A(3,2,−1),B(4,1,1),C(6,2,5) and D(3,3,3) are four points G1,G2,G3 and G4 respectively are the centroids of the triangles BCD, CDA, DAB and ABC. The point of concurrence of the lines AG1,BG2,CG3 and DG4 is
From given, G1 is the centroid of triangle BCD, G1=(
13
3
,
6
3
,
9
3
) G2 is the centroid of triangle CDA, G2=(
12
3
,
7
3
,
7
3
) As the lines AG1,BG2,CG3,DG4 are concurrent, so point of concurrence of these four lines is point of intersection of lines AG1 and BG2 . The equation of line AG1 is,
x−3
4∕3
=
y−2
0
=
z+1
12
3
=r1 So, the point on line AG1 is (3+
4
3
r1,2,−1+
12
3
r1) and line BG2 is
x−4
0
=
y−1
4
5
=
z−1
4
3
=r2 So, the point on line BG2 is (4,1+
4
3
r2,1+
4
3
r2) Let above point to be point of intersection, 3+
4
3
r1=4⇒2=1+
4
3
r2 And −1+
12
3
r1=1+
4
3
r2 From above values, r1=
3
4
,r2=
3
4
So, the required point of concurrence is (4, 2, 2) .