Line in Plane is Straight iff Gradient is Constant

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Theorem

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


Then $\mathcal L$ is a straight line if and only if 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 $p$ with respect to $x$.

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


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.

$\blacksquare$


Sources

For a video presentation of the contents of this page, visit the Khan Academy.