We know that centroid of ∆DEF will be the same that of ∆ABC as ∆DEF is formed by midpoints of ∆ABC.
∴ We know that centroid of a triangle for (x1 , y1), (x2 , y2) and (x3 , y3) is
G(x,y) = \((\frac{x_1+x_2+x_3}3,\frac{y_1+y_2+y_3}3)\)
∴ G(x, y) = \((\frac{4+4-2}3,\frac{5-3+3}3)\)
∴ G(x, y) = ( 2, \(\frac{5}2\))
Hence the centroid is ( 2, \(\frac{5}2\))