For point A, 2x−y−1=0 x−2y+1=0 Solving (1) and (2), we get x=1,y=1 ∴ Point A=(1,1) Altitude from B to line AC is perpendicular to line AC. ∴ Equator of altitude BH is 2x+y+λ=0 It passes through point H(
7
3
,
7
3
) so it satisfy the equation (3).
14
3
+
7
3
+λ=0 ⇒α=−7 ∴ Altitude BH=2x+y−7=0 Solving equation (1) and (4), we get x=2,y=3 ∴ Point B=(2,3) Altitude from C to line AB is perpendicular to line AB. ∴ Equation of altitude CH is x+2y+λ=0 It passes through point H(
7
3
,
7
3
) so it satisfy equation (5).
7
3
+
14
3
+λ=0 ⇒λ=−7 ∴ Altitude CH=x+2y−7=0 Solving equation (2) and (6), we get x=3,y=2 ∴ Point C=(3,2) Centroid G(x,y) of triangle A(1,1),B(2,3) and C(3,2) is x=
1+2+3
3
=2,y=
1+2+3
3
=2 Now dDistance of point G(2,2) from center O(0,0) is OG=√22+22=2√2