In triangle `A B C` , base `B C` and area of triangle are fixed. The locus of the centroid of triangle `A B C` is a straight line that is parallel to side `B C` right bisector of side BC perpendicular to BC inclined at an angle `sin^(-1)((sqrt())/(B C))` to side BC

Options

A. parallel to side BC
B. right bisector of side BC
C. prependicular to BC
D. inclined at an angle `sin^(-1) (sqrtDelta//BC)` to side BC

Answer

A

Explanation

`Delta = (1)/(2) (BC) h`, where h is the distance of vertex A from
`Delta_(GBC) = (Delta)/(3) = ((BC)h)/(6)`, where G is the centroid
`rArr h = (2Delta)/(BC)` = constant
Thus, distance of vertex A from the side is fixed. This, in turns, implies that the distance of centroid from side BC will be fixed.
hence, locus of G will be a line parallel to BC