1. **Problem statement:** (a) A translation $T$ moves point $P(1,2)$ to $P_1(5,3)$. Find the image of $Q(3,4)$ under $T$. (b) Given a frequency distribution of marks with frequencies involving $m$, and mean mark 5, find: (i) the value of $m$, (ii) the median, (iii) the mode. 2. **Translation formula:** A translation adds a fixed vector to every point. If $T$ moves $P(x,y)$ to $P_1(x',y')$, then the translation vector is $$\vec{v} = (x'-x, y'-y)$$ The image of any point $Q(a,b)$ under $T$ is $$Q' = (a + (x'-x), b + (y'-y))$$ 3. **Step (a) Find image of $Q$ under $T$:** - Given $P(1,2)$ to $P_1(5,3)$, translation vector is $$\vec{v} = (5-1, 3-2) = (4,1)$$ - Apply to $Q(3,4)$: $$Q' = (3+4, 4+1) = (7,5)$$ 4. **Step (b) Frequency distribution and mean:** | Marks | 3 | 4 | 5 | 6 | 7 | 8 | | Frequency | 5 | $m$ | $m+1$ | 9 | 4 | 1 | - Total frequency $N = 5 + m + (m+1) + 9 + 4 + 1 = 20 + 2m$ - Mean formula: $$\text{Mean} = \frac{\sum (\text{marks} \times \text{frequency})}{N} = 5$$ - Calculate numerator: $$3 \times 5 + 4 \times m + 5 \times (m+1) + 6 \times 9 + 7 \times 4 + 8 \times 1$$ $$= 15 + 4m + 5m + 5 + 54 + 28 + 8 = 110 + 9m$$ - Set mean equation: $$\frac{110 + 9m}{20 + 2m} = 5$$ 5. **Solve for $m$:** Multiply both sides by denominator: $$110 + 9m = 5(20 + 2m) = 100 + 10m$$ Rearranged: $$110 + 9m = 100 + 10m$$ $$110 - 100 = 10m - 9m$$ $$10 = m$$ 6. **Calculate total frequency $N$ with $m=10$:** $$N = 20 + 2(10) = 40$$ 7. **Find median:** - Median position is at $\frac{N}{2} = 20$th value. - Cumulative frequencies: - Up to 3 marks: 5 - Up to 4 marks: $5 + 10 = 15$ - Up to 5 marks: $15 + 11 = 26$ - The 20th value lies in the 5 marks group. - So, median = 5 8. **Find mode:** - Mode is the mark with highest frequency. - Frequencies: - 3: 5 - 4: 10 - 5: 11 - 6: 9 - 7: 4 - 8: 1 - Highest frequency is 11 at mark 5. - So, mode = 5 **Final answers:** (a) Image of $Q$ under $T$ is $(7,5)$. (b)(i) $m = 10$ (b)(ii) Median = 5 (b)(iii) Mode = 5