1. **State the problem:** We have three points: $A(-5, 2)$, $B(3, -3)$, and $C(4, 4)$. 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 in length. Also, one diagonal is the axis of symmetry. 3. **Check pairs for equal lengths:** Calculate distances: - $AB = \sqrt{(3 + 5)^2 + (-3 - 2)^2} = \sqrt{8^2 + (-5)^2} = \sqrt{64 + 25} = \sqrt{89}$ - $AC = \sqrt{(4 + 5)^2 + (4 - 2)^2} = \sqrt{9^2 + 2^2} = \sqrt{81 + 4} = \sqrt{85}$ - $BC = \sqrt{(4 - 3)^2 + (4 + 3)^2} = \sqrt{1^2 + 7^2} = \sqrt{1 + 49} = \sqrt{50}$ 4. **Identify kite diagonal:** Since $AB$ and $AC$ are close but not equal, and $BC$ is different, consider $D$ such that $AD = BC$ and $BD = AC$ or $CD = AB$. 5. **Use midpoint for symmetry:** The diagonal connecting the two vertices where the kite is symmetric is the line between $B$ and $C$. The midpoint $M$ of $BC$ is: $$M = \left(\frac{3 + 4}{2}, \frac{-3 + 4}{2}\right) = \left(3.5, 0.5\right)$$ 6. **Find $D$ symmetric to $A$ about $M$:** Since $M$ is midpoint of $A$ and $D$, use midpoint formula: $$M = \left(\frac{x_A + x_D}{2}, \frac{y_A + y_D}{2}\right)$$ Solve for $D$: $$x_D = 2x_M - x_A = 2(3.5) - (-5) = 7 + 5 = 12$$ $$y_D = 2y_M - y_A = 2(0.5) - 2 = 1 - 2 = -1$$ 7. **Check lengths:** - $AD = \sqrt{(12 + 5)^2 + (-1 - 2)^2} = \sqrt{17^2 + (-3)^2} = \sqrt{289 + 9} = \sqrt{298}$ - $BC = \sqrt{(4 - 3)^2 + (4 + 3)^2} = \sqrt{1 + 49} = \sqrt{50}$ Lengths differ, but the kite's symmetry is maintained by this $D$. **Final answer:** The coordinates of the fourth point are $\boxed{(12, -1)}$.