Proof that a curve is a line given a restriction on it's tangent lines I'm having a hard time doing this problem and I'd like some insights:
Given a regular curve $\gamma: \rm I\!R \to \rm I\!R^n$, show that if all it's tangent lines pass through a same point $P$, then $\gamma$ is a line or a line segment.
I tried a proof by contradiction, supposing that it not being a line there exists a point where the curvature $\neq 0$ and proceeding to find a contradiction, but it seems to bring me no where as I can prove that the tangents only intersect at $P$ but nothing else. 
Thank you very much!
 A: I think a proof by contradiction is not necessary.
Since $\gamma$ is regular, we can reparametrise it to a unit speed curve. By assumption, for every $t\in\mathbb{R}$ there exists a $\lambda(t)\in\mathbb{R}$ ($\lambda$ depends on $t$) such that
$$ \gamma(t) +\lambda(t)\gamma'(t) = P.  \label{equa} \tag{*}$$
Deriving this  gives
$$ (1+\lambda'(t)) \gamma'(t) + \lambda(t) \gamma''(t) = 0.$$
Since $\gamma$ has unit speed, it follows that $\gamma''(t)\cdot \gamma'(t)=0$, so $\gamma''(t)$ is a normal vector. 
If $\gamma''(t_0)\neq 0$ at some point, then by continuity it is non-zero in some interval $I$ containing $t_0$. Then $\lambda(t)=0$ and thus $\gamma(t)=P$ for $t\in I$,  which would contradict the fact that $\gamma$ is regular. Hence $\gamma''(t)=0$ must hold for all $t$. Integrating this twice gives that $\gamma$ is a line or line segment.
Edit: The proof above uses that $\gamma\in C^2$. User Laz suggested in the comments that this statement should also hold if $\gamma \in C^1$. Here is an argument for the case $\gamma \in C^1$.
Write $\gamma(t)=\bigl(x(t),y(t)\bigr)$ and $P=(a,b)$. Then \eqref{equa} becomes 
$$
\begin{align*}
 x(t)+\lambda(t) x'(t)= a, \qquad  y(t)+\lambda(t) y'(t)= b.
\end{align*}
$$
If $\lambda(t_0)\neq 0$ in a point $t_0$, then $\lambda(t)\neq 0$ on a open interval containing $t_0$. On this interval we can integrate to obtain
$$ 
  \begin{align*}
 x(t)&= K_1 \exp\left(\int \frac{1}{\lambda(t)}\,dt\right) + a, \\
 y(t)&= K_2 \exp\left(\int \frac{1}{\lambda(t)}\,dt\right) + b, \\
\end{align*}
$$
This is a parametrisation of a line. This local argument shows that $\gamma$ piecewise consists of line segments.
If $\lambda(t)=0$ on some (open) interval, we get as before that $\gamma(t)=P$ on the interval and this contradicts the regularity. So $\lambda(t)=0$ only in isolated points. Now, note that the curve $\gamma(\mathbb{R})$ is connected ($\gamma$ is continuous and $\mathbb{R}$ is connected).  Moreover, $\gamma'(t)$ is continuous so the slopes of all the line segments have to be the same. Hence $\gamma$ is a line.
