# directed segment

Let $AB$ a line segment  . The directed segment $\overline{AB}$ is to be taken the segment $AB$ with a direction (similar   to vectors). The defining property is then

 $\overline{AB}=-\overline{BA},$

(which relates to the property of vectors stating that $v$ and $-v$ have opposite direction and same modulus).

 $\overline{AB}+\overline{BA}=0.$

where $0$ represents any segment of the form $\overline{PP}$. If it is stated that we will work with directed segments, it’s customary to omit the overlining and to just write $AB$, convention we will follow now.

Notes.

It can be proved considering cases that no matter the relative position of three points $A,B,C$ on a line, the following equality holds:

 $AB+BC=AC.$

In the above picture $AP+PB=AB$. Notice that $AB$ goes to the left since $AB$ is the segment that starts at $A$ and ends at $B$. Also, taking $A=C$ gives $AB+BA=AA=0$ which is consistent with the earlier remarks.

Just like undirected segments in Euclidean geometry  (and unlike vectors), directed segments can be divided to obtain a ratio. Such ratio is the number obtained dividing the undirected segments, but taking signs int oaccount (ratio of two segments with the same direction is positive, and negative otherwise).

Given two points $A,B$ on a line, we can locate any other point $P$ on the line considering the ratio $AP/PB$. In other words, $P=Q$ if and only if $AP/PB=AQ/QB$. Moreover, to each point $P$ corresponds an extended 11We use extended reals to avoid dealing with separate cases, allowing us to deal also with points at infinity. real $r=AP/PB$ and to each extended real $r$ corresponds a point $P$ such that $r=AP/PB$.

Notice that $AP/PB$ is positive when $AP$ and $PB$ have the same direction, which happens if and only if $p$ is between $A$ and $B$. If $P$ lies outside $AP$, then $AP$ and $PB$ have negative signs and so the ratio will be negative.

Title directed segment DirectedSegment 2013-03-22 14:56:59 2013-03-22 14:56:59 drini (3) drini (3) 9 drini (3) Definition msc 51F99 msc 51M25 CevasTheorem Midpoint    