Question

Difficulty: MediumProperties of Quadrilaterals

In parallelogram ABCDABCD, diagonals ACAC and BDBD intersect at point EE. In the standard (x,y)(x,y) coordinate plane, vertex AA is located at (3,2)(-3, 2) and point EE is located at (2,1)(2, 1). What are the coordinates of vertex CC?

  1. A
    (0.5,1.5)(-0.5, 1.5)
  2. B
    (3,5)(3, -5)
  3. C
    (5,1)(5, -1)
  4. (7,0)(7, 0)Answer
  5. E
    (7,4)(7, 4)

Answer

The coordinates of vertex CC are (7,0)(7, 0).
A fundamental property of any parallelogram is that its diagonals bisect each other. Therefore, the intersection point E(2,1)E(2, 1) must be the midpoint of diagonal ACAC. Using the midpoint formula, 3+xC2=2\frac{-3 + x_C}{2} = 2 gives xC=7x_C = 7, and 2+yC2=1\frac{2 + y_C}{2} = 1 gives yC=0y_C = 0. Thus, the coordinates of vertex CC are (7,0)(7, 0).

Step-by-Step Solution

1
Identify the key geometric property of parallelograms.
The diagonals of a parallelogram bisect each other, which means point EE is the midpoint of diagonal ACAC.
By definition of diagonal bisection in any parallelogram, the intersection point of the diagonals is the midpoint of both diagonal segments.
2
Set up the midpoint formula for segment ACAC with midpoint E(2,1)E(2, 1).
(xA+xC2,yA+yC2)=(2,1)(\frac{x_A + x_C}{2}, \frac{y_A + y_C}{2}) = (2, 1), where xA=3x_A = -3 and yA=2y_A = 2.
The midpoint coordinates are the averages of the endpoint coordinates.
3
Solve for the xx-coordinate of vertex CC.
3+xC2=2    3+xC=4    xC=7\frac{-3 + x_C}{2} = 2 \implies -3 + x_C = 4 \implies x_C = 7.
Multiply by 2 and add 3 to isolate xCx_C.
4
Solve for the yy-coordinate of vertex CC.
2+yC2=1    2+yC=2    yC=0\frac{2 + y_C}{2} = 1 \implies 2 + y_C = 2 \implies y_C = 0.
Multiply by 2 and subtract 2 to isolate yCy_C.

Key Concept

Diagonals of a Parallelogram Bisect Each Other

Alternative Method

Use vector translations: The vector from A(3,2)A(-3, 2) to E(2,1)E(2, 1) is 2(3),12=5,1\langle 2 - (-3), 1 - 2 \rangle = \langle 5, -1 \rangle. Since EE is the midpoint of ACAC, the vector from EE to CC is identical. Adding 5,1\langle 5, -1 \rangle to E(2,1)E(2, 1) gives C(2+5,11)=(7,0)C(2 + 5, 1 - 1) = (7, 0).
Estimated Time:1m 15s
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