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

(c) $\sqrt{2}$

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



$$\begin{gathered} \mathrm{Centroid}\mathrm{of}△\mathrm{ABC}=\left(\frac{0+6+6}{3},\frac{6+0+6}{3}\right) \\ =\left(4,4\right) \end{gathered}$$


$$\begin{gathered} \mathrm{Coordinates}\mathrm{of}\mathrm{N}=\left(\frac{6+6}{2},\frac{6+0}{2}\right) \\ =\left(6,3\right) \\ \mathrm{Coordinates}\mathrm{of}\mathrm{P}=\left(\frac{0+6}{2},\frac{6+6}{2}\right) \\ =\left(3,6\right) \\ \mathrm{Equation}\mathrm{of}\mathrm{MN}\mathrm{is}y=3 \\ \mathrm{Equation}\mathrm{of}\mathrm{MP}\mathrm{is}x=3 \\ \mathrm{As},\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{circumcentre}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}\mathrm{the}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{perpendicular} \\ \mathrm{bisectors}\mathrm{of}\mathrm{any}\mathrm{two}\mathrm{sides}. \\ \mathrm{Therefore},\mathrm{coordinates}\mathrm{of}\mathrm{circumcentre}\mathrm{is}(3,3) \end{gathered}$$


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.

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