How to find the third coordinate of a right triangle given 2 coordinates and lengths of each side  p2
 |\
 |b\
 |  \
A|   \C
 |    \
 |c___a\
p1  B   p3

If given point p1 & p2, side A & B how would you find point p3? I know given this information you can find side C and all of the interior angles.
side C:
C^2 = A^2 + B^2

angle c = 90
angle a = A/SIN(a) = C/SIN(c)
angle b = 180 - (a+c)

But after this, I am trying to find point p3 and I am not sure what direction to take. Any help would be appreciated.
Edit: The triangle will not necessarily be facing upwards along an axis, it will be rotated at angles depending on exterior variables such as position of a mouse on the computer screen.
 A: Let the coordinates of $p_n$ be $(x_n,y_n)$.  Then the slope of $A$ is $m_A=\frac{y_2-y_1}{x_2-x_1}$.  The slope of $B$ is $m_B=\frac{-1}{m_A}=\frac{x_1-x_2}{y_2-y_1}$.  Then $p_3=p_1\pm B(\frac{1}{\sqrt{1+m_B^2}},\frac{m_B}{\sqrt{1+m_B^2}})$ where the sign ambiguity corresponds to two orientations of the triangle.  I have ignored issues when the sides are vertical or horizontal, which can lead to division by zero
A: If your triangle is in space, the given information doesn't determine it yet, because given any such triangle, rotating it around the line joining $p1$ and $p2$ gives you another valid triangle. If your triangle is in the plane, then the information does determine p3 as long as you decide whether the order $p1$, $p2$, $p3$ is clockwise or counterclockwise. Let's say then, the triangle is in the plane and, as shown in your neat ASCII picture, $p1$, $p2$, $p3$ is clockwise.
If you know about complex numbers, then $p3-p1 = -iB/A(p2-p1)$ (because multiplication by $B/A$ changes a length $A$ vector into a length $B$ vector, and multiplication by $-i$ rotates by $90^\circ$ clockwise).
If you don't want to use complex numbers, then say $p1=(x1, y1)$, $p2=(x2, y2)$ and $p3=(x3, y3)$. Since $p1p2p3$ is a right angle, the slopes of $p1p2$ and $p1p3$ multiply to $-1$, that gives you one equation for $x3$ and $y3$. Additionally, you know that the distance from $p1$ to $p3$ is B, which gives you a second equation for $x3$ and $y3$. The system formed by those equations will have two solutions, one corresponding to $p1$, $p2$, $p3$ being clockwise and the other corresponding to counterclockwise.
