Fourth vertex of a parallelogram

We find the fourth vertex $D$ of the parallelogram $ABCD$ whose other vertices are $A(0, 0)$ , $B(3, 0)$ and $C(4, 2)$ ¹.

From the diagonals

In a parallelogram the two diagonals $AC$ and $BD$ bisect each other. The midpoint of $AC$ is $(2, 1)$ . Writing $D(x, y)$ , the midpoint of $BD$ is $\left( \dfrac{3 + x}{2}, \dfrac{y}{2} \right)$ , and setting it equal gives $x = 1$ and $y = 2$ , that is $D(1, 2)$ .

From the parallel sides

The property that opposite sides are parallel and of equal length gives the answer directly. If the step from $A$ to $B$ is $(3, 0)$ , the step from $D$ to $C$ must be the same.

D = C - (3, 0) = (4 - 3,\ 2 - 0) = (1, 2)

Whether by the diagonals or by the parallel sides, the same point is reached.

Checking the parallels

Side	The two points it passes	Slope
$AB$	$(0,0)$ , $(3,0)$	$0$
$DC$	$(1,2)$ , $(4,2)$	$0$
$AD$	$(0,0)$ , $(1,2)$	$2$
$BC$	$(3,0)$ , $(4,2)$	$2$

The four sides do fall into two pairs of parallels.

There are three possible fourth vertices

It is worth noting that three points do not determine a parallelogram uniquely. Depending on which two are taken as the ends of a diagonal, the fourth vertex can be any of three points.

The two points taken as a diagonal	Fourth vertex
$A$ and $C$	$(1, 2)$
$B$ and $C$	$(7, 2)$
$A$ and $B$	$(-1, -2)$

It is the convention of naming the vertices in the order $A \to B \to C \to D$ that pins $D$ down to one of them.

Area

The parallelogram spanned by the direction $(3, 0)$ of the side $AB$ and the direction $(1, 2)$ of the side $AD$ has the following area.

|3 \cdot 2 - 0 \cdot 1| = 6

That it is exactly twice the area $3$ of the triangle $ABC$ stands to reason, since the diagonal $AC$ cuts the parallelogram into two congruent triangles.

The large dots are the four vertices, with $AB$ parallel to $DC$ and $AD$ parallel to $BC$ .

Parallelogram, Wikipedia