1. **Stating the problem:** We need to find the position vector of point D, where lines CB and ON are extended to meet, expressed in terms of vectors $p$ and $q$. 2. **Analyzing the problem:** Points C, B, O, and N have position vectors involving $p$ and $q$. Usually, $p$ and $q$ represent vectors from a common origin. 3. **Using the given expressions:** The problem shows intermediate expressions: - $\frac{9 + p}{2} + \frac{p + 3q}{5}$ - $\frac{10p + 6q}{10}$ 4. **Simplification:** Let's simplify both expressions. First expression: $$\frac{9 + p}{2} + \frac{p + 3q}{5} = \frac{5(9 + p)}{10} + \frac{2(p + 3q)}{10} = \frac{45 + 5p + 2p + 6q}{10} = \frac{45 + 7p + 6q}{10}$$ Second expression: $$\frac{10p + 6q}{10} = p + \frac{3q}{5}$$ 5. **Interpreting results:** The problem likely states that the position vector of D matches these forms based on intersection conditions. 6. **Simplify final form:** The first expression simplifies to: $$\frac{45 + 7p + 6q}{10} = \frac{45}{10} + \frac{7p}{10} + \frac{6q}{10} = 4.5 + 0.7p + 0.6q$$ The second expression is: $$p + 0.6q$$ But since $D$ is a vector, the constant term 4.5 likely indicates an origin offset; assuming vectors start at origin, ignore constants, so position vector of $D$ is: $$\boxed{\frac{10p + 6q}{10} = p + \frac{3q}{5}}$$ Thus, the position vector of $D$ in simplest terms is: $$\mathbf{D} = p + \frac{3}{5}q$$