If the vertices P and Q of a triangle $PQR$ are given by $(2,5)$ and $(4,-11)$ respectively, and the point R moves along the line N: $9x + 7y + 4 =0$, then the locus of the centroid of the $\triangle PQR$ is a straight line which is Parallel to one of these-
$PQ, QR, RP, N$
Find to which line is it Parallel to.
A point $R$ on the line $N$ has coordinates $R\left(t,\;\frac{1}{7} (-9 t-4)\right)$
Centroid $C$ has coordinates which are the average of the coordinates of $PQR$
$$C=\left(\frac{2+4+t}{3};\;\frac{5-11+\frac{1}{7} (-9 t-4)}{3}\right)$$
which leads to the parametric equations of the locus
$$\left(x=\frac{t}{3}+2,y=-\frac{3 t}{7}-\frac{46}{21}\right)$$
Solve the first equation wrt $t$
$t=3x-6$ and plug into the second
$$y=-\frac{3}{7}(3x-6)-\frac{46}{21}\to 27 x+21y=8\to y=\frac{8}{21}-\frac{9 }{7}\,x$$ slope is $m=-\dfrac{9}{7}$ exactly as the slope of line $N$ thus we can say that the locus is parallel to the line $N$
Hope this can be useful
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