I want, by the use of the equation of the line in complex plane, to find the slope and x intercept in x-y plane. Attempt:
$$|z-a|=|z-b|$$ $$y=mx+h$$ $$m=m(ax,ay,bx,by) \quad h=h(ax,ay,bx,by)$$ $$a=ax+iay \quad b=bx+byi$$
How to find $m$ and $h$ here?
You start with $$ |z-a| = |z-b| \iff |z-a|^2 = |z-b|^2 $$ write the squared modulus as $|w|^2 = w\bar w$: $$ (z-a)(\bar z - \bar a) - (z-b) (\bar z-\bar b) = 0 $$ expand $$ z\bar z - a \bar z -\bar a z + a \bar a - z\bar z + b \bar z +\bar b z - b \bar b = 0 $$ cancel the quadratic term and collect $$ (b-a)\bar z +(\bar b - \bar a)z + |a|^2 - |b|^2 = 0 $$ now notice that $2 \Re w = w + \bar w$, so $$ 2\Re ((\bar b-\bar a)z) + |a|^2 - |b|^2 = 0 $$ and by letting $$ a=a_x + i a_y, \qquad b = b_x + i b_y, \qquad z = x + iy $$ you get $$ 2(b_x-a_x) x - 2(b_y-a_y)y + |a|^2 - |b|^2 = 0 $$ and finally, $$ y = \frac{b_x-a_x}{b_y-a_y} x + \frac{|a|^2-|b|^2}{2(b_x-a_x)}. $$