Answer:B
Step-by-step explanation:
Given
[tex]r=-3\theta [/tex]
so [tex]x=-3\theta \cos \theta \quad \ldots(i)[/tex]
and
[tex]y=-3\theta \sin \theta \quad \ldots(ii)[/tex]
squaring and adding [tex](i)[/tex] and [tex](ii)[/tex] we get
[tex]x^2+y^2=(-3\theta \cos \theta)^2+(-3\theta \sin \theta)^2[/tex]
[tex]x^2+y^2=r^2=9\theta ^2\cos ^2 \theta +9\theta ^2\sin ^2 \theta [/tex]
[tex]r^2=9\theta ^2[/tex]
[tex]r=-3\theta [/tex]
for [tex]r[/tex] to be negative both [tex]x[/tex] and [tex]y[/tex] must be negative
