Question

If the points P(6,2) and Q(-2,1) and R are the vertices of a Δ PQR and R is the point on the locus of y=x -3x+4, then find the equation of the locus of centroid of Δ PQR​

Answer 1

Answer:

2x+y=7

Step-by-step explanation:

Given P (6,2) Q(2,-1)

R lies on the line y=x-3x+4

i.e ; y = -2x+4

Let us take (x1,y1) to be R

Then y1= -2x1 + 4

(x1, -2x1+4)

Let G(x,y) be the centroid of  Δ PQR​

(x,y) =(  1/3(6+2+x1) , 1/3(2-1-2x1+4) )

(x,y) = ( (8+x1)/3 ,(5-2x1)/3 )

3x= 8+x1

i.e; x1=3x-8

3y=5-2x1

3y=5-2*(3x-8)

3y=21-6x

Locus of centroid of  Δ PQR​ is

6x+3y=21

2x+y=7