0

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.

nonuser
  • 90,026
user150093
  • 37
  • 4
  • What have you tried so far? I suggest you write the equation for the centroid, then calculate the slopes for all the lines involved in this problem. You are more likely to get answers when you show your work, and let us know where you are stuck – Andrei Nov 01 '17 at 17:37

1 Answers1

0

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

$$...$$

enter image description here

Raffaele
  • 26,371