Question

Let ABC be a triangle and let x, y, z be three arbitrary vectors. For any real number $\lambda >0$, the points M, N, P are chose so that:
$\overrightarrow{AM}=\lambda x,\overrightarrow{BN}=\lambda y,\overrightarrow{CP}=\lambda z$.
Find the locus of the centroid of the triangle MNP

Answer 1

Let G be the centroid of the triangle ABC. We have:
$3\overrightarrow{GQ}=\overrightarrow{GM}+\overrightarrow{GN}+\overrightarrow{GP}$
$=(\overrightarrow{GA}+\lambda x)+(\overrightarrow{GB}+\lambda y)+(\overrightarrow{GC}+\lambda z)$
$=\overrightarrow{0}+\lambda (x+y+z)$.
Setting $x+y+z=v$, the relation $3\overrightarrow{GQ}=\lambda v$ shows that if $v\ne 0$, then Q lies on the passing through the point G, having the direction of the vector v. If $v=0$, then the locus of the point Q reduces to the point G.