lines x = 3 and x = 8 on a floor plan marked in meters. A triangular mat on the floor has corners A(4, 2), B(5, 2) and C(4, 4). A dancer sees the mat reflected in the mirror on x = 3, and that image reflected again in the mirror on x = 8. (a) Find the coordinates of the corners A'', B'' and C'' of this second image. (b) Describe fully the single transformation that maps the mat onto the second image, and show that it would be the same wherever the mat lay between the mirrors.

xy1424x = 3x = 8ABCA'B'C'mirror x = 3: x becomes 6 − xA'(2, 2), B'(1, 2), C'(2, 4)
In the mirror on x = 3 each corner goes as far behind the line as it was in front: x becomes 6 − x.
In the mirror on x = 3, A(4, 2) is 1 m in front of the line, so its image is 1 m behind it, at (2, 2). In the same way B(5, 2) goes to (1, 2) and C(4, 4) goes to (2, 4). In each case x becomes 6 − x.
step 1 of 6

A reflection in a vertical line x = a keeps the y-coordinate and puts the point as far beyond the line as it was in front of it, so its x-coordinate becomes 2a − x. Two reflections, one after the other, can be replaced by one transformation, found by following a general point (x, y) through both of them.

  1. In the mirror on x = 3, A(4, 2) is 1 m in front of the line, so its image is 1 m behind it, at (2, 2). In the same way B(5, 2) goes to (1, 2) and C(4, 4) goes to (2, 4). In each case x becomes 6 − x.
  2. In the mirror on x = 8, x becomes 16 − x: (2, 2) goes to (14, 2), (1, 2) goes to (15, 2) and (2, 4) goes to (14, 4).
  3. (a) The second image has corners A''(14, 2), B''(15, 2) and C''(14, 4).
  4. Each corner has moved 10 m in the x-direction and not at all in the y-direction: A(4, 2) to (14, 2), B(5, 2) to (15, 2) and C(4, 4) to (14, 4). The mat also faces the same way as before, with B'' to the right of A'' and C'' above it, so this is a translation by 100.
  5. For a general point (x, y), the first mirror gives (6 − x, y) and the second gives (16 − (6 − x), y) = (x + 10, y). The 10 does not depend on x or on y, and it is twice the 8 − 3 = 5 m between the mirrors.
  6. (b) The single transformation is the translation by 100: 10 m at right angles to the mirrors, twice the distance between them. It is the same wherever the mat lies, because every point (x, y) goes to (x + 10, y).

answer(a) A''(14, 2), B''(15, 2) and C''(14, 4); (b) the translation by 100, twice the 5 m between the mirrors, the same for every point

techniqueCombined Transformations · Reflection · Translation of Shapes

examsGCSE Higher

Common pitfalls

  • Expecting the second image to be turned over, because each step was a reflection. The first reflection turns the mat over and the second turns it back, so the second image faces the same way as the mat.
  • Giving the translation as 5 m, the distance between the mirrors. The general point moves 10 m: each reflection carries a point to twice its distance from the mirror, and together the two moves add up to twice the gap.
08

A Bank Shot on a Pool Table: Where to Strike the Cushion

methodReflect the Target Ball in the Cushion, Aim a Straight Line at Its Image, and Find Where That Line Crosses the Cushion

A plan of a pool table is drawn on a grid in which one unit is 25 cm. The cushions are the lines x = 0, x = 10, y = 0 and y = 5. The cue ball is at C(1, 2) and a target ball is at T(9, 4). A player plans to roll the cue ball against the cushion y = 5 so that it rebounds onto the target. Assume the ball leaves the cushion at the same angle as it strikes it. (a) Find the coordinates of T', the reflection of the target ball in the cushion y = 5. (b) Find the point on the cushion that the cue ball must strike, and the distance, in meters, that the cue ball rolls to the target.

xy179246cushion y = 5CTT'(9, 6)T(9, 4) is 1 below y = 5so T'(9, 6) is 1 above it
(a) T(9, 4) is 1 unit below the cushion, so its reflection is T'(9, 6), 1 unit above it.
(a) The target T(9, 4) is 1 unit below the cushion y = 5, so its reflection is 1 unit above it, at T'(9, 6).
step 1 of 5

When a ball leaves a cushion at the same angle as it strikes it, its path after the cushion is the reflection, in the cushion, of the straight line it was rolling along. So a ball aimed at the image of the target in the cushion reaches the target itself, and its path is exactly as long as the straight line to the image.

  1. (a) The target T(9, 4) is 1 unit below the cushion y = 5, so its reflection is 1 unit above it, at T'(9, 6).
  2. The straight line from C(1, 2) to T'(9, 6) rises 4 for 8 across, so its gradient is 12 and its equation is y − 2 = 12(x − 1).
  3. The line meets the cushion where y = 5: 5 − 2 = 12(x − 1), so x − 1 = 6 and x = 7. The cue ball must strike the cushion at P(7, 5).
  4. The path from C to P to T is as long as the straight line CT', which is √82 + 42 = √80 ≈ 8.944 units. At 25 cm to a unit, that is 8.944 × 0.25 ≈ 2.24 m.
  5. (b) The cue ball must strike the cushion at (7, 5), and it rolls about 2.24 m. Check: from C to P the ball goes 6 across and 3 up, and from P to T it goes 2 across and 1 down, so both parts of the path slope at 1 in 2 and make the same angle with the cushion.

answer(a) T'(9, 6); (b) the point (7, 5) on the cushion, and the cue ball rolls √80 units, about 2.24 m

techniqueReflection

examsGCSE Higher

Common pitfalls

  • Aiming at the point of the cushion straight above the target, (9, 5). The ball arrives there moving to the right and rebounds still moving to the right, so it leaves the cushion heading away from T, which is straight below (9, 5).
  • Reflecting the target by changing the sign of its y-coordinate, to (9, −4). That is a reflection in the line y = 0, the opposite cushion; the reflection in y = 5 puts the image as far above y = 5 as T is below it.
09

A Robot Arm That Swings a Part Across a Workbench: the Pivot and the Turn

methodDraw the Perpendicular Bisectors of Two Segments Joining Points to Their Images, Meet Them at the Center, Then Compare the Directions from the Center

A robot arm turns about a pivot hidden under its base cover. On a plan of a workbench marked in decimeters, it picks up a triangular part with corners A(6, 1), B(8, 1) and C(6, 2) and puts it down with its corners at A'(4, 5), B'(4, 7) and C'(3, 5). The move is a single rotation about the pivot. (a) Find the coordinates of the pivot, the center of the rotation. (b) Find the angle of the rotation, less than 180°, and its direction.

xy3625ABCA'B'C'AA': midpoint (5, 3), gradient −2bisector: y = 3 + 1/2 (x − 5)
The pivot is as far from A as from A', so it lies on the perpendicular bisector of AA': y = 3 + 12(x − 5).
A(6, 1) goes to A'(4, 5). The midpoint of AA' is (5, 3), and AA' has gradient 5 − 14 − 6 = −2, so its perpendicular bisector has gradient 12 and equation y = 3 + 12(x − 5).
step 1 of 5

In a rotation every point stays the same distance from the center, so the center is equally far from a point and its image and lies on the perpendicular bisector of the segment joining them. The perpendicular bisectors for two points meet at the center. The angle of the rotation is the angle at the center between the line to a point and the line to its image.

  1. A(6, 1) goes to A'(4, 5). The midpoint of AA' is (5, 3), and AA' has gradient 5 − 14 − 6 = −2, so its perpendicular bisector has gradient 12 and equation y = 3 + 12(x − 5).
  2. B(8, 1) goes to B'(4, 7). The midpoint of BB' is (6, 4), and BB' has gradient 7 − 14 − 8 = −32, so its perpendicular bisector has gradient 23 and equation y − 4 = 23(x − 6), which is y = 23x.
  3. (a) The bisectors meet where 23x = 3 + 12(x − 5). Multiplying by 6 gives 4x = 18 + 3x − 15, so x = 3 and y = 23 × 3 = 2. The pivot is at (3, 2).
  4. From the pivot, A is 3−1 away and A' is 13 away. Both have length √10, and their gradients, −13 and 3, multiply to −1, so the lines from the pivot to A and to A' are at right angles.
  5. (b) The direction to A points right and a little down, and the direction to A' points up and a little right, so the turn from one to the other is a quarter turn counterclockwise. The rotation is 90° counterclockwise about (3, 2). Check: C is 30 from the pivot, and a quarter turn counterclockwise makes that 03, which reaches (3, 5), where C' is.

answer(a) the pivot is at (3, 2); (b) 90° counterclockwise

techniqueRotation

examsGCSE Higher

Common pitfalls

  • Taking the midpoint of AA', (5, 3), as the center. The center lies on the perpendicular bisector of AA' but may be anywhere along it; a second pair of points is needed to fix it.
  • Giving the angle without its direction. A quarter turn clockwise about (3, 2) would send A to (2, −1), nowhere near where the part was put down, so the direction is part of the answer.
10

A Hubcap with Five Spokes, Lined Up Again for a Photograph

methodCount the Positions in a Full Turn That Look the Same, Divide 360 Degrees by That Order, and Find the Remainder of the Turn

A car's hubcap has five identical spokes, equally spaced round its center, and no other markings. (a) Find the order of rotational symmetry of the hubcap, and the smallest angle through which it can be turned about its center to look exactly the same. (b) In a photo shoot, the car is rolled forward and the wheel turns through 1000°. Through what smallest further angle, in the same direction, must the wheel turn for the hubcap to look exactly as it did at the start?

spoke 15 identical spokes: the same look5 times in one full turn
Every spoke moves into the place of the next one after a fifth of a turn, and this happens 5 times in a full turn.
As the hubcap turns about its center, every spoke moves into the place of the next one after a fifth of a turn. In one full turn this happens 5 times, the last time with every spoke back in its own place.
step 1 of 5

A shape has rotational symmetry of order n when it looks exactly the same in n positions during one full turn about its center. The smallest angle that maps it onto itself is then 360° divided by n, and a turn through any whole-number multiple of that angle maps it onto itself too.

  1. As the hubcap turns about its center, every spoke moves into the place of the next one after a fifth of a turn. In one full turn this happens 5 times, the last time with every spoke back in its own place.
  2. (a) The order of rotational symmetry is 5, and the smallest angle is 360° ÷ 5 = 72°.
  3. The hubcap looks as it did at the start after any turn that is a whole-number multiple of 72°. Dividing, 1000 ÷ 72 is 13 remainder 64, because 13 × 72 = 936 and 1000 − 936 = 64.
  4. So after 1000° the wheel is 64° past the position it reached after 936°, which looked like the start. The next position that looks like the start comes after 14 × 72 = 1008°.
  5. (b) The wheel must turn a further 1008 − 1000 = 8°. Check: 1008 ÷ 72 = 14 exactly, so after 1008° every spoke is where a spoke was at the start.

answer(a) order 5, and the smallest angle is 72°; (b) 8°

techniqueRotational Symmetry

examsGCSE Higher

Common pitfalls

  • Working with whole turns: 1000 − 2 × 360 = 280, and 360 − 280 = 80° further. That brings each spoke back to its own place, but the hubcap already looks the same whenever each spoke is where another spoke was, every 72°.
  • Giving the remainder, 64°, as the answer. The remainder is how far the wheel has gone past the last position that looked like the start; the further turn is what is left to the next one, 72 − 64 = 8°.
Mr. Chalk Read the guide