Negative-scale-factor enlargement

(a) $(-5, -3)$

1. From the centre to $P$.
   $$(4, 3) - (1, 1) = (3, 2)$$
2. Multiply by $-2$: the image is on the other side of the centre, twice as far.
   $$-2 \times (3, 2) = (-6, -4)$$
3. Add to the centre.
   $$(1, 1) + (-6, -4) = (-5, -3)$$

• M1 The displacement $(3, 2)$ from the centre multiplied by $-2$.
• A1 Correct answer: $(-5, -3)$.

(b) Enlargement, scale factor $-\frac{1}{2}$, centre $(1, 1)$

1. The inverse of scale factor $-2$ is $-\frac{1}{2}$, about the same centre.

• B1 Enlargement with scale factor $-\frac{1}{2}$.
• B1 Centre $(1, 1)$.

(a) $(-4,-9)$

1. $$5-2$$
2. $$3-(-1)$$
3. $$2-2\times 3$$
4. The vector from C to P is (5 $-$ 2, 3 $-$ ($-1$)) = (3, 4).
5. Multiply this vector by $-2$ to give ($-6$, $-8$).
6. Add it to the centre: (2 $-$ 6, $-1$ $-$ 8) = ($-4$, $-9$).

• P1 Establishing $5-2$ or an equivalent valid method.
• P1 Establishing $3-(-1)$ or an equivalent valid method.
• P1 Establishing $2-2\times 3$ or an equivalent valid method.
• A1 Correct answer: $(-4,-9)$

(a) 36 cm

1. Lengths are multiplied by $|-3| = 3$: $12 \times 3 = 36$.

• B1 Correct answer: 36 cm.

(b) $54\text{ cm}^2$

1. Areas are multiplied by the square of the length factor: $3^2 = 9$.
2. $$6 \times 9 = 54$$

• M1 Using the area factor $3^2 = 9$.
• A1 $54\text{ cm}^2$.

(a) $(-2,-5)$

1. Subtract the centre to find the displacement vector.
   $$6-2=4$$
2. Multiply the displacement by the negative scale and add the centre.
   $$x=2-4$$
3. Therefore $(-2,-5)$.

• P1 Subtract the centre to find the displacement vector.
• P1 Multiply the displacement by the negative scale and add the centre.
• A1 Correct answer: $(-2,-5)$

(b) It lies on the same straight line on the opposite side of the centre, because the scale factor is negative.

1. It lies on the same straight line on the opposite side of the centre, because the scale factor is negative.

• C1 Correct conclusion with supporting reasoning: It lies on the same straight line on the opposite side of the centre, because the scale factor is negative.

(a) $(2,-1)$

1. Let the centre be (c, d). For each coordinate, image = centre + scale factor × (original − centre).
   $$-4=c-\frac{3}{2}(6-c)$$
2. Apply the same relationship to the vertical coordinate.
   $$-7=d-\frac{3}{2}(3-d)$$
3. Therefore $(2,-1)$.

• P1 Let the centre be (c, d). For each coordinate, image = centre + scale factor × (original − centre).
• P1 Apply the same relationship to the vertical coordinate.
• A1 Correct answer: $(2,-1)$

(b) $12$ cm

1. Lengths scale by the magnitude of the factor.
   $$8\times (3/2)$$
2. Therefore $12$ cm.

• P1 Lengths scale by the magnitude of the factor.
• A1 Correct answer: $12$ cm

(a) $4$

1. Multiply the scale factors in sequence.
   $$(-2)(-2)$$
2. Therefore $4$.

• M1 Multiply the scale factors in sequence.
• A1 Correct answer: $4$

(b) $(12,-8)$

1. Apply the combined positive scale factor to both coordinates.
   $$3\times 4$$
2. Therefore $(12,-8)$.

• M1 Apply the combined positive scale factor to both coordinates.
• A1 Correct answer: $(12,-8)$

(a) $(2,-1)$

1. Let the centre be (c, d). For each coordinate, image = centre + scale factor × (original − centre).
   $$-10=c-3(6-c)$$
2. Apply the same relationship to the vertical coordinate.
   $$-13=d-3(3-d)$$
3. Therefore $(2,-1)$.

• P1 Let the centre be (c, d). For each coordinate, image = centre + scale factor × (original − centre).
• P1 Apply the same relationship to the vertical coordinate.
• A1 Correct answer: $(2,-1)$

(b) $24$ cm

1. Lengths scale by the magnitude of the factor.
   $$8\times (3)$$
2. Therefore $24$ cm.

• P1 Lengths scale by the magnitude of the factor.
• A1 Correct answer: $24$ cm

(a) Translation by $\begin{pmatrix} 4.5 \\ 0 \end{pmatrix}$

1. Take a general point $(x, y)$. After $E$:
   $$(-2x, -2y)$$
2. After $F$: centre $+ \left(-\tfrac{1}{2}\right) \times$ (point $-$ centre).
   $$(3, 0) - \tfrac{1}{2}\big((-2x, -2y) - (3, 0)\big) = (3, 0) + (x + 1.5,\; y)$$
3. Simplify.
   $$(x + 4.5,\; y)$$
4. Every point moves by the same vector, 4.5 across and 0 up, so the combination is a translation.

• M1 Applying $E$ to a general point: $(-2x, -2y)$.
• M1 Applying $F$ to their image, with the centre $(3, 0)$ used correctly.
• A1 Showing the image is $(x + 4.5, y)$ and concluding it is a translation by the vector with components 4.5 and 0.