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

Given points, $(-2,0)$ , $(-1,\frac{1}{\sqrt{3}})$ and $(\cos \theta ,\sin \theta )$ are collinear.

Three points are collinear if the area is $0$

i.e.,

if $(x1,y1),(x2,y2),(x3,y3)$ are the points

$\frac{1}{2}\left[x1\right(y2-y3)+x2(y3-y1)+x3(y1-y2\left)\right]=0$

Substituting the given points,

$\frac{1}{2}[-2(\frac{1}{\sqrt{3}}-\sin \theta )-1(\sin \theta -0)+\cos \theta (0-\frac{1}{\sqrt{3}}\left)\right]=0$

$\frac{1}{2}\left[\right(\frac{-2}{\sqrt{3}}+2\sin \theta )-\sin \theta -\frac{\cos \theta }{\sqrt{3}})]=0$

$\frac{-1}{\sqrt{3}}+\frac{\sin \theta }{2}-\frac{\cos \theta }{2\sqrt{3}}=0$

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

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

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

No value of $\theta (0\le \theta \le \frac{\pi }{2})$ satisfies the above eqn.