Line in Plane is Straight iff Gradient is Constant

Theorem
Let $\mathcal L$ be a curve which can be embedded in the plane.

Then $\mathcal L$ is a straight line it is of constant gradient.

Proof
Let $L$ be embedded in the cartesian plane.

The slope of $\mathcal L$ at a point $p = \tuple {x, y}$ is defined as being its derivative at $p$ $x$.

$\grad p = \dfrac {\d y} {\d x}$


 * Gradient-of-Straight-Line.png

Let $\mathcal L$ be a straight line.

Let $\triangle ABC$ and $\triangle DEF$ be right triangles constructed so that:


 * $A, B, D, E$ are on $\mathcal L$


 * $AC$ and $DF$ are parallel to the $x$-axis


 * $BC$ and $EF$ are parallel to the $y$-axis.

From Parallelism implies Equal Corresponding Angles:


 * $\angle ABC = \angle DEF$

and:
 * $\angle BAC = \angle EDF$

Also we have that $\angle ACB = \angle DFE$ and are right angles.

Thus $\triangle ABC$ and $\triangle DEF$ are similar.

Thus:
 * $\dfrac {BC} {AC} = \dfrac {EF} {DF}$

That is, the slope of $\mathcal L$ between $A$ and $B$ is the same as the slope of $\mathcal L$ between $D$ and $E$.

The argument reverses.