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

Question

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.

Answer 1

$|z-3|=\sqrt {(x-3)^2+y^2}$

$|z-i|= \sqrt {x^2+(y+1)^2}$

Now ,

$(x-3)^2+y^2=x^2+(y+1)^2$

$x^2+9-6x+y^2=x^2+y^2+2y+1$

$8=6x+2y$

Answer 2

You can also look at the problem geometrically. $\left|z-3\right|=\left|z+i\right|$ is the set of points that are equidistant from $3$ and $-i$ in the complex plane. This is a line that passes through the mid point of $(3,0)$, $(0, -1)$ and is perpendicular to the line that passes through these two points. Find the equation of the line to get the desired result.