Proving that there exists a unique hyperplane that passes through $n$ points I'm trying to prove:

Given $n$ points $x_0,...,x_{n-1}$ such that $x_1-x_0,...,x_{n-1}-x_0$ are linearly independent, then there exists a unique hyperplane that passes through the $n$ points.

The problem is that I can only use the definition provided in the book, which is
$$P=\{x\in \mathbb{R}^n:z\cdot x=c\}$$ for given $ c\in \mathbb{R},z\in \mathbb{R}^n\setminus \{0\}$
I'm at a loss here. Any help would be appreciated. 
 A: Since the vectors $x_1-x_0,\ldots,x_{n-1}-x_0$ are linearly independent, they span a $(n-1)$-dimensional subspace $V=\text{span}\{x_1-x_0,\ldots,x_{n-1}-x_0\}$ of $\mathbb R^n$. Therefore, its orthogonal complement $V^\perp$ is 1-dimensional and can be written as $V^\perp=\text{span}\{z\}$ for some $z\in\mathbb R^n\backslash\{0\}$. Thus, if we define $P:=x_0+V$ and $c:=x_0z\in\mathbb R$, then $P$ is a hyperplane passing through $x_0,x_1,\ldots,x_{n-1}$. Since each $x\in P$ can be written as $x=x_0+v$ for some $v\in V$ we have $xz=x_0z+vz=x_0z=c$, since $vz=0$ by construction. Conversly, if $xz=c$ for $x\in\mathbb R^n$, then $x$ can be written as $x=v+v'$ for some $v\in V$ and $v'\in V^\perp$. But then $(x-x_0)z=xz-x_0z=c-c=0$, which implies $x-x_0\in V$ and hence $x\in P$.
If $Q$ is another hyperplane passing through $x_0,x_1,\ldots,x_{n-1}$, we have that $Q-x_0=V$ and hence $P=x_0+V=x_0+Q-x_0=Q$.
A: Suppose that the hyperplane is defined by $\langle \mathrm a , \mathrm x\rangle = b$, where $\mathrm a$ and $b$ are to be determined. Hence,
$$\begin{bmatrix} — \mathrm x_1^T —\\ — \mathrm x_2^T —\\ \vdots \\ — \mathrm x_n^T —\\ \end{bmatrix} \begin{bmatrix} |\\ \mathrm a\\ | \end{bmatrix} = \begin{bmatrix} b\\ b\\ \vdots \\ b \end{bmatrix}$$
or, more succinctly,
$$\mathrm X^T \mathrm a = b 1_n$$
or, alternatively,
$$\begin{bmatrix} \mathrm X^T & -1_n\end{bmatrix} \begin{bmatrix} \mathrm a\\ b\end{bmatrix} = 0_n$$
which is an underdetermined linear system of $n$ equations in $n+1$ unknowns. If $\mathrm X$ has full rank, then there is $1$ degree of freedom, which stems from the fact that one can multiply $\mathrm a$ and $b$ by the same nonzero scalar, and one will still have the same hyperplane.
