1. **State the problem:** We have a shape ABCD enlarged by a scale factor of $-4$ with center of enlargement at $(1, 2)$ to get shape A'B'C'D'. Given coordinates of A' and B', find coordinates of A and B. 2. **Recall the enlargement formula:** For a point $P(x,y)$ enlarged about center $C(x_c,y_c)$ by scale factor $k$, the image $P'(x',y')$ satisfies: $$x' = x_c + k(x - x_c)$$ $$y' = y_c + k(y - y_c)$$ 3. **Rearrange to find original point:** Given $P'(x',y')$, center $C(1,2)$, and scale factor $k = -4$, solve for $P(x,y)$: $$x = x_c + \frac{x' - x_c}{k} = 1 + \frac{x' - 1}{-4}$$ $$y = y_c + \frac{y' - y_c}{k} = 2 + \frac{y' - 2}{-4}$$ 4. **Calculate coordinates of A:** Given $A'(-8, -6)$, $$x_A = 1 + \frac{-8 - 1}{-4} = 1 + \frac{-9}{-4} = 1 + 2.25 = 3.25$$ $$y_A = 2 + \frac{-6 - 2}{-4} = 2 + \frac{-8}{-4} = 2 + 2 = 4$$ So, $A = (3.25, 4)$. 5. **Calculate coordinates of B:** Given $B'(-11, -1)$, $$x_B = 1 + \frac{-11 - 1}{-4} = 1 + \frac{-12}{-4} = 1 + 3 = 4$$ $$y_B = 2 + \frac{-1 - 2}{-4} = 2 + \frac{-3}{-4} = 2 + 0.75 = 2.75$$ So, $B = (4, 2.75)$. **Final answer:** $$A = (3.25, 4), \quad B = (4, 2.75)$$