Given a line $l$ and two points $p_1$ and $p_2$, identify the point $v$ which is equidistant from $l$, $p_1$, and $p_2$, assuming it exists.
My idea is to: (1) identify the parabolas containing all points equidistant from each point and the line, then (2) intersect these parabolas. As $v$ is equidistant from all three and each parabola contains all points equidistant from $l$ and each point, the intersection of these parabolas must be $v$. However, I have had no luck in finding a way to compute, much less represent, these parabolas.
Best Answer
Assuming such point $Q$ exists it must lie on the Bisector Line b of $P_1$ and $P_2$ i.e. the line through the midpoint of $P_1$ and $P_2$ and orthogonal to the line $\vec {P_1P_2}$.
Thus you can write down a parametric expression for $Q=Q(s)\in b$ and set the following equation for distances:
Firstly note that, without loss of generality, we can assume that the origin coincides with the intersection point between $l$ and $b$. In fact if $b\equiv l$ solution is trivial otherwise if not we can find the intersection point and simply shift the axes (we could also rotate the axes in such way that $l$ or $b$ coincide with an axis but it should be more complex whereas shifting is trivial).
Thus, let's assume:
$P_1=(x_1,y_1), P_2=(x_2,y_2)$,
l: $ ax+by=0$,
b: $cx+dy=0$
NOTE
find c and d is trivial
Finding perpendicular bisector of the line segement joining $ (-1,4)\;\text{and}\;(3,-2)$
Parametric equation of $Q \in b$ is given by:
$Q(t)=Q(d \cdot t,-c \cdot t)$ with $t\in \mathbb{R}$
The distances are given by:
$$\text{d($Ql$)} = \frac{\left | Ax_{0} + By_{0} + C\right |}{\sqrt{A^2 + B^2} }= \frac {\left| ad\cdot t - bc\cdot t \right|}{\sqrt{a^2 + b^2}} $$
$$\text{d($QP_1$)} = \sqrt{(d\cdot t-x_1)^2 + (-c\cdot t-y_1)^2}$$
and thus
$$\frac {\left| adt - bct \right|}{\sqrt{a^2 + b^2}}=\sqrt{(d\cdot t-x_1)^2 + (-c\cdot t-y_1)^2}$$
$$\frac {\left( ad\cdot t - bc\cdot t \right)^2}{{a^2 + b^2}}=(d\cdot t-x_1)^2 + (-c\cdot t-y_1)^2$$
from which "t" and thus "Q" can be easily found.