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# Given three points P,Q,R with P(5,3) and R lies on the x - axis . if equation of RQ is $\;x-2y=2\;$ and PQ is parallel to the x - axis , then the centroid of $\;\bigtriangleup PQR\;$ lies on the line :

$(a)\;2x+y-9=0\qquad(b)\;x-2y+1=0\qquad(c)\;5x-2y=0\qquad(d)\;2x-5y=0$

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