If we have two points, what will happen if we add their coordinates? If point $A$ has coordinates $(1,2)$ and point $B$ has coordinates $(4,1)$, then $A + B = (5,3)$. It seems that there is nothing special about these points, but let us look at their position in the Cartesian coordinates.

Notice that if we add another point on the origin, the four points form a parallelogram. Now the question is, is this always the case? Try plotting two points and add their coordinates several times and see if the four points  form a parallelogram. What do you observe?

To test further our observation above, we use GeoGebra to plot the two points and automatically produce the ‘sum point.’

Instructions

1. Open GeoGebra
2. To plot the points, type the following equations in the input bar and then press the Enter/Return key on the keyboard after each equation: $A = (1,2)$, $B = (4,1)$ , $C = A + B$.
3. To create the fourth point $D$, click the Intersect Two Object tools, click the x-axis, and then click the y-axis.
4. To draw the quadrilateral, click the Polygon tool, and then click points $D, A, C, B, D$ in that order.
5. Now move point $A$ or $B$. Is  the quadrilateral always a parallelogram? What conjecture can be made from your observations?

We now prove our conjecture above.

Theorem:  If you add the coordinates of two points $A$ and $B$, and construct another point $C$ whose coordinates are the sum of the coordinates of points $A$ and $B$, then three points form a parallelogram containing the origin as its fourth vertex.

Proof: Let the coordinates of $A$ and $B$ be $(p,q)$ and $(r,s)$ respectively. Then the coordinates of $C = (p+r, q+s)$.  Let $D$ be the point at the origin. We have to show that $ACBD$ is a  parallelogram.

There are several ways to prove this, but it is sufficient to show that the opposite sides of the $ACBD$ are parallel; that is, $AD$ is parallel to $BC$ and $AC$ is parallel $BD$.

We now compute the slopes $m$ of each side of the parallelogram.

$m_{AD} = \displaystyle\frac{q-0}{p-0} = \frac{q}{p}$

$m_{BC} = \displaystyle\frac{(q+s)-s}{(p+r)-r} = \frac{q}{p}$

$m_{BD} = \displaystyle\frac{s-0}{r-0} = \frac{s}{r}$

$m_{AC} = \displaystyle\frac{(q+s)-q}{(p+r)-p} = \frac{s}{r}$

As we can see, the slopes of $AD$  and $BC$ are equal, which means that they are parallel.  Also,  the slopes of $BD$ and $AC$ are equal, which means that they are also parallel. Therefore, $ADBC$ is a parallelogram.

Note that the parallelogram method above can also represent addition of vectors.
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