1. **Problem 1: Find the centre of enlargement and image of Q(1,1)** under enlargement with scale factor $-3$ that maps $P(3,6)$ to $P_1(7,18)$. 2. The formula for enlargement with centre $C(x_c,y_c)$ and scale factor $k$ is: $$P_1 = C + k(P - C)$$ where $P$ is the original point and $P_1$ is the image. 3. Substitute $P(3,6)$, $P_1(7,18)$, and $k=-3$: $$7 = x_c + (-3)(3 - x_c)$$ $$18 = y_c + (-3)(6 - y_c)$$ 4. Simplify each: $$7 = x_c - 9 + 3x_c = 4x_c - 9$$ $$18 = y_c - 18 + 3y_c = 4y_c - 18$$ 5. Solve for $x_c$ and $y_c$: $$4x_c = 16 \Rightarrow x_c = 4$$ $$4y_c = 36 \Rightarrow y_c = 9$$ 6. So, the centre of enlargement is $C(4,9)$. 7. To find the image of $Q(1,1)$, use the formula: $$Q_1 = C + k(Q - C) = (4,9) + (-3)((1,1) - (4,9))$$ 8. Calculate the vector: $$(1-4, 1-9) = (-3, -8)$$ 9. Multiply by $k=-3$: $$-3 \times (-3, -8) = (9, 24)$$ 10. Add to centre: $$(4,9) + (9,24) = (13,33)$$ **Answer:** - Centre of enlargement: $C(4,9)$ - Image of $Q(1,1)$ under the enlargement: $Q_1(13,33)$ --- 1. **Problem 2: Calculate surface area of larger bar of soap** given masses and surface area of smaller bar. 2. Since the bars are similar, their volumes scale as the cube of the scale factor $k$: $$\frac{V_2}{V_1} = k^3$$ 3. Mass is proportional to volume, so: $$\frac{m_2}{m_1} = k^3$$ 4. Given $m_1=343$, $m_2=1331$: $$k^3 = \frac{1331}{343} = \frac{11^3}{7^3} = \left(\frac{11}{7}\right)^3$$ 5. So, scale factor for length is: $$k = \frac{11}{7}$$ 6. Surface area scales as square of scale factor: $$\frac{A_2}{A_1} = k^2 = \left(\frac{11}{7}\right)^2 = \frac{121}{49}$$ 7. Given $A_1 = 196$ cm², find $A_2$: $$A_2 = A_1 \times \frac{121}{49} = 196 \times \frac{121}{49}$$ 8. Simplify: $$196 \div 49 = 4$$ $$A_2 = 4 \times 121 = 484$$ **Answer:** Surface area of larger bar is $484$ cm².