(−4, 5)
(4, 5)
(−3, 4)
(0, 11)
Parallelogram ABCD with vertices:
The diagonals of a parallelogram bisect each other.
Calculate the midpoint of diagonal AC: \[ \text{Midpoint}_{AC} = \left( \frac{1 + (-2)}{2}, \frac{1 + 8}{2} \right) = \left( \frac{-1}{2}, \frac{9}{2} \right) \]
The midpoint of diagonal BD is: \[ \text{Midpoint}_{BD} = \left( \frac{3 + x}{2}, \frac{4 + y}{2} \right) \] As the diagonals bisect each other, their midpoints are identical: \[ \frac{3 + x}{2} = \frac{-1}{2} \quad \text{and} \quad \frac{4 + y}{2} = \frac{9}{2} \]
Solve the equation for x: \[ \frac{3 + x}{2} = \frac{-1}{2} \Rightarrow 3 + x = -1 \Rightarrow x = -4 \]
Solve the equation for y: \[ \frac{4 + y}{2} = \frac{9}{2} \Rightarrow 4 + y = 9 \Rightarrow y = 5 \]
Final Answer
The coordinates of point \( D \) are: \[ \boxed{(-4, 5)} \]