Describe, geometrically, the kernel of a $4\times4$ matrix with nullity $1$ Let $Ax=0$
\begin{equation}A=\begin{vmatrix}
7 & 5 & 4 & -9 \\
5& 3 & 8 & -2 \\
12 & 8 & 12 & 7 \\
8 & 6 & 2 & -5
\end{vmatrix}\
\end{equation}
I need to find the solution set and describe it geometrically.
I have determined that the RREF is:
\begin{equation}A=\begin{vmatrix}
1 & 0 & 7 & 0 \\
0 & 1 & -9 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{vmatrix}\
\end{equation}
Which makes the solution set:
$(-7t,9t,t,0)$
My question now is in how do I properly explain this geometrically?
I have said that this is a hyperplane with one free variable, but I'm uncertain if this is the right answer. Might it be some sort of hyperline, since the last component is a constant and we only have one free variable, if that is even a thing?
 A: You have mistaken somewhere, solution set is a space of all solutions, not a matrix. You just transformed given matrix, nothing more.... Matrix obviously have one column more than it needs to have, as third is a sum of first and second, so the maximum dimension is three. You need to prove can it go any further before continuing. Transformation of a matrix is important,, but it is not a solution set.
A: the $7$ and $-9$ in the third column of the reduced matrix tells you that the column $3$ of the original matrix is $7$ times column one minus $9$ times column $2.$ in other words, any multiple of $x=(7, -9, -1, 0)^T$ is solution of $Ax = 0.$
A: To address the expanded version of the question:

I have said that this is a hyperplane with one free variable, but I'm uncertain if this is the right answer. Might it be some sort of hyperline, since the last component is a constant and we only have one free variable, if that is even a thing?

The set of vectors of the form $(-7t,9t,t,0)$ is simply a line. To phrase it in purely geometric terms, it is the unique line that passes through both the origin and $(-7,9,1,0)$.
As you say, there is only one free variable. That means the dimension of the solution set is $1$. Vector spaces of dimension $1$ are called lines, and vector spaces of dimension $2$ are called planes. Vector spaces of codimension $1$, meaning dimension one less than the ambient space, are called hyperplanes. In the familiar context of $\mathbb R^3$, a plane is the same as a hyperplane. In the context of $\mathbb R^4$, a hyperplane has dimension $3$. In either case, it wouldn't be correct to call a line a hyperplane. "Hyperline" isn't a commonly used term.
A: The solution set is all $x$ such that $A$ sends $x$ to $0$, which is the kernel, or null space, of the linear map represented by $A$. A quick Google search will give you tons of information on how to interpret this space. For instance, if you get the dimension of the null space you can calculate the rank of $A$ using the Rank-Nullity theorem. 
