Find the circumcenter of the triangle whose $\text{ sides are given }\mathrm{x}+\mathrm{y}+2=0,5\mathrm{x}-\mathrm{y}-2=0\text{ and }$$\mathrm{x}-2\mathrm{y}+5=0$ 

Given sides of triangle are
$\begin{array}{l}\mathrm{x}+\mathrm{y}+2=0 ....\left(1\right)\\ 5\mathrm{x}-\mathrm{y}-2=0 ....\left(2\right)\\ \mathrm{x}-2\mathrm{y}+5=0 ...\left(3\right)\end{array}$
P. I of (1) & (2) is A
$\begin{array}{l}\frac{\mathrm{x}}{-2+2}=\frac{\mathrm{y}}{10+2}=\frac{1}{-1-5}=\frac{\mathrm{x}}{0}=\frac{\mathrm{y}}{12}=\frac{1}{-6}\\ \mathrm{x}=0,\mathrm{y}=-2;\therefore \mathrm{A}(0,-2)\end{array}$
P.I of (1) & (3) is B
$\begin{array}{l}\frac{\mathrm{x}}{5+4}=\frac{\mathrm{y}}{2-5}=\frac{1}{-2-1}=\frac{\mathrm{x}}{9}=\frac{\mathrm{y}}{-3}=\frac{1}{-3}\\ \mathrm{x}=-3,\mathrm{y}=1;\therefore \mathrm{B}=(-3,1)\end{array}$
P.I of (2) & (3) is C
$\begin{array}{l}\frac{\mathrm{x}}{-5-4}=\frac{\mathrm{y}}{-2-25}=\frac{1}{-10+1}=\frac{\mathrm{x}}{-9}=\frac{\mathrm{y}}{-27}=\frac{1}{-9}\\ \mathrm{x}=1,\mathrm{y}=3;\therefore \mathrm{C}=(1,3)\\ \therefore \mathrm{A}(0,-2),\mathrm{B}(-3,1),\mathrm{C}(1,3)\end{array}$
$\text{ Let }\mathrm{S}(\mathrm{\alpha },\mathrm{\beta })\text{ be the circumcentre of ∆ABC}$
$\begin{array}{l}\Rightarrow \mathrm{SA}=\mathrm{SB}=\mathrm{SC}\Rightarrow \mathrm{SA}=\mathrm{SB}\Rightarrow {\mathrm{SA}}^2={\mathrm{SB}}^2\\ \Rightarrow (\mathrm{\alpha }-0{)}^2+(\mathrm{\beta }+2{)}^2=(\mathrm{\alpha }+3{)}^2+(\mathrm{\beta }-1{)}^2\\ \Rightarrow {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+4+4\mathrm{\beta }={\mathrm{\alpha }}^2+9+6\mathrm{\alpha }+{\mathrm{\beta }}^2+1-2\mathrm{\beta }\\ \Rightarrow 4+4\mathrm{\beta }-9-6\mathrm{\alpha }-1+2\mathrm{\beta }=0\\ \Rightarrow -6\mathrm{\alpha }+6\mathrm{\beta }-6=0\\ \Rightarrow \mathrm{\alpha }-\mathrm{\beta }+1=0 ...\left(4\right)\end{array}$
SB = SC
$\begin{array}{l}\Rightarrow {\mathrm{SB}}^2={\mathrm{SC}}^2\\ \Rightarrow (\mathrm{\alpha }+3{)}^2+(\mathrm{\beta }-1{)}^2=(\mathrm{\alpha }-1{)}^2+(\mathrm{\beta }-3{)}^2\\ \Rightarrow {\mathrm{\alpha }}^2+9+6\mathrm{\alpha }+{\mathrm{\beta }}^2+1-2\mathrm{\beta }=\\ {\mathrm{\alpha }}^2+1-2\mathrm{\alpha }+{\mathrm{\beta }}^2+9-6\mathrm{\beta }\\ \Rightarrow 6\mathrm{\alpha }+1-2\mathrm{\beta }-1+2\mathrm{\alpha }+6\mathrm{\beta }=0\\ \Rightarrow 8\mathrm{\alpha }+4\mathrm{\beta }=0\Rightarrow 2\mathrm{\alpha }+\mathrm{\beta }=0\dots \text{...(5) }\end{array}$
Solving (4) &(5) we get $\mathrm{S}(\mathrm{\alpha },\mathrm{\beta })$
$\begin{array}{l}\frac{\mathrm{\alpha }}{0-1}=\frac{\mathrm{\beta }}{2-0}=\frac{1}{1+2}\\ \mathrm{\alpha }=\frac{-1}{3},\mathrm{\beta }=\frac{2}{3}\end{array}$
$\text{ ∴Circumcentre of }\mathrm{△}\mathrm{ABC}\text{ is }\left(\frac{-1}{3},\frac{2}{3}\right)$