If the vertices A and B of a triangle ABC are given by (2, 5) and (4,–11) and C moves along the line  $L_1$ :$9x+7y+4=0$ ,the locus of the centroid of the triangle $ABC$ is a straight line parallel to

Let C be $\left(\alpha ,\beta \right)$ then $9\alpha +7\beta +4=0$ ………………….(1)

Centroid (h, k) of triangle ABC is given by

$h=\frac{2+4+\alpha }{3}\Rightarrow \alpha =3h-6$

$k=\frac{5-11+\beta }{3}\Rightarrow \beta =3k+6$

Substituting in (1) we have $9\left(3h-6\right)+7\left(3k+6\right)+4=0$

$\Rightarrow 3\left(9h+7k\right)-8=0$

$\therefore$  locus of (h, k) is $3\left(9x+7y\right)-8=0$