Uniqueness of Point

Question

I want to proove:

For points $P \neq Q \in \mathbb{R}$ and $a,b >0$ there is exactly one point $X \in [PQ]$ and exactly one point $X' \notin [PQ]$ with ratio $a:b$. For $a=b$ there is only one point $X \in [PQ]$ with $a:b$.

I know that $X$ divides $[PQ]$ in ratio $a:b$ if $$ \frac{d(X,P)}{d(X,Q)}=\frac a b$$ where $d(X,P)$ is the distance between $X$ and $P$.

My first thought was, that I take two points which has both the ratio $a:b$ and then show, that they are the same.

Any hints are welcome.

Answer 1

It can be thought as an analytic geometic solution.

Notice that line segmnet $[P,Q]$ can be parametrized by

$$\vec{f(t)}=\vec{P}+t(\vec{Q}-\vec P)$$ for $t\in [0,1]$ and $f(0)=P \ and \ f(1)=Q$

Set $X=f(t_0)$ and $k$ be the given ratio then

$$d(X,P)=|\vec f(t_0)-\vec f(0)|=t_0(\vec{|P-Q|})$$ and

$$d(X,Q)=|\vec f(t_0)-\vec f(1)|=(1-t_0)\vec{|P-Q|}$$

$$k=\dfrac{d(X,P)}{d(X,Q)}=\dfrac{t_0}{1-t_0}\implies t_0=\dfrac{k}{k+1}$$

If $t_0=1$ then write equation as $$ \dfrac 1 k=\dfrac{d(X,Q)}{d(X,P)}=\dfrac{1-t_0}{t_0}\implies t_0=\dfrac{k}{k+1}$$

You can easliy see that given ratio $k$ uniqly determine $t_0$ by above equation. So such a point is uniqe.

If $X\notin [P,Q]$ then $t_0>1$ in that case we will get similiar formula,

$$k=\dfrac{d(X,P)}{d(X,Q)}=\dfrac{t_0}{t_0-1}\implies t_0=\dfrac{k}{k-1}$$

which can be got in similiar way.

if $a=b$ i.e. $k=1$ it has no solution.

Answer 2

I tried the following (using fgp's hint): $d(X,P)+d(X,Q)=d(P,Q)$

\begin{eqnarray} &&\frac{d(X,P)}{d(X,Q)} = \frac{d(X',P)}{d(X',Q)} \Leftrightarrow \frac{d(P,Q)-d(X,Q)}{d(X,Q)}=\frac{d(P,Q)-d(X',P)}{d(X',Q)} \\ &\Leftrightarrow & \frac{d(P,Q)}{d(X,Q)}-1=\frac{d(P,Q)}{d(X',Q)}-1 \Leftrightarrow \frac{d(P,Q)}{d(X,Q)}=\frac{d(P,Q)}{d(X',Q)} \\ &\Leftrightarrow &d(X',Q)=d(X,Q) \Rightarrow X=X' \end{eqnarray}

Is that ok too?