Barycentric coordinate on the Straight Line

          Let $A$ and $B$ be distinct points. 

     LEMMA  The following statements are equivalent:

                  (i) points $\;X,\;A,\;B\;$ are collinear and $\frac{\overline{XA}}{\overline{XB}}=\kappa$;

                 (ii) $\overrightarrow{OX}=\frac{\overrightarrow{OA}-\kappa \cdot \overrightarrow{OB}}{1-\kappa}$ for a certain point $O$;

                (iii)  $\overrightarrow{OX}=\frac{\overrightarrow{OA}-\kappa \cdot \overrightarrow{OB}}{1-\kappa}$ for any point $O$.

     Corollary  The following statements are equivalent:

          ($\alpha$)  points $\;X,\;A,\;B\;$ are collinear and $\frac{\overline{AX}}{\overline{AB}}=r$;

          ($\beta$)  $\overrightarrow{OX}=(1-r)\cdot \overrightarrow{OA}+r \cdot \overrightarrow{OB}$ for a certain/any point $O$.

 

     Remark  The formula from point ($\beta$) is still writte

$$X=(1-r)\cdot A+r \cdot B.$$

          Proof CiP

          (i)$\Rightarrow$(iii) As oriented segments, we have $\frac{\overline{XA}}{\overline{XB}}=\kappa\;\Rightarrow\;\overrightarrow{XA}=\kappa \cdot \overrightarrow{XB}\;\Rightarrow$

$$\Rightarrow\;\overrightarrow{OA}-\overrightarrow{OX}=\kappa(\overrightarrow{OB}-\overrightarrow{OX})\;\Rightarrow\;(1-\kappa)\overrightarrow{OX}=\overrightarrow{OA}-\kappa \cdot \overrightarrow{OB}$$

where $O$ is an arbitrary point. We get the conclusion.

          (ii)$\Rightarrow$(i) $(1-\kappa)\overrightarrow{OX}=\overrightarrow{OA}-\kappa \cdot \overrightarrow{OB}\;\Rightarrow\;\overrightarrow{OA}-\overrightarrow{OX}=\kappa (\overrightarrow{OB}-\overrightarrow{OX}\;\Rightarrow\;\overrightarrow{XA}=\kappa \cdot \overrightarrow{XB}$

so the vectors $\overrightarrow{XA}$ and $\overrightarrow{XB}$ are collinear and their orientation is according to the sign of $\kappa$, as the figure shows. Hence $\overline {XA}=\kappa \cdot \overline{XB}$.

          (iii)$\rightarrow$(ii)  obvious.

     The Lemma is proved. To prove the corollary we note that

$$\overrightarrow{XA}=\kappa \cdot \overrightarrow{XB}\;\Leftrightarrow\;\overrightarrow{AX}=-\kappa \cdot \overrightarrow{XB}=-\kappa(\overrightarrow{AB}-\overrightarrow{AX})\;\Leftrightarrow \;(1-\kappa)\overrightarrow{AX}=-\kappa \cdot \overrightarrow{AB}$$

hence $r=-\frac{\kappa}{1-\kappa}$.

$\blacksquare \blacksquare$