To find the coordinates of point D that form a parallelogram along with points A(-2, 4), B(1, 3), and C(4, -1), we can use the property of parallelograms that the midpoints of the diagonals are the same. Here's a step-by-step solution:
1. Identify the Midpoints of the Diagonals:
- The diagonals of a parallelogram bisect each other. Therefore, the midpoint of diagonal AC should be the same as the midpoint of diagonal BD.
2. Calculate the Midpoint of Diagonal AC:
- Point A has coordinates (-2, 4).
- Point C has coordinates (4, -1).
The midpoint formula for points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[
\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\][/tex]
Applying the midpoint formula to points A and C:
[tex]\[
\text{Midpoint of AC} = \left( \frac{-2 + 4}{2}, \frac{4 + (-1)}{2} \right) = \left( \frac{2}{2}, \frac{3}{2} \right) = (1, 1.5)
\][/tex]
3. Setup the Midpoint Formula for Diagonal BD:
- Point B has coordinates (1, 3).
- Let point D have coordinates [tex]\((D_x, D_y)\)[/tex].
Using the midpoint formula for points B and D:
[tex]\[
\text{Midpoint of BD} = \left( \frac{1 + D_x}{2}, \frac{3 + D_y}{2} \right)
\][/tex]
4. Equate the Midpoints of AC and BD:
Since the midpoint of AC must be the same as the midpoint of BD:
[tex]\[
\left( \frac{1 + D_x}{2}, \frac{3 + D_y}{2} \right) = (1, 1.5)
\][/tex]
5. Solve for [tex]\(D_x\)[/tex] and [tex]\(D_y\)[/tex]:
- For the x-coordinate:
[tex]\[
\frac{1 + D_x}{2} = 1 \implies 1 + D_x = 2 \implies D_x = 1
\][/tex]
- For the y-coordinate:
[tex]\[
\frac{3 + D_y}{2} = 1.5 \implies 3 + D_y = 3 \implies D_y = 0
\][/tex]
Thus, the coordinates of point D are [tex]\( (1, 0) \)[/tex].
Therefore, the correct answer is:
O D. (1, 0)
