Question

If the points $(-2,0)$, $(-1,\frac{1}{\sqrt{3}})$, and $(\cos \theta ,\sin \theta )$ are collinear, then the number of values of $\theta$ when $0\le \theta \le \frac{\pi }{2}$ is?

• A. $0$
• B. $1$
• C. $2$
• D. Infinite

Answer 1

The correct option is A $0$

Let us consider points $A(-2,0),B(-1,\frac{1}{\sqrt{3}})$ and $C(\cos \theta ,\sin \theta )$
For the points to be collinear $\text{slope of}AB=\text{slope of}BC$

$\text{slope of}AB=\frac{-1+2}{\frac{1}{\sqrt{3}}-0}=\sqrt{3}$

$\text{slope of}BC=\frac{\cos \theta +2}{\sin \theta }$

Now, we can equate both the slopes to give,

$\frac{\cos \theta +2}{\sin \theta }=\sqrt{3}$

$\Rightarrow \cos \theta +2=\sqrt{3}\sin \theta$

$\Rightarrow \sqrt{3}\sin \theta -\cos \theta =2$

Now, we divide $2$ on both sides, we get

$\Rightarrow \frac{\sqrt{3}}{2}\sin \theta -\frac{1}{2}\cos \theta =1$

Here, we know that $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$ and $\sin \frac{\pi }{6}=\frac{1}{2}$

$\Rightarrow \cos \frac{\pi }{6}\sin \theta -\sin \frac{\pi }{6}\cos \theta =1$

$\Rightarrow \sin (\theta -\frac{\pi }{6})=\sin \left(\frac{\pi }{2}\right)$

$\Rightarrow (\theta -\frac{\pi }{6})=n\pi +(-1{)}^n\left(\frac{\pi }{2}\right)$ $(\because \text{if}\sin x=\sin a\Rightarrow x=n\pi +(-1{)}^{n}a,n\in \text{Integers})$

$\Rightarrow \theta =n\pi +(-1{)}^n\left(\frac{\pi }{2}\right)+\frac{\pi }{6}$

If we substitute $n=-1,\theta =-\frac{3\pi }{2}+\frac{\pi }{6}=-\frac{4\pi }{3}\notin [0,\frac{\pi }{2}]$

If we substitute $n=0,\theta =\frac{\pi }{2}+\frac{\pi }{6}=\frac{2\pi }{3}\notin [0,\frac{\pi }{2}]$

If we substitute $n=1,\theta =\frac{\pi }{2}+\frac{\pi }{6}=\frac{2\pi }{3}\notin [0,\frac{\pi }{2}]$

If we substitute any other integral values of $n$ it will still doesn't belong to $[0,\frac{\pi }{2}]$.

Hence, for no values $\theta$ the points would be collinear.