(i) The median through A meet BC at D. So, D is the mid-point of BC. Therefore Coordinates of D = ((6+1)/2, (5+4)/2) = (7/2, 9/2)
(ii) Point P lies on AD, such that AP:PD = 2:1. Therefore (1.4) Coordinates of P = ((2×7/2+1×4)/(2+1), (2×9/2+1×2)/(2+1))
(iii) The median through B meets AC at E. So, E is the mid-point of AC. Therefore, Coordinates of E = ((4+1)/2, (2+4)/2) = (5/2, 3)
(iv) Point Q lies on AE, such that AQ:QE = 2:1.Therefore Coordinates of Q = ((2×5/2+1×6)/(2+1), (2×3+1×5)/(2+1)) = (11/3, 11/3)
The median through C mects AB at F. So, F is the mid-point of AB. Therefore Coordinates of F = ((4+6)/2, (2+5)/2) = (5, 7/2)
Point R lies on CF, such that CR:RF = 2:1.
(v) Therefore, the coordinates of R = ((2×5+1×1)/(2+1), (2×7/2+1×4)/(2+1)) = (11/3, 11/3)
(vi) The coordinates of P, Q and R is same.
(vii)Points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of triangle ABC. Median through A meets BC at D. So, D is the mid-point of BC. Therefore, The coordinates of D = ((x₂+x₃)/2, (y₂+y₃)/2)
Let O be the centroid of the triangle. Point O lies on AD such that A0:0D = 2:1.
Therefore, the coordinates of point O
((2×(x₂+x₃)/2 +1×x₁)/(2+1), (2×(y₂+y₃)/2 +1×y₁)/(2+1)) = ((x₁+x₂+x₃)/2, (y₁+y₂+y₃)/2)

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