1. **State the problem:** We are given a parallelogram ABCD with vector $\overrightarrow{BC} = \binom{-a}{-b}$, where $a > 0$ and $b > 0$. The gradient (slope) of BC is 1, and the magnitude $\left|\overrightarrow{BC}\right| = \sqrt{8}$. We need to find the coordinates of point D. 2. **Recall properties of parallelograms:** In a parallelogram, $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ and $\overrightarrow{AD} = \overrightarrow{BC}$. Also, $\overrightarrow{AD} = \overrightarrow{BC}$ means that vector AD is equal to vector BC. 3. **Use the gradient condition:** The gradient of BC is given by $$\text{slope} = \frac{\Delta y}{\Delta x} = \frac{-b}{-a} = \frac{b}{a} = 1$$ Since slope = 1, we have $$\frac{b}{a} = 1 \implies b = a$$ 4. **Use the magnitude condition:** The magnitude of $\overrightarrow{BC}$ is $$\left|\overrightarrow{BC}\right| = \sqrt{(-a)^2 + (-b)^2} = \sqrt{a^2 + b^2} = \sqrt{8}$$ Substitute $b = a$: $$\sqrt{a^2 + a^2} = \sqrt{8} \implies \sqrt{2a^2} = \sqrt{8}$$ Square both sides: $$2a^2 = 8$$ Divide both sides by 2: $$\cancel{2}a^2 = \cancel{2}4 \implies a^2 = 4$$ Take positive root (since $a > 0$): $$a = 2$$ Then $b = a = 2$. 5. **Find vector $\overrightarrow{BC}$:** $$\overrightarrow{BC} = \binom{-a}{-b} = \binom{-2}{-2}$$ 6. **Find coordinates of D:** Since ABCD is a parallelogram, $\overrightarrow{AD} = \overrightarrow{BC}$. If coordinates of A are $(x_A, y_A)$ and B are $(x_B, y_B)$, then $$\overrightarrow{AB} = \binom{x_B - x_A}{y_B - y_A}$$ Similarly, $D = A + \overrightarrow{BC} = (x_A - 2, y_A - 2)$. Without explicit coordinates for A, B, or C, the coordinates of D relative to A are: $$D = (x_A - 2, y_A - 2)$$ If coordinates of B and C are known, we can find A and then D. But with given info, the vector from A to D is $\binom{-2}{-2}$. **Final answer:** The coordinates of D are $D = (x_A - 2, y_A - 2)$ where $x_A, y_A$ are coordinates of A. If A is at origin $(0,0)$, then $D = (-2, -2)$.