In a $∆ABC \overline{OA}=\overline{a},\overline{OB}=\overline{b},\overline{OC}=\overline{c}.$ If $\left(\overline{a}-\overline{b}\right).\left(\overline{c}-\overline{b}\right)=0,$ then P.V of the circumcentre is

$$\begin{gathered} \left(\overline{\mathrm{a}}-\overline{\mathrm{b}}\right)·(\overline{\mathrm{c}}-\overline{\mathrm{b}})=0 \Rightarrow \overline{\mathrm{BA}}·\overline{\mathrm{BC}}=0 \\ \Rightarrow \overline{\mathrm{BA}} \perp \overline{\mathrm{BC}} \Rightarrow \overline{)\underline{\mathrm{B}}}={90}^0 \end{gathered}$$

$for right angle triangle circum centre is midpoint of hypotenuse$

$\therefore \mathrm{S}=\left(\frac{\overline{\mathrm{a}}+\overline{\mathrm{c}}}{2}\right)$