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The vertices of a triangle are (6, 0), (0, 6) and (6, 6). Find the distance between its circumcenter and centroid.

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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

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