Given 3 random points, what is the probability of these two situations involving a perpendicular bisector and distances? Suppose we're given 3 random points $p_0=(x_0,y_0),p_1=(x_1,y_1),p_2=(x_2,y_2)$ from a two-dimensional continuous uniform distribution $\{U(a,b)\}^2$, for some $(a\in\mathbb{R})\lt (b\in\mathbb{R})$, and it is known that $y_0\lt y_1\lt y_2$.
Also, let $B(x):\mathbb{R}\rightarrow\mathbb{R}$ be the function which graphs the perpendicular bisector of $p_0$ and $p_1$, such that $(x,B(x))$ intercepts the perpendicular bisector.
Finally, let $p_b=(x_2,B(x_2))$, let $d_1=dist(p_1,p_b)$, and let $d_2=dist(p_2,p_b)$, where $dist(\alpha,\beta):\mathbb{R^2,R^2}\rightarrow\mathbb{R}$ is the Euclidean distance between the points $\alpha$ and $\beta$.


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*Assuming $x_2\lt x_1$, what is the probability that $d_2\lt d_1$?

*Assuming $x_2\gt x_1$, what is the probability that $d_2\gt d_1$?



As a side note, in case anyone can point me to a source which may have information related to what I'm looking for, these questions are part of my attempting to do some analysis and optimization of Fortune's algorithm, which used to generate Voronoi diagrams, given a set of points.


First, let's find the perpendicular bisector of the line $p_0p_1$.
We'll call the midpoint of $p_0p_1$ $M$, and its slope $s$.  Also, $s'$ will be the slope of the bisector, $s' = -1/s$.
$$y = s'(x - M_x) + M_y = -\frac{x - M_x}{s} + M_y$$
Now, let's find the location of a $p_2$ that happens to be exactly the same distance from $p_b$ as $p_1$ is (it'll also be the same distance as $p_0$, since $p_b$ is on the bisector).
$$y_2 = y_b - \sqrt{(y_b - y_1)^2+(x_b-x_1)^2}$$
But $y_b$ is on the line, so we can find its location based on $x_b$, and actually $x_b = x_2$, so
$$y_2 = s'(x_2-M_x)+M_y - \sqrt{(s'(x_2-M_x)+M_y - y_1)^2+(x_2-x_1)^2}$$
Everything below this has $d_2 > d_1$.
From here you'll have to integrate to find the probabilities you need.  I'm not entirely sure if that is nicely integrable though.
