Line in Plane is Straight iff Gradient is Constant

Theorem

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

Then $\LL$ 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 $\LL$ at a point $p = \tuple {x, y}$ is defined as being its derivative at $p$ with respect to $x$:

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

Let $\LL$ be a straight line.

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

$A, B, D, E$ are on $\LL$

$AC$ 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 $\LL$ between $A$ and $B$ is the same as the slope of $\LL$ between $D$ and $E$.

The argument reverses.

$\blacksquare$

Sources

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