Question

If the points $\left(-2,0\right)$, $\left(-1,1/\sqrt{3}\right)$ and $\left(cos\theta ,sin\theta \right)$ are collinear, then the number of values of $\theta \in \left[0,2\pi \right]$ is

• A. $0$
• B. $1$
• C. $2$
• D. None of these

Answer 1

Answer: (B)

Let points are $A\left(-2,0\right)$, $B\left(-1,\frac{1}{\sqrt{3}}\right)$ and $C\left(cos\theta ,sin\theta \right)$
$\because$ The given points are collinear
$\Rightarrow$ Slope of $AB=$ Slope of $BC$
i.e., $\frac{\frac{1}{\sqrt{3}}-0}{-1-\left(-2\right)}=\frac{sin\theta -\frac{1}{\sqrt{3}}}{cos\theta -\left(-1\right)}$
$\Rightarrow \frac{1}{1}=\frac{\sqrt{3}sin\theta -1}{cos\theta +1}$
$\Rightarrow \sqrt{3}sin\theta -cos\theta =2$
$\Rightarrow \frac{\sqrt{3}}{2}sin\theta -\frac{1}{2}cos\theta =1$
$\Rightarrow sin\left(\theta -\frac{\pi }{6}\right)=1$
$\Rightarrow sin\left(\theta -\frac{\pi }{6}\right)=sin\left(\frac{\pi }{2}\right)$
$\Rightarrow \theta =n\pi +{\left(-1\right)}^n\frac{\pi }{2}+\frac{\pi }{6}\forall n\in Z$