For a vector $x = (x_1, \ldots, x_p)$, and corresponding $y = b_0 + \sum_{j=1}^{p} x_j b_j$, the set of tuples $(x,y)$ is a hyperplane. Suppose I have a vector $x = (x_1, \ldots, x_p) $ where $x_i \in \mathbb{R}$ for $1 \leq i \leq p$. 
Define $y = b_0 + \sum_{j=1}^{p} x_j b_j$ for some real numbers $b_i \in \mathbb{R}$ for $0 \leq i \leq p$.
Then an author of a textbook I am reading claims that the $(p+1)$ dimensional input-output space of tuples $(x,y)$ represents a hyperplane.
From what I understand, a hyperplane for a vector space of dimension $n$ is a subspace of dimension $n-1$ and can be written as $\{x \in R^n | x \cdot n = t\}$ where $n \in \mathbb{R}^n$ is the normal of the hyperplane, and $t \in \mathbb{R}$.
Why is the described space a hyperplane?
 A: Rewrite the original equation in the form 
$$
-b_0 =  (\sum_{j=1}^{p} x_j b_j) + (-1) y.
$$
Let $n = (b_1, \ldots, b_p, -1)$.
Then the equation above can be read as 
$$
-b_0 = n \cdot (x_1, x_2, \ldots, x_p, y) 
$$
Letting $t = -b_0$, this gives you 
$$
t = n \cdot (x_1, x_2, \ldots, x_p, y) 
$$
which is the equation of a hyperplane in the space of $(x_1, \ldots, x_p, y)$ tuples, which is just $\Bbb R^{p+1}$, slightly disguised. 
A: First consider the special case $b_0=0$. Then
$$
U_b := \{(x, b\cdot x) : x\in R^p\}\subseteq R^{p+1}.
$$
is the image of the linear transformation
$$
f_b:R^p\to R^{p+1},\quad f(x) = (x, b\cdot x).
$$
It is clear that if $f_b(x)=0$ then $x=0$, making $f_b$ injective. Therefore, 
$$\dim U_b = \dim (\operatorname{image} f_b) = \dim R^p = p.$$
You can write down an equation cutting out $U$ as follows:
As $(-b, 1)\cdot(x, b\cdot x) = 0$, the vector $(-b, 1)$ is a nonzero element of $U^\perp$. Since $U$ is a $p$-dimensional subspace of $R^{p+1}$, $U^\perp$ is $1$-dimensional. Therefore, $(-b, 1)$ is a basis of $U^\perp$. But then
$$
U_b = (U_b^\perp)^\perp (-b, 1)^\perp = \{(x, y)\in R^{p+1} : (x, y)\cdot (-b, 1)=0\}
$$
Now let $b_0$ be arbitrary. Then
$$
U_{b_0, b} := \{(x, b_0 + b\cdot x) : x\in R^p\}
= U_b + (0, b_0).
$$
It follows that $(x, y)\in U_{b, b_0}$ if and only if
$$
(x, y)\cdot (-b, 1) = (0, b_0)\cdot (-b, 1) = b_0.
$$
In other words,
$$
U_{b_0,b} = \{(x,y)\in R^{p+1} : (x,y)\cdot(-b, 1) = b_0\}.
$$
