The distance between the circumcentre and orthocentre of the triangle whose vertices are (0, 0), (6,8) and (4,3) $\text{ is }L\text{ then the value of }\frac{L}{\sqrt{5}}\text{ is }$

Given vertices of triangle are O(0,0), 8(6,8) and C(4,3). 

$\begin{array}{r}\text{ Slope of }OB=\frac{8}{6}=\frac{4}{3}\\ \text{ Slope of }OC=-\frac{3}{4}\\ \therefore \mathrm{\angle }BOC=\frac{\pi }{2}\\ \mathrm{\Delta }OBC\text{ is right angled at }O\text{ . }\end{array}$

$\text{ Circumcentre }=\text{ mid point of hypotenuse }BC=\left(1,\frac{11}{2}\right)$

$\begin{array}{l}\text{ Orthocentre }=\text{ vertex }O(0,0)\\ \text{ Required distance }=\sqrt{\left(1+\frac{121}{4}\right)}=\frac{5\sqrt{5}}{2}\text{ unit }\end{array}$