Question

The vertices of a triangle are $(6,6),(0,6)$ and $(6,0)$. 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 the triangle be $ABC$ with $A=(x_1,y_1)=(6,6),B=\left(x_2,y_2\right)=(0,6)$ and $C=(x_3,y_3)=(6,0)$

Since $y_1=y_3\Rightarrow AC\parallel Y$-axis and

$x_1=x_2\Rightarrow AB\parallel X$-axis

$\therefore BA\perp CA$

$\therefore \Delta AB$ is a right one with $\angle A=1$ right angle.

So, its circumcentre is mid-point of the hypoteneuse $BC$ which is $S=(\frac{x_2-x_3}{2},\frac{y_2-y_3}{2})=(\frac{0+6}{2},\frac{6+0}{2})=(3,3)$

Its centroid is $G=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})=(\frac{6+0+6}{3},\frac{6+6+0}{3})=(4,4)$

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