Equation of Straight Line in Plane/Two-Point Form/Proof 4

Theorem

Let $P_1 := \tuple {x_1, y_1}$ and $P_2 := \tuple {x_2, y_2}$ be points in a cartesian plane.

Let $\LL$ be the straight line passing through $P_1$ and $P_2$.

Then $\LL$ can be described by the equation:

$\dfrac {y - y_1} {x - x_1} = \dfrac {y_2 - y_1} {x_2 - x_1}$

or:

$\dfrac {x - x_1} {x_2 - x_1} = \dfrac {y - y_1} {y_2 - y_1}$

Proof

Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$ be the position vectors of the points $A$ and $B$ embedded in the complex plane.

Let $z = x + i y$ be the position vector of an arbitrary point $P$ on the straight line $AB$.

From the diagram:

\(\ds OA + AP\)

\(=\)

\(\ds OP\)

\(\ds \leadsto \ \ \)

\(\ds z_1 + AP\)

\(=\)

\(\ds z\)

\(\ds \leadsto \ \ \)

\(\ds AP\)

\(=\)

\(\ds z - z_1\)

and:

\(\ds OA + AB\)

\(=\)

\(\ds OB\)

\(\ds \leadsto \ \ \)

\(\ds z_1 + AB\)

\(=\)

\(\ds z_2\)

\(\ds \leadsto \ \ \)

\(\ds AB\)

\(=\)

\(\ds z_2 - z_1\)

Then:

\(\ds \exists t \in \R: \, \)

\(\ds AP\)

\(=\)

\(\ds t AB\)

as $A$, $P$ and $B$ are collinear

\(\ds \leadsto \ \ \)

\(\ds z - z_1\)

\(=\)

\(\ds t \paren {z_2 - z_1}\)

\(\ds \leadsto \ \ \)

\(\ds z\)

\(=\)

\(\ds \paren {1 - t} z_1 + t z_2\)

\(\ds \leadsto \ \ \)

\(\ds x - x_1\)

\(=\)

\(\ds t \paren {x_2 - x_1}\)

\(\ds y - y_1\)

\(=\)

\(\ds t \paren {y_2 - y_1}\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac {x - x_1} {x_2 - x_1}\)

\(=\)

\(\ds \dfrac {y - y_1} {y_2 - y_1}\)

$\blacksquare$

Sources

• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $11$