1. **State the problem:** Determine if the quadrilateral formed by points F(4,1), G(3,-1), H(-3,2), I(-2,4) is a rectangle using the distance formula. 2. **Recall the distance formula:** The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ 3. **Calculate the lengths of all sides:** - $FG = \sqrt{(3-4)^2 + (-1-1)^2} = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$ - $GH = \sqrt{(-3-3)^2 + (2+1)^2} = \sqrt{(-6)^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}$ - $HI = \sqrt{(-2+3)^2 + (4-2)^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$ - $IF = \sqrt{(4+2)^2 + (1-4)^2} = \sqrt{6^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}$ 4. **Check if opposite sides are equal:** - $FG = HI = \sqrt{5}$ - $GH = IF = 3\sqrt{5}$ 5. **Calculate the slopes of adjacent sides to check for right angles:** - Slope $FG = \frac{-1 - 1}{3 - 4} = \frac{-2}{-1} = 2$ - Slope $GH = \frac{2 - (-1)}{-3 - 3} = \frac{3}{-6} = -\frac{1}{2}$ 6. **Check if slopes of adjacent sides are negative reciprocals:** - $2$ and $-\frac{1}{2}$ are negative reciprocals, so $FG$ is perpendicular to $GH$. 7. **Since opposite sides are equal and adjacent sides are perpendicular, the figure is a rectangle.**