If G(– 2, 1) is the centroid of a ΔABC and two of its vertices are A(1, – 6) and B(– 5, 2), find the third vertex of the triangle.
The figure is shonw as:
– 2 = (m1x2 + m2x1)/ m1 + m2
– 2 = (2 × x + 1x 1)/2 + 1
– 2 = (2x + 1) /3
– 6 = 2x + 1
– 7 = 2x
⇒ x = – 7/2
1 = (m1y2 + m2y1)/ m1 + m2
1 = (2x y + 1x (– 6))/3
1 = (2y – 6)/2
2 = 2y – 6
8 = 2y
⇒ y = 4
D(– 7/2,4)
Now for BC
– 7/2 = (m1x2 + m2x1)/ m1 + m2
– 7/2 = (1 × x + 1x (– 5))/1 + 1
– 7/2 = (x – 5) /2
– 7 = x – 5
– 7 + 5 = x
⇒ x = – 2
4 = (m1y2 + m2y1)/ m1 + m2
4 = (1 × y + 1x 2)/2
4 = (y + 2)/2
8 = y + 2
⇒ y = 6
Hence, C(– 2, 6)