Question

The centroid of a triangle formed by the points $(0,0),(\cos \theta ,\sin \theta )$ and $(\sin \theta ,-\cos \theta )$ lie on the line $y=2x$; then $\theta$ is.

• A. ${\tan }^{-1}2$
• B. ${\tan }^{-1}\frac{1}{3}$
• C. ${\tan }^{-1}\frac{1}{2}$
• D. ${\tan }^{-1}-3$

Answer 1

The correct option is D ${\tan }^{-1}-3$

$D$ is the mid-point of $AB$

$\therefore$ $D$ $=$ $[\frac{\sin \theta +\cos \theta }{2},\frac{\sin \theta -\cos \theta }{2}]$

If $P$ is centroid of a $△OAB,$ then $P$ divides the median $OD$ in the ratio $2:1$

$\therefore$ $P=[0+\frac{2}{3}\left[\frac{\sin \theta +\cos \theta }{2}\right],0+\frac{2}{3}\left[\frac{\sin \theta -\cos \theta }{2}\right]]$

$P=[\frac{\sin \theta +\cos \theta }{3},\frac{\sin \theta -\cos \theta }{3}]$

It is given that centroid lies on line $y=2x$

$\therefore$ $\frac{\sin \theta -\cos \theta }{3}=2\left[\frac{\sin \theta +\cos \theta }{3}\right]$

$\sin \theta -\cos \theta =2\sin \theta +2\cos \theta$

$-\sin \theta =3\cos \theta$

$\tan \theta =-3$

$\theta ={\tan }^{-1}(-3)$