(−4, 5)
(4, 5)
(−3, 4)
(0, 11)
Parallelogram ABCD with vertices:
In a parallelogram, diagonals bisect each other.
Find the midpoint of diagonal AC: \[ \text{Midpoint}_{AC} = \left( \frac{1 + (-2)}{2}, \frac{1 + 8}{2} \right) = \left( \frac{-1}{2}, \frac{9}{2} \right) \]
Midpoint of diagonal BD is: \[ \text{Midpoint}_{BD} = \left( \frac{3 + x}{2}, \frac{4 + y}{2} \right) \] Since diagonals bisect each other: \[ \frac{3 + x}{2} = \frac{-1}{2} \quad \text{and} \quad \frac{4 + y}{2} = \frac{9}{2} \]
First equation: \[ \frac{3 + x}{2} = \frac{-1}{2} \Rightarrow 3 + x = -1 \Rightarrow x = -4 \]
Second equation: \[ \frac{4 + y}{2} = \frac{9}{2} \Rightarrow 4 + y = 9 \Rightarrow y = 5 \]
Final Answer
Therefore, the coordinates of point \( D \) are: \[ \boxed{(-4, 5)} \]
In a parallelogram, the diagonals bisect each other.
Therefore, the midpoint of diagonal AC must equal the midpoint of diagonal BD.
\[ \text{Midpoint}_{AC} = \left( \frac{1 + (-2)}{2}, \frac{1 + 8}{2} \right) = \left( \frac{-1}{2}, \frac{9}{2} \right) \]
\[ \text{Midpoint}_{BD} = \left( \frac{3 + x}{2}, \frac{4 + y}{2} \right) \]
Equating midpoints: \[ \frac{3 + x}{2} = \frac{-1}{2}, \quad \frac{4 + y}{2} = \frac{9}{2} \]
Solve the first equation: \[ 3 + x = -1 \Rightarrow x = -4 \] Solve the second equation: \[ 4 + y = 9 \Rightarrow y = 5 \]
Therefore, the coordinates of point \( D \) are: \[ \boxed{(-4, 5)} \]