Answer
$y = \frac{\sqrt{3}~x}{3}$ and $x \leq 0$
Work Step by Step
A bearing of $240^{\circ}$ makes an angle of $30^{\circ}$ below the west axis.
$tan~30^{\circ} = \frac{1}{\sqrt{3}}$
$\frac{y}{x} = \frac{1}{\sqrt{3}}$
Then: $y = \frac{\sqrt{3}~x}{3}$ and $x \leq 0$
