= Solution
The required <general linear position> condition is that every $n+1$ of the $n+2$ representing vectors are <linearly independent>. Such an ordered configuration is a <projective frame>.
Choose representatives $p_0,\ldots,p_n$ of the first $n+1$ <projective points>; they form a <basis>. Write a representative of the last point as $p_{n+1}=\sum_i a_ip_i$. Every $a_i$ is nonzero: otherwise the last point together with the other $n$ basis points would fail the independence condition. Rescale the basis representatives to $\widetilde p_i=a_ip_i$, so that $p_{n+1}=\sum_i\widetilde p_i$.
Do the same for the target configuration, obtaining representatives $\widetilde q_i$ with $q_{n+1}=\sum_i\widetilde q_i$. The unique <linear map> sending the basis $\widetilde p_i$ to $\widetilde q_i$ is invertible and also sends $p_{n+1}$ to $q_{n+1}$. Its induced <projective transformation> sends every point to its prescribed partner. Thus \b[any two ordered <projective frames> are projectively equivalent].
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