1. **Stating the problem:** We have points $A(-4,-2)$, $B(-4,3)$, $C(3,5)$, and $D(3,-2)$ forming rectangle $ABCD$. We want to find the symmetric rectangle of $ABCD$ with respect to the diagonal line $AC$. 2. **Understanding the problem:** The diagonal $AC$ connects points $A(-4,-2)$ and $C(3,5)$. Reflecting the rectangle $ABCD$ about line $AC$ means finding the image of each vertex after reflection over $AC$. 3. **Equation of line $AC$:** Slope $m = \frac{5 - (-2)}{3 - (-4)} = \frac{7}{7} = 1$. Using point-slope form with point $A(-4,-2)$: $$y - (-2) = 1(x - (-4)) \Rightarrow y + 2 = x + 4 \Rightarrow y = x + 2$$ 4. **Reflection formula:** For a point $P(x_0,y_0)$ reflected over line $y = mx + b$, the reflected point $P'(x',y')$ is given by: $$d = \frac{(x_0 + (y_0 - b)m)}{1 + m^2}$$ $$x' = 2d - x_0$$ $$y' = 2dm - y_0 + 2b$$ Here, $m=1$, $b=2$. 5. **Reflect points $B$ and $D$ (since $A$ and $C$ lie on the line, their reflections are themselves):** - For $B(-4,3)$: $$d = \frac{-4 + (3 - 2) \times 1}{1 + 1^2} = \frac{-4 + 1}{2} = \frac{-3}{2} = -1.5$$ $$x' = 2 \times (-1.5) - (-4) = -3 + 4 = 1$$ $$y' = 2 \times (-1.5) \times 1 - 3 + 2 \times 2 = -3 - 3 + 4 = -2$$ So, $B' = (1, -2)$. - For $D(3,-2)$: $$d = \frac{3 + (-2 - 2) \times 1}{2} = \frac{3 - 4}{2} = \frac{-1}{2} = -0.5$$ $$x' = 2 \times (-0.5) - 3 = -1 - 3 = -4$$ $$y' = 2 \times (-0.5) \times 1 - (-2) + 4 = -1 + 2 + 4 = 5$$ So, $D' = (-4, 5)$. 6. **New rectangle vertices after reflection:** $$A' = A = (-4,-2)$$ $$B' = (1,-2)$$ $$C' = C = (3,5)$$ $$D' = (-4,5)$$ 7. **Conclusion:** The symmetric rectangle with respect to diagonal $AC$ has vertices $A'(-4,-2)$, $B'(1,-2)$, $C'(3,5)$, and $D'(-4,5)$. **Final answer:** $$A'(-4,-2), B'(1,-2), C'(3,5), D'(-4,5)$$