Let the vertices of the triangle are A(x1, y1) = (6, 0), B(x2, y2) = (0, 6) and C(x3, y3) = (6, 6).
Let GD, GE and GF be the right bisectors of BC,
CA and AB respectively with circumcenter G(h, k).
∴ Co-ordinates of D = midpoint of BC
= \(\big(\frac{6 + 0}{2}, \frac{6 + 6}{2}\big)\) = (3, 6)
Co-ordinate of E = midpoint of
AC = \(\big(\frac{6 + 6}{2}, \frac{6 + 0}{2}\big)\) = (6, 3)
Since GD ⊥ BC, so
Slope of GD × slope of BC = – 1
⇒ \(\frac{6 − k}{3 − h} × \frac{6 − 6}{6 − 0}\) = − 1
⇒ 6(3 – h) × (– 1) = 0
⇒ h = 3
And, slope of GE × slope of AC = – 1
⇒ \(\frac{3 − k}{6 − h} × \frac{6 − 0}{6 − 6}\) = − 1
(6 – k) 6 = 1
⇒ 6(3 – h) × (– 1) = 0
⇒ h = 3
And, slope of GE × slope of AC = – 1
⇒ \(\frac{3 − k}{6 − h} × \frac{6 − 0}{6 − 6}\) = − 1
⇒ 6(3 – k) = 0
K = 3.
And centroid of ∆ABC has co-ordinates
\(\big(\frac{6+0}{3},\frac{6+6+0}{3}\big)\) = (4, 4)
∴ Distance between circumcenter and centroid
= \(\sqrt{(4 − 3)^2 + (4 − 3)^2}\) = \(\sqrt{2}\) unit.
AC2 = 72, BC2 + AB2 = 36 + 36 = 72
∴ AC2 = AB2 + BC2
∴ ABC is a right triangle.
Hence AC is the diameter of circumcircle
∴ centre of circumcircle = \(\big(\frac{6 + 0}{2}, \frac{0 + 6}{2}\big)\)
= (3, 3)
Centroid of ∆ABC = \(\big(\frac{x_1 + x_2+x_3}{3}, \frac{y_1 + y_2+y_3}{3}\big)\)
= \(\big(\frac{6 + 0 + 6}{3}, \frac{6 + 0 + 6}{3}\big)\)
= (4, 4)
Distance between centre and centroid of ∆ABC