Condition for Collinearity of Points in Complex Plane/Formulation 1

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Theorem

Let $z_1$, $z_2$ and $z_3$ be points in the complex plane.

Then $z_1$, $z_2$ and $z_3$ are collinear if and only if:

$\dfrac {z_1 - z_3} {z_3 - z_2} = \lambda$

where $\lambda \in \R$ is a real number.


If this is the case, then $z_3$ divides the line segment in the ratio $\lambda$.

If $\lambda > 0$ then $z_3$ is between $z_1$ and $z_2$, and if $\lambda < 0$ then $z_3$ is outside the line segment joining $z_1$ to $z_2$.


Proof

By Geometrical Interpretation of Complex Subtraction:

$z_1 - z_3$ can be represented as the line segment from $z_3$ to $z_1$
$z_3 - z_2$ can be represented as the line segment from $z_2$ to $z_3$.

Thus we have that $z_1$, $z_2$ and $z_3$ are collinear if and only if $z_1 - z_3$ is parallel to $z_3 - z_2$, when expressed as line segments.


Let $\dfrac {z_1 - z_3} {z_3 - z_2} = \lambda$ for $\lambda \in \C$.

That is:

$z_1 - z_3 = \lambda \paren {z_3 - z_2}$


By Complex Multiplication as Geometrical Transformation‎:

$\map \arg {z_1 - z_3} = \map \arg {z_3 - z_2} \iff \map \arg {z_3 - z_2} \arg \lambda \in \R_{>0}$

and:

$\map \arg {z_1 - z_3} = -\map \arg {z_3 - z_2} \iff \map \arg {z_3 - z_2} \arg \lambda \in \R_{<0}$

where $\arg$ denotes the argument of a complex number.


Also by Complex Multiplication as Geometrical Transformation‎, $z_1 - z_3$ is $\lambda$ the length of $z_3 - z_2$.

The result follows.

$\blacksquare$


Sources