1. **Problem statement:** Given points $P(-4, -1)$, $Q(6, 3)$, $R(6, b)$, and $S(-4, -3)$, we need to: a) Determine the gradient (slope) of line segment $PQ$. b) If $PQ$ is parallel to $SR$, find the value of $b$. 2. **Formula for gradient:** The gradient $m$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ 3. **Calculate gradient of $PQ$:** Using $P(-4, -1)$ and $Q(6, 3)$: $$m_{PQ} = \frac{3 - (-1)}{6 - (-4)} = \frac{3 + 1}{6 + 4} = \frac{4}{10} = \frac{2}{5}$$ 4. **Gradient of $SR$:** Points $S(-4, -3)$ and $R(6, b)$: $$m_{SR} = \frac{b - (-3)}{6 - (-4)} = \frac{b + 3}{10}$$ 5. **Condition for parallel lines:** Two lines are parallel if their gradients are equal: $$m_{PQ} = m_{SR}$$ Substitute values: $$\frac{2}{5} = \frac{b + 3}{10}$$ 6. **Solve for $b$:** Cross-multiply: $$2 \times 10 = 5 \times (b + 3)$$ $$20 = 5b + 15$$ Subtract 15 from both sides: $$20 - 15 = 5b$$ $$5 = 5b$$ Divide both sides by 5: $$b = 1$$ **Final answers:** - Gradient of $PQ$ is $\frac{2}{5}$. - Value of $b$ such that $PQ$ is parallel to $SR$ is $1$.