Distribution of a sample of uniformly distributed points in the 2D Let there be a rectangle in the plane and a set of points distributed in the rectangle by a uniform distribution. I select a random point on the top and right border and draw the red line. The blue triangle is formed after drawing the line. I would like to know if the distribution of points inside the blue triangle is still uniform (with different parameters)? Or it is changed to something else?

 A: The picture is very nice.  The evenness of the blue color is an accurate visual metaphor for the evenness of the probability represented by the color.  A uniform distribution has perfectly even probability, so even when you cut out ("truncate") a piece of the sample space, the probability remains even--and therefore is uniform.

You seem to want a rigorous solution, too, so here goes.  We need two definitions: uniform probability and truncation.  After that, the demonstration is short and simple.
Uniform probability
In any space $S$ where we can measure the "area" or, more generally, the "size" $|\mathcal R|$ of regions $\mathcal R\subset S,$ a probability distribution $\mathbb P$ defined on $S$ is said to be uniform when probabilities are proportional to size.  Because the probability of $S$ itself must be $1,$ we deduce
$$\mathbb{P}(\mathcal R) = \frac{|\mathcal R|}{|\mathcal S|}.$$
(Notice this implies $S$ has finite size!)
Truncation
Your act of slicing off a portion of the rectangle $S$ amounts to fixing a subregion $\mathcal T\subset S$ (the triangle).  Doing that is called truncation (from the Latin for "cutting off").  No matter what the original probability distribution might have been (uniform or not), it determines a new probability distribution on $\mathcal T:$ all you have to do is normalize the original one to make the total probability of $\mathcal T$ equal to $1.$  Therefore, when $\mathcal R \subset \mathcal T,$ the new probability is
$$\mathbb{P}_{\mathcal T}(\mathcal R) = \frac{\mathbb P(\mathcal R)}{\mathbb P(\mathcal T)}.$$
(Notice this can be defined only when $\mathcal T$ has nonzero probability!)
Solution
With these definitions in place, the solution amounts to a basic fact of arithmetic.  When $\mathbb P$ is uniform and $\mathcal R\subset \mathcal T,$ we plug the first equation (uniform probability) into the second (truncation) to find
$$\mathbb{P}_{\mathcal T}(\mathcal R) =  \frac{\mathbb P(\mathcal R)}{\mathbb P(\mathcal T)} = \frac{|\mathcal R|/|\mathcal S|}{|\mathcal T|/|\mathcal S|} = \frac{|\mathcal R|}{|\mathcal T|},$$
because the common nonzero factors of $1/|S|$ in numerator and denominator cancel.  But the resulting formula is exactly the uniform distribution on $\mathcal T,$ which is what we needed to show.
