Question

$A\equiv (\cos \theta ,\sin \theta ),B=(\sin \theta ,-\cos \theta )$ are two points. The locus of the centroid of $\Delta OAB$, where ${}^{\prime }{O}^{\prime }$ 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)

From figure, we observe that,
$\theta =\frac{\pi }{4}$
$\therefore A=\left(\cos \frac{\pi }{4},\sin \frac{\pi }{4}\right)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$
$B=\left(\sin \frac{\pi }{4},-\cos \frac{\pi }{4}\right)=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$

and $0=(0,0)=$ origin
$\therefore$ Centroid of $\Delta ABO$
$=\left\{\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+0}{3},\frac{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+0}{3}\right\}$
$=\left(\frac{2}{3\sqrt{2}},0\right)=\left(\frac{\sqrt{2}}{3},0\right)$
Let $(h,k)$ be centroid of $\Delta OAB$.
Then, $(h,k)=\left(\frac{\sin \theta +\cos \theta }{3},\frac{\sin \theta -\cos \theta }{3}\right)$
$\Rightarrow \sin \theta +\cos \theta =3h\dots (i)$
and $\sin \theta -\cos \theta =3k...(ii)$
Now, on adding Eqs. (i) and (ii) after squaring, we get
$(3h{)}^2+(3k{)}^2=(\sin \theta +\cos \theta {)}^2+(\sin \theta -\cos \theta {)}^2=2$
$\Rightarrow n^2+k^2=\frac{2}{9}$
Hence, required locus is
$x^2+y^2=\frac{2}{9}$ or $9x^2+9y^2=2$