Question 825317
This has four different possible solutions.  Try using (0,1) and (0,6) to be the segment of one of the legs.  Another point may be (a, 6), for {{{a>0}}}.  Draw this so you see it.


The lengths of the two legs are:

5, and a.


The length of the hypotenuse is:

{{{sqrt((6-1)^2+(a-0)^2)}}}
{{{sqrt(25+a^2)}}}


Can we find a without using the hypotenuse?
Area formula gives {{{(1/2)a*5=10}}};
{{{a=(2/1)(1/5)10=4}}}.


That one possible point is therefore, (4, 6).