1. **State the problem:** We are given three points: $A(-3,4)$, $B(2,1)$, and $C(-4,-5)$. We need to find a fourth point $D(x,y)$ such that $ABCD$ forms a kite. 2. **Recall kite properties:** A kite has two pairs of adjacent sides equal and one diagonal that is the axis of symmetry. The diagonals intersect at right angles, and one diagonal is bisected by the other. 3. **Approach:** We try to find $D$ so that $ABCD$ has two pairs of adjacent equal sides. We can try to pair $A$ with $D$ and $B$ with $C$ or vice versa. 4. **Calculate lengths:** - $AB = \sqrt{(2 - (-3))^2 + (1 - 4)^2} = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$ - $BC = \sqrt{(-4 - 2)^2 + (-5 - 1)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72}$ - $AC = \sqrt{(-4 - (-3))^2 + (-5 - 4)^2} = \sqrt{(-1)^2 + (-9)^2} = \sqrt{1 + 81} = \sqrt{82}$ 5. **Set $AD = AB$ and $DC = BC$ for kite sides:** We want $AD = AB = \sqrt{34}$ and $DC = BC = \sqrt{72}$. 6. **Let $D = (x,y)$, find $D$ such that:** $$AD^2 = (x + 3)^2 + (y - 4)^2 = 34$$ $$DC^2 = (x + 4)^2 + (y + 5)^2 = 72$$ 7. **Solve system:** Subtract the two equations: $$(x + 3)^2 + (y - 4)^2 - (x + 4)^2 - (y + 5)^2 = 34 - 72 = -38$$ Expand: $$(x^2 + 6x + 9) + (y^2 - 8y + 16) - (x^2 + 8x + 16) - (y^2 + 10y + 25) = -38$$ Simplify: $$6x + 9 - 8y + 16 - 8x - 16 - 10y - 25 = -38$$ $$(6x - 8x) + (-8y - 10y) + (9 + 16 - 16 - 25) = -38$$ $$-2x - 18y - 16 = -38$$ Add 16 to both sides: $$-2x - 18y = -22$$ Divide by -2: $$x + 9y = 11$$ 8. **Express $x$ in terms of $y$:** $$x = 11 - 9y$$ 9. **Substitute into $AD^2 = 34$:** $$(11 - 9y + 3)^2 + (y - 4)^2 = 34$$ $$(14 - 9y)^2 + (y - 4)^2 = 34$$ Expand: $$(14 - 9y)^2 = 196 - 252y + 81y^2$$ $$(y - 4)^2 = y^2 - 8y + 16$$ Sum: $$196 - 252y + 81y^2 + y^2 - 8y + 16 = 34$$ $$82y^2 - 260y + 212 = 34$$ Subtract 34: $$82y^2 - 260y + 178 = 0$$ 10. **Simplify quadratic:** Divide entire equation by 2: $$41y^2 - 130y + 89 = 0$$ 11. **Use quadratic formula:** $$y = \frac{130 \pm \sqrt{(-130)^2 - 4 \times 41 \times 89}}{2 \times 41} = \frac{130 \pm \sqrt{16900 - 14596}}{82} = \frac{130 \pm \sqrt{2304}}{82}$$ $$= \frac{130 \pm 48}{82}$$ 12. **Calculate roots:** - $$y_1 = \frac{130 + 48}{82} = \frac{178}{82} = \frac{89}{41} \approx 2.17$$ - $$y_2 = \frac{130 - 48}{82} = \frac{82}{82} = 1$$ 13. **Find corresponding $x$ values:** - For $y_1 = \frac{89}{41}$: $$x = 11 - 9 \times \frac{89}{41} = 11 - \frac{801}{41} = \frac{451}{41} - \frac{801}{41} = -\frac{350}{41} \approx -8.54$$ - For $y_2 = 1$: $$x = 11 - 9 \times 1 = 2$$ 14. **Check which $D$ forms a kite:** Try $D(2,1)$ which is point $B$ already, so discard. Try $D(-\frac{350}{41}, \frac{89}{41}) \approx (-8.54, 2.17)$. 15. **Conclusion:** The fourth point $D$ is approximately $(-8.54, 2.17)$ to form a kite with the given points. **Final answer:** $$D = \left(-\frac{350}{41}, \frac{89}{41}\right)$$