How to Translate two Equations for a "+/-" For a National Board Exam Review:

Find the Equation for the Asymptotes of a Hyperbola ${ (y-x)^2 - (x+5)^2 = 36 }$

Answer is ${ y-5 = \pm (x+5) }$
I've already solved the equations: here they are:
$${ y = x+10 }$$
$${ y = -x }$$
My problem is how to translate it into this " ${ y-5 = \pm (x+5) }$ " ? I know that if you reverse engineer the equation by doing seperate equations for each you could end up with my answer... But I want to know if there is a method for methodically translating my answer to the one with the ${\pm}$ sign on it... 
 A: You already got the geometric answer, here is the algebraic one:
You have the equations:
$$y=ax + b\\
y=-ax - c$$
You can rearange the second one into:
$$y=-(ax + \frac{b+c}{2} + c - \frac{b+c}{2})$$
Which rearanges into $$y-\frac{b-c}{2} = -(ax+\frac{b+c}{2})$$
Similarly, the second equation rearanges from
$$y=ax + \frac{b+c}{2} + b - \frac{b+c}{2}$$
Into $$y-\frac{b-c}{2} = ax + \frac{b+c}{2}$$
A: Just draw the lines in a coordinate system; they will intersect at $(x,y)=(-5,5)$, so in terms of the translated coordinate system centered at that point, $(u,v)=(x+5,y-5)$, their equations will be $u = \pm v$.
A: Using
$$b=\frac{b+c}{2}+\frac{b-c}{2}\quad\text{and}\quad c=\frac{b+c}{2}-\frac{b-c}{2},$$
we can see that
$$y=ax+b$$
$$y=-ax+c$$
can be written as
$$y=ax+\frac{b+c}{2}+\frac{b-c}{2}\Rightarrow y-\frac{b+c}{2}=ax+\frac{b-c}{2}$$
$$y=-ax+\frac{b+c}{2}-\frac{b-c}{2}\Rightarrow y-\frac{b+c}{2}=-ax-\frac{b-c}{2},$$
i.e.
$$y-\frac{b+c}{2}=\pm\left(ax+\frac{b-c}{2}\right)$$
Your case is $(a,b,c)=(1,10,0)$.
