Find the coordinate of two line segments that are perpendicular How can the x,y position (actually will need in 3d but for simplicity asking in 2d) of a point be found if it is the intersection of two perpendicular line segments. 
I have two points, $p_1$ and $p_2$ that are known $x,y$ locations.  I also have line segments with known lengths $a$ and $b$.  I made a diagram below to illustrate the problem. 

 A: Sometimes a figure is worth a 1000 words:

In three dimensions:

A: If you draw a circle with diameter $\sqrt{a^2+b^2}$ then $p_3$ can be either of two points on that circle.
In two dimensions...
Let $\vec i$ be the unit vector in the direction of the 
vector $\overrightarrow{p_1p_2}=\dfrac{p_2-p_1}{|p_2-p_1|}$ Let $\vec j$ be a unit vector perpendicular to $\vec i$. (If $\vec i = \langle u,v\rangle$, then, for example, $\vec j = \langle-v, u\rangle$.) Then 
$$\overrightarrow{p_1p_3} = 
   \dfrac{a^2}{\sqrt{a^2+b^2}} \ \vec i \pm
   \dfrac{ab}{\sqrt{a^2+b^2}} \ \vec j$$
See here for details.
In three dimesions, there will be a circle of possible points.
$$\overrightarrow{p_1p_3} = 
   \dfrac{a^2}{\sqrt{a^2+b^2}} \ \vec i +
   \dfrac{ab}{\sqrt{a^2+b^2}}\cos(\theta) \ \vec j + 
   \dfrac{ab}{\sqrt{a^2+b^2}}\sin(\theta) \ \vec k$$
A: *

*Let $\ell$ be the distance between $p_1$ and $p_2$, and let $p_3$ be the point where the two line segments will meet. 

*The three points $p_1, p_2, p_3$ form a triangle whose side lengths we know: $a, b, \ell$.

*We can find out where $p_3$ is by measuring properties of this triangle.  Draw a perpendicular altitude from $p_3$ down to the side of the triangle between $p_1$ and $p_2$.  Call the intersection point $p_4$.

*If we knew the length $h$ of this altitude, we would know where to put $p_3$. First, we could find out where $p_4$ is.  By the Pythagorean theorem, the distance between $p_1$ and $p_4$ is $\sqrt{a^2 - h^2}$. So
$$p_4 = p_1 + \frac{\sqrt{a^2-h^2}}{\ell} (p_2 - p_1)$$
Next, we would use $p_4$ to find out where to put $p_3$.  We start from $p_4$ and travel a distance $h$ in a direction perpendicular to the $p_1$ $p_2$ base of the triangle.  
In two dimensions, we have two choices of perpendicular direction. In three dimensions, we actually have an infinite number of perpendicular directions[*], so you'll have to pick one. Pick a unit vector $\widehat{u}$ in a perpendicular direction.
Then $p_3 = p_4 + h\cdot \widehat{u}$, which gives you the answer you want.

*But how do we find $h$? We can find $h$ if we know the area of the triangle. Heron's formula lets you calculate the area of the triangle when you know the length of all the sides. $A = \sqrt{s(s-a)(s-b)(s-\ell)}$, where $s$ is half the perimeter $s = (a+b+\ell)/2$. Once you have the area, you can calculate the height of the altitude: $A = \frac{1}{2}\text{base}\times\text{height}$, so $h = 2A/\ell$.
[*] In 3D, you have an infinite number of perpendicular directions. Imagine $p_1=\langle 0,0\rangle$ and $p_2 = \langle 1,0\rangle$, and $a$ and $b$ are some numbers. There's a satisfactory point $p_3$ in the x-y plane. But if you spin $p_3$ around the $x$-axis, you find an infinite number of other satisfactory points too.
A: As I mentioned in my comment, there is no solution unless $a^2+b^2=\lvert P_1P_2\rvert^2$. Let $Q$ be the point $(x_3,y_3)$, $R$ the foot of the perpendicular from $Q$ to $\overline{P_1P_2}$, and $c = \lvert P_1P_2\rvert$. $\triangle{P_1QP_2}$, $\triangle{P_1RQ}$ and $\triangle{QRP_2}$ are all similar, therefore $\lvert P_1R\rvert = a^2/c$, which means that $$R = \left(1-{a^2\over c^2}\right)P_1+{a^2\over c^2} P_2.$$ Similarly, $\lvert QR\rvert = (ab)/c$ and a unit vector perpendicular to $\overline{P_1P_2}$ is $\frac1c(y_1-y_2,x_2-x_1)$, therefore the two solutions are $$Q=R\pm{ab\over c^2}(y_1-y_2,x_2-x_1).$$ 
In three dimensions, there is an infinite number of solutions that lie on a circle of radius $(ab)/c$ centered at $R$ and in a plane perpendicular to $\overline{P_1P_2}$.
