Printable · GCSE Higher · ages 14-16
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 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
- 2.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.
- 3.Triangle T has an area of 6 cm². It is enlarged by scale factor −2 about the origin, and the image is then rotated 90° about the origin. Work out the area of the final image, and state whether it is the same way round as T or a mirror image of it.
- 4.A game designer places a coin at (5, 3) on a grid. The game rotates the coin 90° clockwise about the point (2, 3) each time the player presses a button, and this is applied twice in a row. Work out the coordinates of the coin after the button is pressed twice.
- 5.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′?
- 6.Point P has coordinates (2, 1). Transformation A reflects a point in the x-axis. Transformation B translates a point by the vector (0, 4). Work out the coordinates of the image of P when A is applied first, followed by B.
- 7.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?
- 8.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?
- 9.A trapezium has vertices (0, 2), (4, 2), (3, 5) and (1, 5). It is rotated 180° about the point (2, 2), and the image is then translated by the vector (1, −5). Work out the coordinates of the image of (0, 2).
- 10.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
- 11.Triangle S has vertices (2, 2), (5, 2) and (2, 5). It is mapped onto triangle S′ with vertices (2, 5), (5, 5) and (2, 2). Which single composition of two transformations maps S onto S′?
- 12.A trapezium has vertices (2, 1), (6, 1), (5, 4) and (3, 4). It is rotated 180° about the vertex (2, 1). Work out the number of points on the trapezium — including its vertices, edges and interior — that are invariant under this rotation.
- 13.A shape is reflected in the line y = −x. Which of these points is invariant under this reflection?y = −x
- 14.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.
- 15.Triangle D has vertices (3, 1), (3, 5) and (6, 1). It is reflected in the line x = 3. Write down the number of vertices of the triangle that are invariant under this reflection.
Answer key
- (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°.
- (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).
- (d) 24 cm²; same way round as T — An enlargement scales area by the square of the scale factor, whatever its sign: area factor = (−2)² = 4, so the image's area is 6 × 4 = 24 cm². A negative scale factor is equivalent to an enlargement by the positive scale factor together with a rotation of 180°, so it does not create a mirror image — it is still a direct transformation, and combining it with a further rotation cannot create one either, since rotations never change a shape's orientation. Using the scale factor itself rather than its square gives area 6 × 2 = 12 cm², which is too small. Believing that a negative scale factor reflects the shape gives 'a mirror image of T' alongside the correct area of 24 cm², and combining both mistakes gives 12 cm² together with 'a mirror image of T'.
- (d) (−1, 3) — Method: two 90° rotations about the SAME centre, applied one after another, combine into a single 180° rotation about that same centre: use the shortcut (x, y) → (2a − x, 2b − y) for a half-turn about (a, b). Working: with centre (2, 3), doubling each coordinate gives 2 × 2 = 4 and 2 × 3 = 6, so the rule is (x, y) → (4 − x, 6 − y). Applying it to (5, 3) gives 4 − 5 = −1 and 6 − 3 = 3, so the coin ends at (−1, 3). Answer: (−1, 3). Rotate about the centre (2, 3) stated in the game, not about the origin, and remember the button is pressed TWICE: stopping after one press, or rotating about the wrong centre, both leave the coin somewhere else.
- (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.
- (b) (2, 3) — Applying A first: reflecting (2, 1) in the x-axis gives (2, −1). Applying B to that image: translating (2, −1) by (0, 4) gives (2, −1 + 4) = (2, 3). Applying the transformations in the opposite order — B first, then A — gives a different result: (2, 1) translates to (2, 5), which then reflects to (2, −5); this shows that the order genuinely matters here. Applying only A and stopping there, without the translation, gives (2, −1). Applying only B and stopping there, without the reflection, gives (2, 5). Do both transformations, in the order A then B, and the image of P is (2, 3).
- (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.
- (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.
- (b) (5, −3) — Method: rotate the point about the given centre first, then translate the image, in the stated order. Working: rotating (0, 2) by 180° about (2, 2) uses the rule (x, y) → (4 − x, 4 − y), since the centre doubles in each coordinate. This gives 4 − 0 = 4 and 4 − 2 = 2, so (0, 2) maps to (4, 2). Translating (4, 2) by the vector (1, −5) gives 4 + 1 = 5 and 2 − 5 = −3, so the final image is (5, −3). Answer: (5, −3). Rotate about the centre GIVEN in the question, (2, 2), not about the origin, and translate the rotated image afterwards, in that order: rotating about the wrong centre, swapping the order, or stopping after one step all give a different point.
- (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).
- (a) Reflect in the x-axis, then translate by (0, 7). — Reflecting in the x-axis sends (x, y) to (x, −y); applied to S's vertices (2, 2), (5, 2) and (2, 5) this gives (2, −2), (5, −2) and (2, −5). Translating this image by the vector (0, 7) adds 7 to every y-coordinate, giving (2, 5), (5, 5) and (2, 2), which matches S′ exactly. Using the correct reflection but translating by (7, 0) instead moves the image sideways rather than upwards, giving (9, −2), (12, −2) and (9, −5) — nowhere near S′. Reflecting in the y-axis instead of the x-axis changes the sign of the x-coordinate rather than the y-coordinate, so translating that image by (0, 7) gives (−2, 9), (−5, 9) and (−2, 12), the wrong shape entirely. Rotating 180° about the origin instead of reflecting sends every coordinate to its negative, so translating by (0, 7) gives (−2, 5), (−5, 5) and (−2, 2) — the y-coordinates match S′ but the x-coordinates do not.
- (b) 1 point — A rotation about a point P always leaves P itself unchanged, and — unless the shape has rotational symmetry about that exact point — no other point of the shape maps onto itself. The centre of rotation here is the vertex (2, 1), so exactly one point of the trapezium, that vertex, is invariant. Thinking that a rotation always has no invariant points ignores the centre of rotation itself, which is always fixed, and gives 0 points. Assuming every vertex of the shape is invariant confuses a rotation with the identity transformation and gives 4 points, the total number of vertices. Believing that the point diametrically opposite the centre is also fixed applies point symmetry of the whole coordinate grid rather than checking whether that point actually lies on this particular trapezium, and gives 2 points.
- (a) (4, −4) — A point is invariant under a reflection only if it lies exactly on the mirror line. The line y = −x consists of every point where the y-coordinate is the negative of the x-coordinate: (4, −4) satisfies this, since −4 = −(4), so it is invariant. (5, 5) lies on the line y = x, a different line altogether, not y = −x. (4, 4) has equal coordinates, but that alone does not put it on y = −x; it would need y = −4, not 4. (−4, −4) also has equal coordinates and lies on y = x, not y = −x — its coordinates would need opposite signs to sit on the given mirror line. Only a point whose coordinates are negatives of each other stays fixed under this reflection.
- (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) 2 — Method: a point is invariant under a reflection exactly when it lies on the mirror line, so check each vertex's coordinate against the line's equation. Working: the line of reflection is x = 3, so a vertex is invariant only if its x-coordinate equals 3. (3, 1) has x = 3, so it is invariant. (3, 5) has x = 3, so it is also invariant. (6, 1) has x = 6, so it is not invariant, since 6 is not equal to 3. Two of the three vertices are invariant. Answer: 2. Check EVERY vertex against the mirror line's equation rather than assuming a shape either keeps all its vertices fixed or none of them: a vertex is invariant only when it sits exactly on the line, and here two do and one does not.
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