Question: Consider, the equation 

$ |z+i| = |z-i|$ 

Now the interpretation of this equation is that it is the locus of points which are the perpendicular bisector of the line connecting $i$ and $-i$. But how would we justify this reason/ motivate it?

Answer: 
If $|z-a|=|z-b|$ then the distance from $z$ to $a$ is equal to the distance from $z$ to $b$. Therefore the triangle with vertices $z,a,b$ is isosceles, so the line from $z$ to the mid-point between $a$ and $b$ is perpendicular to the line from $a$ to $b$. Conversely, any point on that line is equidistant from $a$ and $b$.
In general, $|z-a|=|z-b|$ iff $z$ is on the perpendicular bisector of the line segment with endpoints $a,\,b$. To see this, note first that $|z-w|$ is a translationally and rotationally invariant distance in the complex plane, and the resulting distances' ratios are also invariant under scaling. So without loss of generality, assume $a=-1,\,b=1,\,z=x+iy,\,x,\,y\in\Bbb R$ , whence$$|z-a|^2=(x+1)^2+y^2=x^2+y^2+1+2x,$$and similarly $|z-b^2|=x^2+y^2+1-2x$. These are equal iff $x=0$, i.e. $z$ is on the bisector.