Describing and Combining Transformations

Has the triangle been turned, flipped over, slid or changed size?

Look for a line that is exactly halfway between each point of A and its matching point on B.

Pick one vertex of A and count how far it moves to reach the matching vertex of B.

Compare a side of B with the matching side of A.

To describe a rotation fully you need three things.

The mirror line is exactly halfway between each vertex of A and its matching vertex of B.

1. The vertex (2,1) of A matches the vertex (2,-3) of B.
2. Halfway between y=1 and y=-3 is y=-1. The vertex (4,3) and its match (4,-5) give the same: halfway is y=-1.
3. The mirror line is horizontal, so its equation is y=-1.

B is A turned upside down. Find the point that is exactly halfway between a vertex of A and its matching vertex of B.

1. B is the same size as A and has been turned through a half turn, so it is a rotation of 180°.
2. The right angle of A is at (2,1) and the right angle of B is at (-2,-3). Halfway between them is (0,-1).
3. Check another pair: (5,1) and (-5,-3) also have (0,-1) halfway between them.
4. The transformation is a rotation of 180° about (0,-1).

Compare matching sides for the scale factor. For the centre, think of lines through matching vertices.

Part (a)

1. The base of A is 2 squares long and the base of B is 4 squares long.
2. 4\div2=2, so the scale factor is 2.

Part (b)

1. The line through (3,5) and (2,4) goes 1 left and 1 down, so carry it on to (1,3).
2. The line through (7,5) and (4,4) goes 3 left and 1 down, so carry it on to (1,3) as well.
3. The centre of enlargement is (1,3).

Think about what someone would need to know to draw the image exactly.

Part (a)

1. A translation slides a shape. The column vector says how far across and how far up or down it moves.
2. So you give the word translation and the column vector.

Part (b)

1. An enlargement changes the size of a shape from a fixed point.
2. So you give the word enlargement, the scale factor and the centre of enlargement.

Is B bigger or smaller than A? The description must take A to B, not B to A.

1. B is smaller than A. The base of A is 4 squares and the base of B is 2 squares.
2. 2\div4=\frac{1}{2}, so the scale factor from A to B is \frac{1}{2}.
3. Each vertex of B is half as far from the origin as the matching vertex of A, for example (6,2) and (3,1), so the centre is (0,0).
4. The transformation is an enlargement, scale factor \frac{1}{2}, centre (0,0).

Do the reflection first, then the translation.

1. Reflecting in the x-axis keeps the x-coordinate and changes the sign of the y-coordinate: (2,3) goes to (2,-3).
2. Translate by moving 4 left and 1 up: 2-4=-2 and -3+1=-2.
3. The final position is (-2,-2).

Use the grid: turn Q a quarter turn clockwise about the origin, then translate it.

Part (a)

1. Q is 1 right and 4 up from the origin.
2. A quarter turn clockwise takes 'up' to 'right' and 'right' to 'down'.
3. So the image is 4 right and 1 down: (4,-1).

Part (b)

1. Move (4,-1) 2 left and 3 up.
2. 4-2=2 and -1+3=2.
3. The final position is (2,2).

Follow the vertex R one step at a time.

Part (a)

1. R is at (2,4).
2. Reflecting in the y-axis keeps the y-coordinate and changes the sign of the x-coordinate.
3. The matching vertex of B is (-2,4).

Part (b)

1. A rotation of 180° about (0,0) changes the sign of both coordinates.
2. (-2,4) goes to (2,-4).

Find Q first. Then count from Q to R.

Part (a)

1. Reflecting (3,2) in the x-axis gives Q at (3,-2).
2. From x=3 to x=1 is 2 left, so c=-2.

Part (b)

1. From y=-2 to y=3 is 5 up, so d=5.
2. The translation is \begin{pmatrix}-2\\5\end{pmatrix}.

Work backwards from C, undoing the last step first.

Part (a)

1. The translation moved B 2 right and 3 down to reach C.
2. So B is 2 left and 3 up from C: -1-2=-3 and -5+3=-2.
3. B is (-3,-2).

Part (b)

1. A rotation of 180° about (0,0) changes the sign of both coordinates.
2. Undoing it does the same, so A is (3,2).

Reflect A in the y-axis first. Then count how far the reflected triangle has to move to reach B.

Part (a)

1. Reflecting in the y-axis keeps the y-coordinate and changes the sign of the x-coordinate.
2. P goes to (-5,2).

Part (b)

1. After the reflection the triangle has vertices (-5,2), (-2,2) and (-2,4), with the long side at the bottom and the right angle on the right.
2. B has its right angle at (4,-4). The reflected triangle has its right angle at (-2,2).
3. From x=-2 to x=4 is 6 right, so c=6.

Part (c)

1. From y=2 to y=-4 is 6 down, so d=-6.
2. The translation is \begin{pmatrix}6\\-6\end{pmatrix}.

Undo the steps in reverse order: undo the reflection first, then the translation.

1. Undo the reflection in the x-axis: (2,4) goes back to (2,-4).
2. Undo the translation by moving 3 left and 1 up: 2-3=-1 and -4+1=-3.
3. The vertex of T is at (-1,-3).
4. Check: (-1,-3) translates to (2,-4), which reflects in the x-axis to (2,4).

Rotate each vertex first, then translate it. A quarter turn clockwise about the origin takes 'up' to 'right' and 'right' to 'down'.

Part (a)

1. P is 1 right and 1 up from the origin.
2. After the rotation it is 1 right and 1 down: (1,-1).
3. Translate 5 left and 5 up: 1-5=-4 and -1+5=4.
4. P′ is (-4,4).

Part (b)

1. R is 4 right and 3 up from the origin.
2. After the rotation it is 3 right and 4 down: (3,-4).
3. Translate 5 left and 5 up: 3-5=-2 and -4+5=1.
4. R′ is (-2,1).

Check each step separately. How far is P from the line x=1?

Part (a)

1. P is 3 to the right of the line x=1, so its reflection is 3 to the left of the line, at x=-2. Ali's step 1 is wrong: he reflected in the y-axis instead.
2. Step 2 is done correctly for the point he had: -4+3=-1 and 2-5=-3.

Part (b)

1. P reflected in x=1 is (-2,2).
2. Translate 3 right and 5 down: -2+3=1 and 2-5=-3.
3. The correct final position is (1,-3).

Rotate A first. Then compare the rotated triangle with B.

Part (a)

1. Rotating A 180° about (0,0) gives vertices (-2,-1), (-5,-1) and (-2,-3).
2. The right angle moves from (-2,-1) to (-2,4) in B: 0 across and 5 up.
3. The vector is \begin{pmatrix}0\\5\end{pmatrix}.

Part (b)

1. Rotating 180° about (0,0) changes the sign of both coordinates: (2,4) goes to (-2,-4).
2. Translate 0 across and 5 up: (-2,-4+5)=(-2,1).
3. The final position of K is (-2,1).