Magnetic field of off center triangle - Biot Savart Law

[asadpasat]

PICTURE INCLUDED
1. Homework Statement
A piece of wire is bent into an isosceles right triangle whose shorter sides have length a The wire carries current I. Calculate the magnetic field for point P. Point P is located on the Y-axis ( 0, √2a). Two corners of the triangle are are located at (0,0), (√2a,0).

Homework Equations

I have to use the definition of the magnetic field for each side.
dB = (u/4π) (Ids x r(unit vector))/r^2

The Attempt at a Solution

My problem is more conceptual. Can I split the triangle into three different parts and have different coordinate systems for them? I was thinking of setting one side to be along x axis, and rotate the other side. Pic included. Can I add the magnetic field for each side if they had different coordinate systems?

 

[kuruman]

All segments produce fields in the same direction, into the screen. One coordinate system is sufficient for all. As you pointed out, one segment contributes nothing. Note that both of the other two segments produce fields that are at a perpendicular distance from their end that is equal to their length. One length is $a$ and the other $\sqrt{2}a$. So basically you need to find the field from the end of a segment of distance L at perpendicular distance L from its end, then add two contributions one with $L=a$ and one with $L=\sqrt{2}a$.