Question

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

Answer 1

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`
`h= (6+a)/3` and `k = (-46-9a)/21`
`a = 3h-6` and `a = (-21k+46)/9`
Thus,`(-21k+46)/9 = 3h-6`
`=>27h+21k-8=0`
`=>9h+7k-8/3=0` that is clearly parallel to line N. So, option D is the correct option.