This question is inspired by an answer to Nitrogen's answer to my Are the path connected components of $\Omega S_1$ contractible? .
Here we are asked whether the space of paths in $\mathbb{R}$ sending $0$ to $0$, $1$ to $n$, with $[0,n] \subset \mathbb{R}$, is contractible? The answer is yes because $\mathbb{R}$ is convex, i.e. there is a natural choice of a path(namely a line segment) from the tail endpoint of a path, to $n$. In more detail, we have a deformation retract from $P_{0 \to 0} ([0,N])$ of basepoint fixing paths in $[0,N]$ to the space $P_{0 \to 0, 1 \to N} ([0,N])$. The retraction is given by sending any path $\gamma_a(t)$ with $\gamma_a(t)=x$ to the path $\gamma_b$ that does the line segment from $x$ to $n$ on $[(n-x)/n, 1]$ and a shrinked version of $\gamma_a$ on $[0,(n-x)/n]$. Thus $P_{0 \to 0} ([0,N])$ which is contractible is homotopy equivalent to $P_{0 \to 0, 1 \to N} ([0,N])$.
Is the following generalization of this true: Let $A$ be a contractible topological space and let $x_1$ and $x_2 \in A$ be given. Is the space of paths with the indicated endpoints $P_{0 \to x_1, 1 \to x_2}(A)$ contractible?
To elaborate on Qiaochu's comment: Suppose $H:A \times I \to A$ is a homotopy from the identity to the constant map on $x_2$. Then $\lambda=H(x_1,-)$ is a path from $x_1$ to $x_2$.
I claim that $P_{0 \to x_1, 1 \to x_2}(A)$ is homotopy equivalent to $\Omega A$, the based loops at $x_1$, which is contractible. Consider the maps $$f:P_{0 \to x_1, 1 \to x_2}(A) \to \Omega A, \ \ g:\Omega A \to P_{0 \to x_1, 1 \to x_2}(A)$$ given by $f(\alpha)=\alpha \cdot \bar{\lambda}$ and $g(\alpha)=\alpha \cdot \lambda$. With a little bit of techinique, using $H$, you can show that these maps are homotopy inverse to each other.