The vertices of a $∆ABC$ are $A(0, 0), B(0, 2)$ and $C(2, 0).$ Find the distance between the circum-centre and orthocentre.

Clearly, it is a right-angled triangle. As we know that in case of a right angled triangle, the circumcentre is the mid-point of the hypotenuse and the orthocentre is at right angle.

Thus, Circumcentre $=\left(\frac{0+2}{2},\frac{2+0}{2}\right)=(1, 1)$

and Orthocentre$=(0, 0)$

Hence, the required distance $=\sqrt{(1-0{)}^2+(1-0{)}^2}=\sqrt{2}$