1. **State the problem:** We have points A(-5, -2), B(-3, -2), and D(-5, -4) in the third quadrant. We need to find point C to form a square ABCD. 2. **Find point C:** Since ABCD is a square, sides are equal and adjacent sides are perpendicular. - Vector AB = B - A = (-3 + 5, -2 + 2) = (2, 0) - Vector AD = D - A = (-5 + 5, -4 + 2) = (0, -2) Since AB and AD are perpendicular and equal length (both length 2), point C is at B + AD = (-3 + 0, -2 - 2) = (-3, -4). So, C = (-3, -4). 3. **Confirm square:** Sides AB, BC, CD, DA all have length 2 and adjacent sides are perpendicular. 4. **Rotation to fourth quadrant:** The square is rotated so that vertex C' is at (4, -3). - Original C is at (-3, -4), rotated C' at (4, -3). 5. **Find center of rotation:** The center of the square is midpoint of A and C or B and D. - Midpoint M = ((-5 + -3)/2, (-2 + -4)/2) = (-4, -3) 6. **Rotation rule:** Rotation about point M by angle $\theta$ transforms point $(x,y)$ to $(x',y')$: $$ \begin{cases} x' = x_0 + (x - x_0)\cos\theta - (y - y_0)\sin\theta \\ y' = y_0 + (x - x_0)\sin\theta + (y - y_0)\cos\theta \end{cases} $$ where $(x_0,y_0)$ is center of rotation. 7. **Find angle $\theta$:** Rotate C(-3,-4) about M(-4,-3) to C'(4,-3). - Vector MC = (-3 + 4, -4 + 3) = (1, -1) - Vector MC' = (4 + 4, -3 + 3) = (8, 0) is incorrect, must be relative to M: Actually, MC' = (4 - (-4), -3 - (-3)) = (8, 0) 8. **Calculate $\theta$:** Vector MC rotated by $\theta$ equals MC'. - Length MC = $\sqrt{1^2 + (-1)^2} = \sqrt{2}$ - Length MC' = 8 Lengths differ, so scale is not preserved, but rotation preserves length. This suggests a mistake: the square must be rotated and possibly translated. 9. **Check if rotation about origin:** Try rotation about origin. - Rotate C(-3,-4) to C'(4,-3). - Lengths: $\sqrt{(-3)^2 + (-4)^2} = 5$, $\sqrt{4^2 + (-3)^2} = 5$ equal. - Angle between vectors: $$\cos\theta = \frac{(-3)(4) + (-4)(-3)}{5 \times 5} = \frac{-12 + 12}{25} = 0$$ So $\theta = 90^\circ$ or $270^\circ$. - Check direction: Rotate (-3,-4) by 90° CCW: $$ (x', y') = (-y, x) = (4, -3) $$ Matches C'. 10. **Coordinate rule:** Rotation by 90° counterclockwise about origin: $$ (x,y) \to (-y, x) $$ **Final answer:** The coordinate rule to rotate the square from the third to the fourth quadrant with vertex C' at (4,-3) is rotation by 90° counterclockwise about the origin, given by $$ (x,y) \to (-y, x) $$