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 $(3,1)$?

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 $$\frac{\text{Required Area}}{\text{Total Area}}=\frac{1/2\times1/3\times 1+4/3+1/3}{2}=\frac{11}{12}$$
