If $\left |z-3\right |=\left |z+i\right |$, where $z=x+iy$, prove that $3x+y=4$ 
If $\left |z-3 \right |=\left |z+i\right |$, where $z=x+iy$, prove that $3x+y=4$.

I have got to the point where I have $\left |z \right |= \sqrt{x^2+(y+1)^2} = \sqrt{(x-3)^2+y^2}$
But really don't know where to go from here or if my start is anywhere near right :(
Can someone help me.

$|z-3|=\sqrt {(x-3)^2+y^2}$
$|z-i|= \sqrt {x^2+(y+1)^2}$
Now , 
$(x-3)^2+y^2=x^2+(y+1)^2$
$x^2+9-6x+y^2=x^2+y^2+2y+1$
$8=6x+2y$
