Equation of line in form of determinant

The equation of a line in complex plane is $az+b\bar{z}+c=0$. For a line passes through points $z_1$ and $z_2$ we cal find $a$, $b$ and $c$. so $$az_1+b\bar{z_1}+c=0\hspace{1cm}and\hspace{1cm}az_2+b\bar{z_2}+c=0$$ we find $a(z_1-z_2)=b(\bar{z_2}-\bar{z_1})=k$ thus $$a=\frac{k}{z_1-z_2},\,b=\frac{k}{\bar{z_2}-\bar{z_1}},\,c=\frac{k(\bar{z_1}z_2-z_1\bar{z_2})}{(z_1-z_2)(\bar{z_2}-\bar{z_1})}$$ thus $$\frac{k}{z_1-z_2}z+\frac{k}{\bar{z_2}-\bar{z_1}}\bar{z}+\frac{k(\bar{z_1}z_2-z_1\bar{z_2})}{(z_1-z_2)(\bar{z_2}-\bar{z_1})}=0$$ after deletion $k$ we have $$z(\bar{z_2}-\bar{z_1})+\bar{z}(z_1-z_2)+(\bar{z_1}z_2-z_1\bar{z_2})=0$$ or $$ -z\left|\begin{array}{rr}\bar{z_1}&1\\\bar{z_2}&1\end{array}\right| +\bar{z}\left|\begin{array}{rr}z_1&1\\z_2&1\end{array}\right| +\left|\begin{array}{rr}\bar{z_1}&z_1\\\bar{z_2}&z_2\end{array}\right| =0 $$ which can be write as $$\color{red}{ \left|\begin{array}{rrr} z&\bar{z}&1\\z_1&\bar{z_1}&1\\z_2&\bar{z_2}&1 \end{array}\right|=0} $$

Write the ordinary equation of a line through two points:

$$\Delta=\left|\begin{matrix}x&y&1\\x_1&y_1&1\\x_2&y_2&1\\\end{matrix}\right|=0.$$

You can multiply the second column by $i$ and add it to the first without invalidating the equality:

$$i\Delta=\left|\begin{matrix}x+iy&iy&1\\x_1+iy_1&iy_1&1\\x_2+iy_2&iy_2&1\\\end{matrix}\right|=0.$$

Then you can multiply the second by $-2$ and add the first:

$$-2i\Delta=\left|\begin{matrix}x+iy&x-iy&1\\x_1+iy_1&x_1-iy_1&1\\x_2+iy_2&x_2-iy_2&1\\\end{matrix}\right|=0.$$

The condition given is clearly equivalent to the alignment of $z=x+iy$ with $z_1=x_1+iy_1$ and $z_2=x_2+iy_2.$

Let us give the names $D$ and $E$ to the following determinants:

$$D=\left|\begin{array}{ccc} z&\bar{z}&1\\z_1&\bar{z_1}&1\\z_2&\bar{z_2}&1 \end{array}\right| \ \ \text{and} \ \ E=\begin{vmatrix}x&y&1\\x_1&y_1&1\\x_2&y_2&1\end{vmatrix}$$

Result: $$\tag{1}D=-2iE.$$

Proof of $(1)$.

$$D=\left|\begin{array}{ccc} x+iy&x-iy&1\\x_1+iy_1&x_1-iy_1&1\\x_2+iy_2&x_2-iy_2&1 \end{array}\right|=$$

can be expanded (multilinearity property of determinants) in the following way:

$$\tag{2}D=\underbrace{\left|\begin{array}{ccc} x&x&1\\x_1&x_1&1\\x_2&x_2&1 \end{array}\right|}_{ \ = \ 0}+ \underbrace{\left|\begin{array}{ccc} x&-iy&1\\x_1&-iy_1&1\\x_2&-iy_2&1 \end{array}\right|}_{= \ -iE}+\underbrace{\left|\begin{array}{ccc} iy&x&1\\iy_1&x_1&1\\iy_2&x_2&1 \end{array}\right|}_{= \ -iE}+\underbrace{\left|\begin{array}{ccc} iy&-iy&1\\iy_1&-iy_1&1\\iy_2&-iy_2&1 \end{array}\right|}_{= \ -i^2 \times 0}.$$

(The first and fourth determinants are zero because 2 columns are proportional).

ending the proof of $(1)$.

Now, what is the interest of relationship $(1)$ ?

$E=0$ is a classical alignment constraint (see (http://mathworld.wolfram.com/Collinear.html)).

Thus, $D=0$ is a as well an alignment constraint.