Answer
$y = -\sqrt{3}~x$ and $x \geq 0$
Work Step by Step
A bearing of $150^{\circ}$ makes an angle of $60^{\circ}$ below the east axis.
$tan~(-60^{\circ}) = -\frac{\sqrt{3}}{1}$
$\frac{y}{x} = -\frac{\sqrt{3}}{1}$
Then: $y = -\sqrt{3}~x$ and $x \geq 0$
