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?


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.
