The vertices of a triangle are (6, 0), (0, 6) and (6, 6). The distance between its circumcentre and centroid is

#### Options

\[2\sqrt{2}\]

2

\[\sqrt{2}\]

1

#### Solution

\[\sqrt{2}\] Let A(0, 6), B(6, 0) and C(6, 6) be the vertices of the given triangle.

\[\text { Centroid of } \bigtriangleup \text { ABC } = \left( \frac{0 + 6 + 6}{3}, \frac{6 + 0 + 6}{3} \right)\]

\[ = \left( 4, 4 \right)\]

\[\text { Coordinates of N } = \left( \frac{6 + 6}{2}, \frac{6 + 0}{2} \right)\]

\[ = \left( 6, 3 \right)\]

\[\text { Coordinates of P } = \left( \frac{0 + 6}{2}, \frac{6 + 6}{2} \right)\]

\[ = \left( 3, 6 \right)\]

Equation of MN is y = 3

Equation of MP is x = 3

As , we know that circumcentre of a triangle is the intersection of the perpendicular

bisectors of any two sides .

Therefore, coordinates of circumcentre is (3, 3)

Thus, the coordinates of the circumcentre are (3, 3) and the centroid of the triangle is (4,4).

Let d be the distance between the circumcentre and the centroid.

\[\therefore d = \sqrt{\left( 4 - 3 \right)^2 + \left( 4 - 3 \right)^2} = \sqrt{2}\]