If the vertices `Pa n dQ` of a triangle `P Q R` are given by (2, 5) and `(4,-11)` , respectively, and the point `R` moves along the line `N` given by `9x+7y+4=0` , then the locus of the centroid of triangle `P Q R` is a straight line parallel to PQ (b) QR (c) RP (d) N

Explanation

Please refer to video for the diagram.

Let co-ordinates of point R is(a,b).

If, (h,k) the centroid, then we can say that

`h = (2+4+a)/3 and k = (5-11+b)/3 ->(1)`
As point, R is passing through line N fiven by `9x+7y+4=0`, we can have,

`b=(-4-9a)/7`

Now, putting value of b in (1),

`h= (6+a)/3` and `k = (-6-((4+9a))/7)/3`

...