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., $\frac{\frac{1}{\sqrt{3}}-0}{-1-\left(-2\right)}=\frac{sin\,\theta-\frac{1}{\sqrt{3}}}{cos\,\theta-\left(-1\right)}$
$\Rightarrow \frac{1}{1}=\frac{\sqrt{3}\,sin\,\theta-1}{cos\,\theta+1}$
$\Rightarrow \sqrt{3}\,sin\,\theta-cos\,\theta=2$
$\Rightarrow \frac{\sqrt{3}}{2}sin\,\theta-\frac{1}{2}\,cos\,\theta=1$
$\Rightarrow sin\left(\theta-\frac{\pi}{6}\right)=1$
$\Rightarrow sin\left(\theta-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{2}\right)$
$\Rightarrow \theta=n\pi+\left(-1\right)^{n} \frac{\pi}{2}+\frac{\pi}{6}\,\forall\,n \in Z$