In ∆ PQR ,Q=(0,0),PQ=2, Angle PQR=π/3 and the midpoint of QR has coordinate (2,0) then centroid of ∆ is.

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We have:Q0,0 and mid point of QRis 2,02,0=0+xR2,0+yR2=xR2,yR2xR2=2 and yR2=0xR=4 and yR=0Hence coordinates of R are 4,0On X-axisLet coordinates of P be h,kAs PQR=π3 Q is origin and R lies on X-axis, hence tanπ3 will beslope of PQslope of PQ=tanπ3k-0h-0=3k=h3PQ=2h-02+k-02=2h2+k2=4h2+h32=4h2+3h2=44h2=4h=±1Angle of π3 means P lies in first quadrant, henceh=1k=h3=3Hence we have: P1,3, Q0,0 and R4,0Hence centroid is 1+0+43,3+0+03=53,33=53,13

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