1. **Problem Statement:** Given points $P(-4,-1)$, $Q(6,3)$, $R(6,b)$, and $S(-4,-3)$, we need to solve several parts related to gradients, parallelism, lengths, midpoints, and properties of quadrilateral $PQRS$. 2. **Formula for Gradient:** The gradient (slope) between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by: $$\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1}$$ 3. **a) Determine the gradient of $PQ$:** $$\text{gradient}_{PQ} = \frac{3 - (-1)}{6 - (-4)} = \frac{4}{10} = \frac{2}{5}$$ 4. **b) If $PQ$ is parallel to $SR$, determine $b$:** Parallel lines have equal gradients. Calculate gradient of $SR$: $$\text{gradient}_{SR} = \frac{b - (-3)}{6 - (-4)} = \frac{b + 3}{10}$$ Set equal to gradient of $PQ$: $$\frac{b + 3}{10} = \frac{2}{5}$$ Multiply both sides by 10: $$b + 3 = 4$$ Solve for $b$: $$b = 1$$ 5. **c) Show that $PQ = SR$:** Distance formula between points $(x_1,y_1)$ and $(x_2,y_2)$: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ Calculate $PQ$: $$PQ = \sqrt{(6 - (-4))^2 + (3 - (-1))^2} = \sqrt{10^2 + 4^2} = \sqrt{100 + 16} = \sqrt{116}$$ Calculate $SR$ with $b=1$: $$SR = \sqrt{(6 - (-4))^2 + (1 - (-3))^2} = \sqrt{10^2 + 4^2} = \sqrt{116}$$ Thus, $PQ = SR$. 6. **d) Is quadrilateral $PQRS$ a parallelogram?** A quadrilateral is a parallelogram if both pairs of opposite sides are parallel. We have $PQ \parallel SR$ (given). Check gradient of $QR$ and $PS$: $$\text{gradient}_{QR} = \frac{b - 3}{6 - 6} = \frac{1 - 3}{0} = \text{undefined (vertical line)}$$ $$\text{gradient}_{PS} = \frac{-3 - (-1)}{-4 - (-4)} = \frac{-2}{0} = \text{undefined (vertical line)}$$ Since both are vertical lines, $QR \parallel PS$. Therefore, $PQRS$ is a parallelogram. 7. **e) Calculate midpoints:** Midpoint formula: $$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$$ (i) Midpoint of $PR$: $$P(-4,-1), R(6,1)$$ $$M_{PR} = \left( \frac{-4 + 6}{2}, \frac{-1 + 1}{2} \right) = (1, 0)$$ (ii) Midpoint of $SQ$: $$S(-4,-3), Q(6,3)$$ $$M_{SQ} = \left( \frac{-4 + 6}{2}, \frac{-3 + 3}{2} \right) = (1, 0)$$ 8. **f) Deduction about diagonals of a parallelogram:** Since midpoints of diagonals $PR$ and $SQ$ are the same, the diagonals bisect each other. 9. **g) Show $PQRS$ is not a rhombus:** A rhombus has all sides equal. Calculate length $QR$: $$QR = \sqrt{(6 - 6)^2 + (1 - 3)^2} = \sqrt{0 + (-2)^2} = 2$$ Calculate length $PQ$ (from step 5): $$PQ = \sqrt{116} \approx 10.77$$ Since $PQ \neq QR$, $PQRS$ is not a rhombus. **Final answers:** - Gradient $PQ = \frac{2}{5}$ - $b = 1$ - $PQ = SR = \sqrt{116}$ - $PQRS$ is a parallelogram - Midpoints $M_{PR} = M_{SQ} = (1,0)$ - Diagonals bisect each other - $PQRS$ is not a rhombus