Reversal of Order of Vertices of Triangle causes Reversal of Sign of Area

Theorem

Let $\triangle ABC$ be a triangle embedded in the plane.

Let $\Area \triangle ABC = \AA$.

Then:

$\Area \triangle CBA = -\AA$.

Proof

$\triangle CBA$ is the same as $\triangle ABC$ but with its vertices in the reverse order.

We have that:

if $\triangle ABC$ is traversed anticlockwise going $AB \to BC \to CA$, then $\triangle CBA$ is traversed clockwise going $CB \to BA \to AC$

if $\triangle ABC$ is traversed clockwise going $AB \to BC \to CA$, then $\triangle CBA$ is traversed anticlockwise going $CB \to BA \to AC$.

The result follows by definition of sign of $\Area \triangle ABC$

$\blacksquare$

Sources

• 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text I$. Coordinates: $7$. Sign of an area