Show that the equation of a line passing through points z_1…
Show that the equation of a line passing through points $z_1$ and $z_2$ has the form $$ \begin{vmatrix} z & \bar z & 1 \\ z_1 & \bar z_1 & 1 \\ z_2 & \bar z_2 & 1 \end{vmatrix} = 0. $$
Explanation of Method
Expand the determinant along the first row to get a linear equation of the form $Az+\bar A\bar z+C=0$ with $C$ purely imaginary — exactly the general equation of a line in the complex plane. Then check that $z=z_1$ and $z=z_2$ satisfy the equation (their rows repeat), so this line passes through both points, and since two distinct points determine a unique line, this must be it.
Step 1 · Expand along the first row
$$ \begin{vmatrix} z & \bar z & 1 \\ z_1 & \bar z_1 & 1 \\ z_2 & \bar z_2 & 1 \end{vmatrix} = z(\bar z_1-\bar z_2) - \bar z(z_1-z_2) + (z_1\bar z_2-z_2\bar z_1) $$
Step 2 · Write in the standard form $Az-\bar A\bar z+C=0$
Let $A=\bar z_1-\bar z_2$, so that $\bar A=z_1-z_2$, and let $C=z_1\bar z_2-z_2\bar z_1$. The expanded determinant from Step 1 reads exactly:
$$ Az - \bar A\,\bar z + C = 0 $$
Moreover $C$ is purely imaginary, since:
$$ \bar C = \overline{z_1\bar z_2-z_2\bar z_1} = \bar z_1 z_2-\bar z_2 z_1 = -C $$
An equation of the form $Az-\bar A\bar z+C=0$ with $C$ purely imaginary is precisely the general equation of a straight line in the complex plane (writing $z=x+iy$ turns it into a real linear equation in $x,y$).
Step 3 · $z_1$ and $z_2$ satisfy the equation
Substituting $z=z_1$ makes the first and second rows of the determinant identical, so the determinant is $0$. Substituting $z=z_2$ makes the first and third rows identical, again giving $0$. So both $z_1$ and $z_2$ lie on the locus defined by the equation.
Step 4 · Uniqueness
The equation is linear in $z,\bar z$ (equivalently linear in $x,y$ after substituting $z=x+iy$), hence its solution set is a line (or the whole plane / empty set in degenerate cases, which don't occur here since $z_1\neq z_2$ makes $A=\bar z_1-\bar z_2\neq0$). Since a line is determined uniquely by two distinct points, and both $z_1,z_2$ satisfy this equation, it must be the line through $z_1$ and $z_2$.
The determinant vanishes exactly when $z$ is collinear with $z_1$ and $z_2$.
Step 5 · Worked example
Let $z_1=1$, $z_2=i$. The determinant equation expands to $z(1+i)-\bar z(1-i)-2i=0$. Writing $z=x+iy$:
$$ (x+iy)(1+i)-(x-iy)(1-i)-2i=0 \quad\Longrightarrow\quad -2y+2ix-2i=0 \quad\Longrightarrow\quad x-1=y $$
which is the line $y=x-1$, correctly passing through $(1,0)$ and $(0,1)$.
Result
is the equation of the unique line through $z_1$ and $z_2$: it expands to a linear relation $Az-\bar A\bar z+C=0$ with $C$ purely imaginary (the defining form of a line), and both $z_1,z_2$ satisfy it by construction.