The circumcentre of the triangle formed by the lines 3x-y-5=0, x+3y-5=0, x = y is

Given lines

$\begin{array}{l}3x-y-5=0--\left(1\right)\Rightarrow Slope=\frac{-3}{-1}=3\\ x+3y-5=0\Rightarrow \left(2\right)\Rightarrow Slope=\frac{-1}{3}\\ x=y--\left(3\right)\Rightarrow Slope=1\\ \left(1\right)\perp r\left(2\right)\\ B=P.Iof\left(1\right)of\left(3\right)\\ Susy=xin\left(1\right)\Rightarrow 3x-x-5=0\Rightarrow 2x=5\Rightarrow x=5/2\\ \therefore y=5/2\\ B=\left(\frac{5}{2},\frac{5}{2}\right)\end{array}$

C = P.I of (2)  & ( 3 )

Put y = x in (2)

$\begin{array}{l}\Rightarrow x=3x-5=0\\ \Rightarrow 4x=5\Rightarrow x=\frac{5}{4}\\ \therefore y=\frac{5}{4}\\ \therefore C=\left(\frac{5}{4},\frac{5}{4}\right)\end{array}$

Circumcentre = Mid point of CB

$=\left(\frac{\frac{5}{2}+\frac{5}{4}}{2},\frac{\frac{5}{2}+\frac{5}{4}}{2}\right)\Rightarrow \left(\frac{15}{8},\frac{15}{8}\right)$