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Combinations of transformations and invariance worksheet — GCSE Higher
Fifteen questions on "combinations of transformations and invariance" — DfE statement G8. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Combinations of transformations and invariance worksheet — GCSE Higher
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- 1.A shape is reflected in the line x = 1, and the image is then reflected in the line x = 5. Which single transformation is equivalent to this combination, for every point?
- 2.Line l1 has equation x = 2, and line l2 has equation y = x − 1. They intersect at a single point. A shape is reflected in l1, and the image is then reflected in l2. Which point is invariant under this combined transformation?y = x − 1
- 3.A designer plots a logo point at (3, 4) on a grid. She reflects it in the line y = 1 and then rotates the image 90° clockwise about the point (1, 1) to create a repeating tile pattern. Work out the coordinates of the point after both transformations.
- 4.Point P has coordinates (4, 2). P is rotated 90° clockwise about the point (1, 1), and the image is then reflected in the line x = 1. Work out the coordinates of the final image of P.
- 5.Point P has coordinates (1, 3). P is reflected in the line x = 4, then the image is reflected in the line y = 1, and then that image is translated by the vector (2, −3). Work out the coordinates of the final image of P.
- 6.A shape is translated by the vector (5, −3), and the image is then translated by the vector (−5, 3). Which single transformation has the same overall effect as this combination, for every point?
- 7.A shape is reflected in the line y = 1, and the image is then reflected in the line y = 4. Which single transformation is equivalent to this combination, for every point?
- 8.A shape is reflected in the x-axis and the image is then reflected in the y-axis. Which single transformation is equivalent to this combination for every point?
- 9.Point P has coordinates (6, 4). P is rotated 90° clockwise about the point (1, 2), and the image is then translated by the vector (−3, 5). Work out the coordinates of the final image of P.
- 10.Square B has vertices (1, 1), (4, 1), (4, 4) and (1, 4). It is rotated 90° clockwise about the origin, and the image is then translated by the vector (2, −1). Work out the coordinates of the image of (4, 1).
- 11.Point Q has coordinates (3, 5). It is translated by the vector (1, −2), and the image is then enlarged by scale factor 3, centre the origin. Work out the coordinates of the final image of Q.
- 12.Triangle T has vertices (1, 1), (3, 1) and (1, 4). It is mapped onto triangle T′ with vertices (5, −1), (3, −1) and (5, −4). Which single composition of two transformations maps T onto T′?
- 13.Triangle A has vertices (2, 1), (5, 1) and (2, 4). It is translated by the vector (−3, 2) and the image is then reflected in the x-axis. Work out the coordinates of the image of (5, 1).
- 14.A shape is reflected in the line y = x, and the image is then reflected in the line y = 0. These two lines meet at the origin. Work out the single rotation, centre and angle, that is equivalent to this combination of two reflections, for every point.y = x
- 15.A rotation of 180° about the point (3, 2) can be written as a reflection in the line x = 3, followed by a reflection in a second line, for every point in the plane. Which line is the second line?
Answer key
- (c) Translation by the vector (8, 0) — Method: reflecting twice in two parallel vertical lines is always equivalent to a single translation, at right angles to the lines, of twice the distance between them. Working: the two lines are 5 − 1 = 4 units apart, so the translation is 2 × 4 = 8 units in the positive x-direction. Answer: translation by the vector (8, 0). Using just the gap itself, without doubling it, gives (4, 0); translating in the negative x-direction, from the second line back towards the first, gives (−8, 0); and describing the combination as a single reflection in the line halfway between them, x = 3, confuses this combination with the effect of a single reflection — two reflections in parallel lines are always equivalent to a translation, never to another reflection. Always double the gap between the lines, and translate in the direction from the first line towards the second.
- (c) (2, 1) — A point that lies on both mirror lines is fixed by each reflection individually, and so is fixed by the combination of the two — it is the intersection point of l1 and l2 that is invariant. Substituting x = 2 into y = x − 1 gives y = 2 − 1 = 1, so the intersection point is (2, 1). Forgetting the '− 1' in l2's equation and using y = x instead gives (2, 2). Making a sign error and computing y = x − (−1) = x + 1 instead gives (2, 3). Solving for x from an assumed y = 0 instead of substituting the given x = 2 gives (1, 0). Substitute x = 2 into l2's equation correctly, and the invariant point is (2, 1).
- (b) (−2, −1) — Method: apply the reflection to the point first, then rotate the image about the given centre, in the order the question states them. Working: reflecting (3, 4) in the line y = 1 keeps x = 3 and puts the image as far below the line as the point is above it: 4 is 3 units above y = 1, so the image is 3 units below, at 1 − 3 = −2 (the same as 2 × 1 − 4 = −2). The reflected point is (3, −2). Rotating (3, −2) by 90° clockwise about (1, 1): subtracting the centre gives 3 − 1 = 2 and −2 − 1 = −3, the clockwise rule swaps and negates these to give −3 and −2, and adding the centre back gives 1 + (−3) = −2 and 1 + (−2) = −1. The final image is (−2, −1). Answer: (−2, −1). Reflect before you rotate, exactly as the design process is described, and rotate about the CENTRE (1, 1) given in the question rather than the origin: either mistake, or reversing the two steps, sends the tile to a different point.
- (d) (0, −2) — To rotate (4, 2) by 90° clockwise about (1, 1), first find its position relative to the centre: (4 − 1, 2 − 1) = (3, 1). A 90° clockwise rotation sends (a, b) to (b, −a), so (3, 1) becomes (1, −3); adding the centre back gives (1 + 1, 1 − 3) = (2, −2). Reflecting (2, −2) in the line x = 1 gives (2 × 1 − 2, −2) = (0, −2). Doing the two transformations in the opposite order, reflecting first and then rotating, gives a different result, (2, 4), which shows the order matters. Stopping after the rotation and forgetting the reflection gives (2, −2). Stopping after only reflecting P in x = 1 and forgetting the rotation entirely gives (−2, 2). Rotate first, then reflect, in that order, and the final image is (0, −2).
- (d) (9, −4) — Reflecting (1, 3) in the line x = 4 gives (2 × 4 − 1, 3) = (7, 3). Reflecting (7, 3) in the line y = 1 gives (7, 2 × 1 − 3) = (7, −1). Translating (7, −1) by the vector (2, −3) gives (7 + 2, −1 + (−3)) = (9, −4). Stopping after the two reflections and forgetting the translation gives (7, −1). Applying the translation's y-component with the wrong sign, adding 3 instead of subtracting it, gives (9, 2). Forgetting that the two reflections are centred on (4, 1) rather than the origin, and instead rotating (1, 3) by 180° about the origin to (−1, −3) before translating, gives (1, −6). Do all three steps in order, each one correctly, and the final image is (9, −4).
- (d) No transformation — every point stays exactly where it was — The two vectors (5, −3) and (−5, 3) are opposites, so adding them gives (0, 0): every point ends up exactly where it started, and there is no transformation at all. Misreading the second vector's signs and effectively adding (5, −3) to itself instead of to its opposite gives a translation by the vector (10, −6). Assuming two translations must combine into a reflection gives a reflection in the x-axis — but a reflection reverses orientation, and translations never do. Assuming that two opposite vectors must mean a half turn gives a rotation of 180° about the origin — but a 180° rotation moves every point except its own centre, whereas this pair of translations leaves every single point exactly where it was. Two translations by opposite vectors always cancel exactly, leaving every point unmoved.
- (a) Translation by the vector (0, 6) — Method: reflecting twice in two parallel horizontal lines is always equivalent to a single translation, at right angles to the lines, of twice the distance between them. Working: the two lines are 4 − 1 = 3 units apart, so the translation is 2 × 3 = 6 units in the positive y-direction. Answer: translation by the vector (0, 6). Using just the gap itself, without doubling it, gives (0, 3); translating in the negative y-direction, from the second line back towards the first, gives (0, −6); and describing the combination as a single reflection in the line halfway between them, y = 2.5, confuses this combination with the effect of a single reflection — two reflections in parallel lines are always equivalent to a translation, never to another reflection. Always double the gap between the lines, and translate in the direction from the first line towards the second.
- (a) A rotation of 180° about the origin — Method: composing two reflections in lines that cross is always a single rotation about the point where the lines meet, through twice the angle between them. Working: the x-axis and y-axis meet at the origin at an angle of 90°, so the combined transformation is a rotation about the origin through 2 × 90 = 180 degrees. Answer: a rotation of 180° about the origin. The rotation angle is TWICE the angle between the mirror lines, not the angle itself, and the centre is always where the two lines cross, not some other point, and the result of two reflections in intersecting lines is a rotation, never another reflection.
- (b) (0, 2) — Method: to rotate about a point that is not the origin, first subtract the centre's coordinates, apply the rotation rule to the shifted point, then add the centre's coordinates back on; only after that do you apply the translation, in the order the question states them. Working: shifting P relative to the centre gives (6 − 1, 4 − 2) = (5, 2); rotating 90° clockwise sends (x, y) to (y, −x), giving (2, −5); adding the centre back on gives (2 + 1, −5 + 2) = (3, −3); applying the translation (−3, 5) gives (3 − 3, −3 + 5) = (0, 2). Answer: (0, 2). Applying the translation BEFORE the rotation, reversing the order the question gives them in, gives (8, 0); stopping after the rotation and forgetting the translation altogether gives (3, −3); and rotating anticlockwise instead of clockwise, using (x, y) → (−y, x), gives (−4, 12). Always carry out the two transformations in the order stated — rotate about the given centre first, then translate — and check each step before moving to the next.
- (b) (3, −5) — Method: apply the rotation to the point first, then translate the image, in the order the question gives them. Working: rotating (4, 1) by 90° clockwise about the origin sends (x, y) to (y, −x), so (4, 1) becomes (1, −4). Translating (1, −4) by the vector (2, −1) gives 1 + 2 = 3 and −4 − 1 = −5, so the final image is (3, −5). Answer: (3, −5). Use the CLOCKWISE rule, (x, y) → (y, −x), not the anticlockwise one, and apply the rotation before the translation, exactly as the question states them: reversing the order or the direction of turn both land on a different point.
- (b) (12, 9) — Apply the transformations in the order given: first translate, then enlarge. Translating (3, 5) by the vector (1, −2) gives (3 + 1, 5 − 2) = (4, 3). Enlarging this by scale factor 3 about the origin multiplies both coordinates by 3: (4 × 3, 3 × 3) = (12, 9). Enlarging first and translating afterwards reverses the order and gives (3 × 3 + 1, 5 × 3 − 2) = (10, 13), a different point because the two transformations do not commute. Enlarging the original point by scale factor 3 while forgetting to translate it at all gives (3 × 3, 5 × 3) = (9, 15). Reversing the signs of the translation vector before applying it gives (3 − 1, 5 + 2) = (2, 7), which then enlarges to (2 × 3, 7 × 3) = (6, 21).
- (b) Rotate 180° about the origin, then translate by (6, 0). — Rotating 180° about the origin sends (x, y) to (−x, −y); applied to T's vertices (1, 1), (3, 1) and (1, 4) this gives (−1, −1), (−3, −1) and (−1, −4). Translating this image by the vector (6, 0) adds 6 to every x-coordinate, giving (5, −1), (3, −1) and (5, −4), which matches T′ exactly. Reflecting in the x-axis first changes the sign of the y-coordinate only, and translating that image by (6, 0) gives (7, −1), (9, −1) and (7, −4) — the wrong triangle. Using the correct rotation but translating by (4, 0) instead of (6, 0) gives (3, −1), (1, −1) and (3, −4), shifted 2 units too far left. Reflecting in the y-axis first changes the sign of the x-coordinate only, so translating that image by (6, 0) leaves every y-coordinate positive, giving (5, 1), (3, 1) and (5, 4) — the correct x-coordinates but the wrong sign throughout on y.
- (c) (2, −3) — Method: apply the transformations in the order given, translation first, then reflection, to the point itself. Working: translating (5, 1) by the vector (−3, 2) moves the x-coordinate by −3 and the y-coordinate by 2, giving 5 − 3 = 2 and 1 + 2 = 3, so the image after the translation is (2, 3). Reflecting this point in the x-axis leaves the x-coordinate unchanged and reverses the sign of the y-coordinate, giving (2, −3). Answer: (2, −3). Do the two maps in the stated order and carry the point all the way through both of them: swapping the order, or stopping after the first map, lands you on a different point altogether.
- (a) 90° clockwise about (0, 0) — Two reflections in lines through a common point compose to a single rotation about that point, through an angle equal to twice the angle between the two lines, in the direction from the first line to the second. The line y = x makes a 45° angle with the line y = 0, so the resulting rotation turns through 2 × 45° = 90°; testing the point (1, 0) — which reflects to (0, 1) in y = x, then to (0, −1) in y = 0 — shows the turn is clockwise, about the origin where the two lines cross. Taking the rotation anticlockwise instead reverses the direction the two reflections actually compose in. Using 45° directly, without doubling the angle between the lines, gives an angle equal to only half the true rotation. Treating any pair of reflecting lines as perpendicular, and so always giving a 180° rotation, ignores that these two lines actually meet at 45°, not 90°.
- (d) y = 2 — Two reflections in perpendicular lines that cross at a point combine to a 180° rotation about that point. The line x = 3 is vertical, so the second line must be horizontal, and it must pass through the centre of rotation (3, 2) — that line is y = 2. Taking the y-coordinate of the centre but writing it against the wrong letter gives y = 3. Assuming the second line must also be vertical, like the first one, and just swapping in the other coordinate gives x = 2. Reaching for the standard mirror line y = x without checking that it actually passes through (3, 2) gives y = x — it does not pass through that point at all. The line that is both perpendicular to x = 3 and through (3, 2) is y = 2.
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