1. **State the problem:** Determine the most descriptive name for quadrilateral PQRS using the slope and distance formulas. 2. **Recall formulas:** - Slope between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by: $$\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}$$ - Distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. **Calculate slopes of sides:** - $PQ$: $\frac{6 - (-3)}{-4 - (-1)} = \frac{9}{-3} = -3$ - $QR$: $\frac{10 - 6}{8 - (-4)} = \frac{4}{12} = \frac{1}{3}$ - $RS$: $\frac{1 - 10}{11 - 8} = \frac{-9}{3} = -3$ - $SP$: $\frac{-3 - 1}{-1 - 11} = \frac{-4}{-12} = \frac{1}{3}$ 4. **Calculate lengths of sides:** - $PQ$: $\sqrt{(-4 + 1)^2 + (6 + 3)^2} = \sqrt{(-3)^2 + 9^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10}$ - $QR$: $\sqrt{(8 + 4)^2 + (10 - 6)^2} = \sqrt{12^2 + 4^2} = \sqrt{144 + 16} = \sqrt{160} = 4\sqrt{10}$ - $RS$: $\sqrt{(11 - 8)^2 + (1 - 10)^2} = \sqrt{3^2 + (-9)^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10}$ - $SP$: $\sqrt{(-1 - 11)^2 + (-3 - 1)^2} = \sqrt{(-12)^2 + (-4)^2} = \sqrt{144 + 16} = \sqrt{160} = 4\sqrt{10}$ 5. **Analyze slopes and lengths:** - Opposite sides $PQ$ and $RS$ have equal slopes ($-3$) and equal lengths ($3\sqrt{10}$). - Opposite sides $QR$ and $SP$ have equal slopes ($\frac{1}{3}$) and equal lengths ($4\sqrt{10}$). - Adjacent sides are not perpendicular since the product of slopes is $-3 \times \frac{1}{3} = -1$, which means they are perpendicular. 6. **Conclusion:** Since opposite sides are parallel and equal in length, and adjacent sides are perpendicular, quadrilateral PQRS is a **rectangle**. **Final answer:** Quadrilateral PQRS is a rectangle.