What would be the Mean shortest distance from random points in the right angled triangle to the Hypotenuse. The problem is to find the average shortest distance of uniformly random points from the hypotenuse in a right angled rectangle.
The distance d shows the shortest distance to the hypotenuse from a random point N (x1,y1). I want to find the average (mean) shortest distance from N random points inside the triangle.
What I have in mind is to integrate the distance formula of point distance to the hypotenuse.
Let P be (0,0) => Q (a,0) and R (a,b)
Then slope of hypotenuse will be $m = \frac{b-0}{a-0} = \frac{b}{a}$
The equation of hypotenuse be $bx -ay = 0$.
The distance d would be $$d = \frac{|b*x_1 -a*y_1|}{\sqrt{a^2 +b^2}}$$
I thought integrating it over x and y for the given range would provide me with the mean shortest distance -
$$D_{mean} = \int_{0}^{a} \int_{0}^{b} d.dx.dy = \int_{0}^{a} \int_{0}^{b} \frac{|b*x_1-a*y_1|}{\sqrt{a^2+b^2}} dx.dy$$
I am stuck in solving it... (given that my approach is the correct one.)


Take $P$ to be the origin.  Then the equation of the hypotenuse is
$ y = \dfrac{b}{a} x , \hspace{15pt} x \in [0, a] $
Now pick a point $ (x_1, y_1) $ such that $ y_1 \in [0, \dfrac{b}{a} x_1 ], then its perpendicular distance from the hypotenuse is
$d (x_1) = \dfrac{ \dfrac{b}{a} x_1 - y_1 } {\sqrt{1 + \left(\dfrac{b}{a}\right)^2} } = \dfrac{ b x_1 - a y_1 }{\sqrt{a^2 + b^2}}$
So the average distance is
$ \overline{d} = \displaystyle \dfrac{1}{A}  \int_{x_1=0}^{a} \int_{y_1 = 0 }^{\dfrac{b}{a} x_1} \dfrac{ b x_1 - a y_1}{\sqrt{a^2 + b^2}} dy_1 dx_1  $
where $A = \displaystyle \int_{x_1=0}^{a} \int_{y_1 = 0 }^{\dfrac{b}{a} x_1} dy_1 dx_1 = \dfrac{1}{2} a b  $
Integrating with respect to $y_1$,
$\overline{d} = \dfrac{2}{ab \sqrt{a^2 + b^2}} \displaystyle \int_{x_1=0}^{a} \dfrac{b^2}{2a} x_1^2 dx_1 = \left(\dfrac{b}{a^2 \sqrt{a^2 + b^2}}\right) \left(\dfrac{a^3}{3}\right)  =\dfrac{ab}{3 \sqrt{a^2 + b^2}} $
