Let AB≡x−2y+1=0 and AC≡2x−y−1=0
So, A(1,1)
Now, altitude from B which is perpendicular to line 2x−y−1=0 and which passes through (37,37) is BH=x+2y−7=0,
Now intersection of BH&AB will give point B(3,2)
Similarly, altitude from C is CH=2x+y−7=0⇒C(2,3)
Centroid of \triangle ABC=E(2,2)&OE=2\sqrt{2}
where O is origin.