Square Rotation B3C753

1. **Problem Statement:** We have a square ABCD with vertices $A(3,1)$, $B(3,6)$, $C(8,1)$, and $D(8,6)$. We want to rotate this square 90 degrees about the point $A(3,1)$. 2. **Rotation Formula:** To rotate a point $(x,y)$ about a pivot point $(x_0,y_0)$ by 90 degrees counterclockwise, use: $$ \begin{cases} x' = x_0 - (y - y_0) \\ y' = y_0 + (x - x_0) \end{cases} $$ 3. **Apply rotation to each vertex:** - For $A(3,1)$ (pivot), rotation keeps it the same: $$A' = (3,1)$$ - For $B(3,6)$: $$x' = 3 - (6 - 1) = 3 - 5 = -2$$ $$y' = 1 + (3 - 3) = 1 + 0 = 1$$ So, $B' = (-2,1)$ - For $C(8,1)$: $$x' = 3 - (1 - 1) = 3 - 0 = 3$$ $$y' = 1 + (8 - 3) = 1 + 5 = 6$$ So, $C' = (3,6)$ - For $D(8,6)$: $$x' = 3 - (6 - 1) = 3 - 5 = -2$$ $$y' = 1 + (8 - 3) = 1 + 5 = 6$$ So, $D' = (-2,6)$ 4. **Summary of rotated vertices:** $$A'(3,1), B'(-2,1), C'(3,6), D'(-2,6)$$ 5. **Explanation:** The rotation moves points around $A$ by swapping and negating the relative coordinates according to the 90-degree rotation formula. 6. **Final answer:** The square after rotation has vertices at $A'(3,1)$, $B'(-2,1)$, $C'(3,6)$, and $D'(-2,6)$.