Question

$A\equiv (cos\theta ,sin\theta ),B\equiv (sin\theta ,-cos\theta )$ are two points. The locus of the centroid of $\Delta OAB$, where 'O' is the origin is _____.

• A. $x^2+y^2=3$
• B. $9x^2+9y^2=2$
• C. $2x^2+2y^2=9$
• D. $3x^2+3y^2=2$

Answer 1

Answer: (B)

$9x^2+9y^2=2$

Let centroid of the triangle be $(h,k)$

So $(h,k)=(\frac{\cos \theta +\sin \theta +0}{3},\frac{\sin \theta -\cos \theta +0}{3})$

$\Rightarrow \cos \theta +\sin \theta =3h$ and $\sin \theta -\cos \theta =3k$

Now square and add both

$(\cos \theta +\sin \theta {)}^2+(\sin \theta -\cos \theta {)}^2=(3h{)}^2+(3k{)}^2$

$\Rightarrow 1+2\sin \theta \cos \theta +1-2\sin \theta \cos \theta =9h^2+9k^2$, since ${\sin }^{2}x+{\cos }^{2}x=1$

$\Rightarrow 2=9h^2+9k^2$

Hence the locus is $9x^2+9y^2=2$