Probability of point inside of rectangle $(0,0),(2,0),(2,1),(0,1)$ closer to $(0,0)$ than $(3,1)$. A point $P$ is randomly selected from the rectangular region with vertices $(0,0)$, $(2,0)$, $(2,1)$, $(0,1)$. What is the probability that $P$ is closer to the origin than it is to the point $(3,1)$?
 A: Visual hint, this is an exercise that can be solved by basic geometry:

A: Call the vertices of the rectangle $A,B,C,D$ in the order in which you have listed them, and let $E=(3,1)$.  Let $M$ be the midpoint of $AE$, and let $P,Q$ be the points at which the perpendicular to $AE$ through $M$ intersects the sides of the rectangle $CD$ and $AB$ respectively.  It is clear that $P$ is in fact on $CD$; and $Q$ is on $AB$ since the distance $AB$ is greater than $BE$.  The area of $AQPD$ is the height $AD$ times the perpendicular distance from $AD$ to $M$, which is $1.5$.  So the probability is $\frac{1.5}2=\frac34$.
A: Distance from $(0,0) \le$Distance from $(3,1)$
i.e. let the point be $(x,y)$ then
$$\sqrt{x^2+y^2}\le \sqrt{(x-3)^2+(y-1)^2}$$
Squaring both sides canceling out returns following:-
$$y\le5-3x \tag{1}$$
The point should be on the left side of line $5-3x$

So we calculate the area bounded by inequality $(1)$ and rectangle and divide it by total area.
Line $5-3x$ cuts rectangle at $(\frac{4}{3},1)$ and $(\frac{5}{3},0)$. So $$\frac{\text{Required Area}}{\text{Total Area}}=\frac{1/2\times1/3\times 1+4/3+1/3}{2}=\frac{11}{12}$$
A: Using the visual clue, the line connecting $(0,0)$ and $(3,1)$ has slope $\frac 13$ so the perpendicular bisector has slope $-3$ and passes through the point $(\frac 32,\frac12)$. 
The the equation of the perpendicular bisector is $y=-3x+5$. So the perpendicular bisector intersects the rectangle at $(0,\frac 53)$ and $(1,\frac 43)$. 
The trapezoid of points that contain all the points closer to the $(0,0)$ has area $\frac 32$. So the probability is $\frac 34$.
