4.1 Addition of Points

From now on I will denote points in the plane by lower case boldface letters, e.g. $\mbox{{\bf a}},\mbox{{\bf b}},\cdots$. If I specify a point $\mbox{{\bf a}}$ and do not explicitly write down its components, you should assume $\mbox{{\bf a}}=(a_1,a_2),\; \mbox{{\bf b}}=(b_1,b_2),\; \cdots , \mbox{{\bf k}}=(k_1,k_2)$, etc. The one exception to this rule is that I will always take

$${\bf x} = (x,y).$$

4.1   Definition (Addition of Points) If $\mbox{{\bf a}}$ and $\mbox{{\bf b}}$ are points in $\mbox{{\bf R}}^2$ and $t\in\mbox{{\bf R}}$, we define

$$\begin{eqnarray*} \mbox{{\bf a}}+ \mbox{{\bf b}}& = & (a_1,a_2)+(b_1,b_2)=(a_1+b... ...a_1-b_1,a_2-b_2)\\ t\mbox{{\bf a}}& = & t(a_1,a_2)=(ta_1,ta_2). \end{eqnarray*}$$

If $t\neq 0$, we will write $${ {\mbox{{\bf a}}\over t}}$$ for $${ {1\over t}\mbox{{\bf a}}}$$; i.e., $${{\mbox{{\bf a}}\over t} = \left( {{a_1}\over t}, {{b_1}\over t}\right)}$$. We will abbreviate $(-1)\mbox{{\bf a}}$ by $-\mbox{{\bf a}}$, and we will write $\mbox{{\bf0}} =(0,0)$.

4.2   Theorem. Let $\mbox{{\bf a}},\; \mbox{{\bf b}},\; \mathbf{{\bf c}}$ be arbitrary points in $\mbox{{\bf R}}^2$ and let $s,t$ be arbitrary numbers. Then we have:

Addition is commutative,

$$\mbox{{\bf a}}+\mbox{{\bf b}}=\mbox{{\bf b}}+\mbox{{\bf a}}.$$

Addition is associative,

$$(\mbox{{\bf a}}+\mbox{{\bf b}})+\mathbf{{\bf c}}=\mbox{{\bf a}}+(\mbox{{\bf b}}+\mathbf{{\bf c}}).$$

We have the following law that resembles the associative law for multiplication:

$$s(t\mbox{{\bf a}})=(st)\mbox{{\bf a}}.$$

We have the following distributive laws:

$$(s+t)\mbox{{\bf a}}=s\mbox{{\bf a}}+ t\mbox{{\bf a}},$$ (4.3)

$$s(\mbox{{\bf a}}+\mbox{{\bf b}})=s\mbox{{\bf a}}+s\mbox{{\bf b}}.$$ (4.4)

Also,

$$1\mbox{{\bf a}}=\mbox{{\bf a}}, \;\; 0 \mbox{{\bf a}}=\mbox{{... ... \mbox{ and } \mbox{{\bf a}}+ (-\mbox{{\bf a}})=\mbox{{\bf0}} .$$

All of these properties follow easily from the corresponding properties of real numbers. I will prove the commutative law and one of the distributive laws, and omit the remaining proofs.

Proof of Commutative Law: Let $\mbox{{\bf a}},\; \mbox{{\bf b}}$ be points in $\mbox{{\bf R}}^2$. By the commutative law for $\mbox{{\bf R}}$,

$$a_1 +b_1=b_1 +a_1 \mbox{ and } a_2+b_2=b_2+a_2.$$

Hence

$$\begin{eqnarray*} \mbox{{\bf a}}+\mbox{{\bf b}}& = & (a_1,a_2)+(b_1,b_2)=(a_1+b_... ...a_2)\\ & = & (b_1,b_2)+(a_1,a_2)=\mbox{{\bf b}}+\mbox{{\bf a}}. \end{eqnarray*}$$

and hence $\mbox{{\bf a}}+\mbox{{\bf b}}=\mbox{{\bf b}}+\mbox{{\bf a}}$.

Proof of (4.3): Let $s,t\in\mbox{{\bf R}}$ and let $\mbox{{\bf a}}\in\mbox{{\bf R}}^2$. By the distributive law for $\mbox{{\bf R}}$ we have

$$(s+t)a_1=sa_1+ta_1 \mbox{ and } (s+t)a_2=sa_2+ta_2.$$

Hence,

$$\begin{eqnarray*} (s+t)\mbox{{\bf a}}& = & (s+t)(a_1,a_2)=\left( (s+t)a_1,(s+t)a... ...=s(a_1,a_2)+t(a_1,a_2)\\ & = & s\mbox{{\bf a}}+t\mbox{{\bf a}}, \end{eqnarray*}$$

i.e,

$$(s+t)\mbox{{\bf a}}=s\mbox{{\bf a}}+t\mbox{{\bf a}}.\mbox{ $\diamondsuit$}$$

4.5   Notation (Lines in $\mbox{{\bf R}}^2$.) If $\mbox{{\bf a}},\mbox{{\bf b}}$ are distinct points in $\mbox{{\bf R}}^2$, I will denote the (infinite) line through $\mbox{{\bf a}}$ and $\mbox{{\bf b}}$ by $\mbox{{\bf a}}\mbox{{\bf b}}$, and I will denote the line segment joining $\mbox{{\bf a}}$ to $\mbox{{\bf b}}$ by $[\mbox{{\bf a}}\mbox{{\bf b}}]$. Hence $[\mbox{{\bf a}} \mbox{{\bf b}}]=[\mbox{{\bf b}}\mbox{{\bf a}}]$.

Remark: Let $\mbox{{\bf a}}$, $\mbox{{\bf b}}$ be points in $\mbox{{\bf R}}^2$ such that $\mbox{{\bf0}},\;\mbox{{\bf a}}$ and $\mbox{{\bf b}}$ are not all in a straight line. Then $\mbox{{\bf a}}+\mbox{{\bf b}}$ is the vertex opposite $\mbox{{\bf0}}$ in the parallelogram whose other three vertices are $\mbox{{\bf b}},\;\mbox{{\bf0}}$ and $\mbox{{\bf a}}$.

Proof: In this proof I will suppose $a_1\neq 0$ and $b_1\neq 0$, so that neither of $\mbox{{\bf0}}\mbox{{\bf a}},\mbox{{\bf0}}\mbox{{\bf b}}$ is a vertical line. (I leave the other cases to you.) The slope of line $\mbox{{\bf0}}\mbox{{\bf a}}$ is $${ {{a_2-0}\over {a_1-0}}={{a_2}\over {a_1}}}$$, and the slope of $\mbox{{\bf b}}(\mbox{{\bf a}}+\mbox{{\bf b}})$ is $${ {{(a_2+b_2)-b_2}\over {(a_1+b_1)-b_1}}={{a_2}\over {a_1}}}$$. Thus the lines $\mbox{{\bf0}}\mbox{{\bf a}}$ and $\mbox{{\bf b}}(\mbox{{\bf a}}+\mbox{{\bf b}})$ are parallel.

The slope of line $\mbox{{\bf0}}\mbox{{\bf b}}$ is $${ {{b_2-0}\over {b_1-0}}={{b_2}\over {b_1}}}$$, and the slope of $\mbox{{\bf a}}(\mbox{{\bf a}}+\mbox{{\bf b}})$ is
$${ {{(a_2+b_2)-a_2}\over {(a_1+b_1)-a_1}}={{b_2}\over {b_1}}}$$. Thus the lines $\mbox{{\bf0}}\mbox{{\bf b}}$ and $\mbox{{\bf a}}(\mbox{{\bf a}}+\mbox{{\bf b}})$ are parallel. It follows that the figure $\mbox{{\bf0}}\mbox{{\bf a}}(\mbox{{\bf a}}+\mbox{{\bf b}})\mbox{{\bf b}}$ is a parallelogram, i.e., $\mbox{{\bf a}}+\mbox{{\bf b}}$ is the fourth vertex of a parallelogram having $\mbox{{\bf0}},\;\mbox{{\bf a}},$ and $\mbox{{\bf b}}$ as its other vertices. $\diamondsuit$

4.6   Example. In the figure you should be able to see the parallelograms defining $\mbox{{\bf a}}+\mbox{{\bf b}},\; (\mbox{{\bf a}}+\mbox{{\bf b}})+\mathbf{{\bf c}},\; \mbox{{\bf b}}+\mathbf{{\bf c}}$ and $\mbox{{\bf a}}+(\mbox{{\bf b}}+\mathbf{{\bf c}})$. Also you should be able to see geometrically that $(\mbox{{\bf a}}+\mbox{{\bf b}})+\mathbf{{\bf c}}=\mbox{{\bf a}}+(\mbox{{\bf b}}+\mathbf{{\bf c}})$. What is the point marked x in the figure?

4.7   Exercise. In figure a), $\mbox{{\bf a}},\;\mbox{{\bf b}},\;\mathbf{{\bf c}},\;\mbox{{\bf d}},\;\mbox{{\bf e}}$, and ${\mathbf f}$ are the vertices of a regular hexagon centered at $\mbox{{\bf0}}$. Sketch the points $\mbox{{\bf a}}+\mbox{{\bf b}},\; (\mbox{{\bf a}}+\mbox{{\bf b}})+\mathbf{{\bf c}}$, $(\mbox{{\bf a}}+\mbox{{\bf b}}+\mathbf{{\bf c}})+\mbox{{\bf d}},\; (\mbox{{\bf a}}+\mbox{{\bf b}}+\mathbf{{\bf c}}+\mbox{{\bf d}})+\mbox{{\bf e}}$, and $(\mbox{{\bf a}}+\mbox{{\bf b}}+\mathbf{{\bf c}}+\mbox{{\bf d}}+\mbox{{\bf e}})+{\mathbf f}$ as accurately as you can.

In figure b), $\mbox{{\bf a}},\;\mbox{{\bf b}},\;\mathbf{{\bf c}},\;\mbox{{\bf d}},\;\mbox{{\bf e}}$ and $\mbox{{\bf f}}$ are the vertices of a regular hexagon with $\mbox{{\bf f}}=\mbox{{\bf0}}$. Sketch the points $\mbox{{\bf a}}+\mbox{{\bf b}},\; (\mbox{{\bf a}}+\mbox{{\bf b}})+\mathbf{{\bf c}},\; (\mbox{{\bf a}}+\mbox{{\bf b}}+\mathbf{{\bf c}})+\mbox{{\bf d}}$, and $(\mbox{{\bf a}}+\mbox{{\bf b}}+\mathbf{{\bf c}}+\mbox{{\bf d}})+\mbox{{\bf e}}$ as accurately as you can. (This problem should be done geometrically. Do not calculate the coordinates of any of these points.)

4.8   Example (Line segment) We will now give an analytical description for a non-vertical line segment $[\mbox{{\bf a}}\mbox{{\bf b}}]$, $(a_1\neq b_1)$. Suppose first that $a_1 < b_1$. The equation for the line through $\mbox{{\bf a}}$ and $\mbox{{\bf b}}$ is

$$y = a_2 + {b_2-a_2 \over b_1 - a_1}(x - a_1).$$

Hence a point $(x,y)$ is in $[\mbox{{\bf a}}\mbox{{\bf b}}]$ if and only if there is a number $x \in [a_1,b_1]$ such that

$$\begin{eqnarray*} (x,y) &=& \Big(x,a_2 + {b_2-a_2 \over b_1 - a_1}(x - a_1)\Big)... ...{{\bf a}}+{x-a_1 \over b_1-a_1}(\mbox{{\bf b}}- \mbox{{\bf a}}). \end{eqnarray*}$$

Now

$$\begin{eqnarray*} x \in [a_1,b_1] &\mbox{$\Longleftrightarrow$}& a_1 \leq x \le... ...x{$\Longleftrightarrow$}& 0 \leq {x-a_1 \over b_1 - a_1} \leq 1. \end{eqnarray*}$$

Thus

$$[\mbox{{\bf a}}\mbox{{\bf b}}]= \{\mbox{{\bf a}}+ t(\mbox{{\bf b}}-\mbox{{\bf a}}) : 0 \leq t \leq 1\}.$$

If $b_1 < a_1$ then

$$\begin{eqnarray*}[\mbox{{\bf a}}\mbox{{\bf b}}]&=& [\mbox{{\bf b}}\mbox{{\bf a}}... ...f a}}+ (1-t)(\mbox{{\bf b}}- \mbox{{\bf a}}): 0 \leq t \leq 1\}. \end{eqnarray*}$$

Now as $t$ runs through all values in $[0,1]$, we see that $1-t$ also takes on all values in $[0,1]$ so we get the same description for $[\mbox{{\bf a}}\mbox{{\bf b}}]$ when $b_1 < a_1$ as we do when $a_1 < b_1$. Note that this description is exactly what you would expect from the pictures, and that it also works for vertical segments.