Is it possible to define a hyperplane as all points that are equidistant of two initial points? Generally, as I understand it, it is neccesary to define a plane by the same number of points as  the dimension of the plane. However, an exception to this is defining a plane in 3 dimensions as all the points that are equidistant between two initial points. I.e. a plane that is perpendicular to the line between the two points and passes the midpoint between them.
My question is, does this hold true in higher dimensions? Could you define a 4D plane based on all points equidistant between 2 points? Would you need 3? I can picture it working in 2D and 3D but can't picture how it would work in 4D or ND. The use case is working out how many data points minimum I need to generate a full constrained hyperplane for a support vector machine.
edit 1: Corrected spelling
 A: Let's solve the problem. If we consider two points $p,q\in\mathbb R^n$ and now we look for the points equidistant to these two, we have
$$d(p,x)=d(q,x)\Rightarrow d(p,x)^2=d(q,x)^2$$
$$(p_1-x_1)^2+\cdots+(p_n-x_n)^2=(q_1-x_1)^2+\cdots+(q_n-x_n)^2$$
Expanding the squares and cancelling the terms $x_i^2$ at both sides, we get
$$p_1^2-2x_1p_1+\cdots+p_n^2-2x_np_n=q_1^2-2x_1q_1+\cdots+q_n^2-2x_nq_n$$
$$2(q_1-p_1)x_1+\cdots+2(q_n-p_n)x_n=d(q,0)^2-d(p,0)^2$$
Which is a linear expression that defines a hyperplane. Then for any dimension $D$, the bisector of two points defines a hyperplane.
To complete the answer, I will say that in $D$ dimensions $D$ points are necessary to define a single hyperplane that passes through them all. If you have only $n$ points, there are $D-n$ extra degrees of freedom to select a hyperplane. From this we can see that the need for the points to be equidistant to all points in a hyperplane is much more stronger than the need for a hyperplane to contain the points.
