Square Vertices 992D27

1. **State the problem:** We need to show that the points $(0,1)$, $(1,4)$, $(4,3)$, and $(3,0)$ are vertices of a square. 2. **Formula and rules:** To prove points form a square, we check: - All sides are equal in length. - The diagonals are equal in length. - Adjacent sides are perpendicular (dot product = 0). 3. **Calculate side lengths:** Use distance formula between points $A(x_1,y_1)$ and $B(x_2,y_2)$: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ Label points as $A(0,1)$, $B(1,4)$, $C(4,3)$, $D(3,0)$. Calculate sides: - $AB = \sqrt{(1-0)^2 + (4-1)^2} = \sqrt{1 + 9} = \sqrt{10}$ - $BC = \sqrt{(4-1)^2 + (3-4)^2} = \sqrt{9 + 1} = \sqrt{10}$ - $CD = \sqrt{(3-4)^2 + (0-3)^2} = \sqrt{1 + 9} = \sqrt{10}$ - $DA = \sqrt{(0-3)^2 + (1-0)^2} = \sqrt{9 + 1} = \sqrt{10}$ All sides are equal to $\sqrt{10}$. 4. **Calculate diagonals:** - $AC = \sqrt{(4-0)^2 + (3-1)^2} = \sqrt{16 + 4} = \sqrt{20}$ - $BD = \sqrt{(3-1)^2 + (0-4)^2} = \sqrt{4 + 16} = \sqrt{20}$ Diagonals are equal. 5. **Check perpendicularity of adjacent sides:** Vectors: - $\overrightarrow{AB} = (1-0, 4-1) = (1,3)$ - $\overrightarrow{BC} = (4-1, 3-4) = (3,-1)$ Dot product: $$\overrightarrow{AB} \cdot \overrightarrow{BC} = 1 \times 3 + 3 \times (-1) = 3 - 3 = 0$$ Since dot product is zero, $AB$ is perpendicular to $BC$. Similarly check $\overrightarrow{BC} \cdot \overrightarrow{CD}$: - $\overrightarrow{CD} = (3-4, 0-3) = (-1,-3)$ Dot product: $$3 \times (-1) + (-1) \times (-3) = -3 + 3 = 0$$ Perpendicular. 6. **Conclusion:** All sides are equal, diagonals are equal, and adjacent sides are perpendicular, so the points form a square.