Question

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

• A. $2\sqrt{2}$
• B. 2
• C. $\sqrt{2}$
• D. 1

Answer 1

The correct option is C

$\sqrt{2}$

Let A(0, 6), B(6, 0) and C(6, 6) be the vertices of the given triangle.

Centroid of $\Delta ABC=(\frac{0+6+6}{3},\frac{6+0+6}{3})$

$=(4,4)$

Coordinates of $N=(\frac{6+6}{2},\frac{6+0}{2})$

$=(6,3)$

Coordinates of $P=(\frac{0+6}{2},\frac{6+6}{2})$

$=(3,6)$

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 beween the circumcentre and the centroid.

$\therefore d=\sqrt{(4-3{)}^2+(4-3{)}^2}=\sqrt{2}$