1. **Problem Statement:** Prove that the points (4, 3), (6, 4), (5, 6), and (3, 5) are the vertices of a square. 2. **Formula and Rules:** To prove a quadrilateral is a square, we need to show: - All four sides are equal in length. - The diagonals are equal in length. - Adjacent sides are perpendicular (form right angles). Distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ Slope formula to check perpendicularity: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Two lines are perpendicular if the product of their slopes is $-1$. 3. **Calculate side lengths:** - $AB = \sqrt{(6-4)^2 + (4-3)^2} = \sqrt{2^2 + 1^2} = \sqrt{5}$ - $BC = \sqrt{(5-6)^2 + (6-4)^2} = \sqrt{(-1)^2 + 2^2} = \sqrt{5}$ - $CD = \sqrt{(3-5)^2 + (5-6)^2} = \sqrt{(-2)^2 + (-1)^2} = \sqrt{5}$ - $DA = \sqrt{(4-3)^2 + (3-5)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{5}$ All sides are equal to $\sqrt{5}$. 4. **Calculate diagonal lengths:** - $AC = \sqrt{(5-4)^2 + (6-3)^2} = \sqrt{1^2 + 3^2} = \sqrt{10}$ - $BD = \sqrt{(3-6)^2 + (5-4)^2} = \sqrt{(-3)^2 + 1^2} = \sqrt{10}$ Diagonals are equal. 5. **Check perpendicularity of adjacent sides:** - Slope $AB = \frac{4-3}{6-4} = \frac{1}{2}$ - Slope $BC = \frac{6-4}{5-6} = \frac{2}{-1} = -2$ Product of slopes $= \frac{1}{2} \times (-2) = -1$, so $AB \perp BC$. Similarly, check $BC$ and $CD$: - Slope $CD = \frac{5-6}{3-5} = \frac{-1}{-2} = \frac{1}{2}$ Product $= -2 \times \frac{1}{2} = -1$, so $BC \perp CD$. 6. **Conclusion:** Since all sides are equal, diagonals are equal, and adjacent sides are perpendicular, the quadrilateral with vertices (4,3), (6,4), (5,6), and (3,5) is a square.