Coordinate Geometry Question 121
Question: If $ A(\cos \alpha ,\sin \alpha ),\ B(\sin \alpha ,-\cos \alpha ),C(1,2) $ are the vertices of a $ \Delta ABC $ , then as $ \alpha $ varies, the locus of its centroid is
Options:
A) $ x^{2}+y^{2}-2x-4y+1=0 $
B) $ 3(x^{2}+y^{2})-2x-4y+1=0 $
C) $ x^{2}+y^{2}-2x-4y+3=0 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ (h,k) $ be the centroid of the triangle, then $ h=\frac{\cos \alpha +\sin \alpha +1}{3} $ and $ k=\frac{\sin \alpha -\cos \alpha +2}{3} $
$ \Rightarrow 3h-1=\cos \alpha +\sin \alpha $ and $ 3k-2=\sin \alpha -\cos \alpha $
$ \Rightarrow {{(3h-1)}^{2}}+{{(3k-2)}^{2}}=2 $ , (squaring and adding)
$ \Rightarrow 9(h^{2}+k^{2})-6h-12k+3=0 $
$ \Rightarrow 3(h^{2}+k^{2})-2h-4k+1=0 $
$ \therefore $ Locus of $ (h,k) $ is $ 3(x^{2}+y^{2})-2x-4y+1=0 $ .
BETA
