If the vertices of a triangle are \((-2,3),(6,-1)\) and \((4,3)\), then the co-ordinates of the circumcentre of the triangle are

Here, \(\mathrm{A}(-2,3), \mathrm{B}(6,-1), \mathrm{C}(4,3)\) are the vertices of \(\triangle \mathrm{ABC}\).
Let \(\mathrm{F}\) be the circumcentre of \(\triangle \mathrm{ABC}\).
Let \(\mathrm{FD}\) and \(\mathrm{FE}\) be the perpendicular bisectors of the sides \(\mathrm{BC}\) and \(\mathrm{AC}\) respectively.
\(\therefore \quad \mathrm{D}\) and \(\mathrm{E}\) are the midpoints of side \(\mathrm{BC}\) and \(\mathrm{AC}\) respectively.
\(\begin{array}{ll}\therefore & D \equiv\left(\frac{6+4}{2}, \frac{-1+3}{2}\right) \\ \therefore & D=(5,1) \\ & \text { and } E \equiv\left(\frac{-2+4}{2}, \frac{3+3}{2}\right)\end{array}\)
\(\therefore \quad \mathrm{E}=(1,3)\)

Now, slope of \(B C=\frac{3-(-1)}{4-6}=\frac{4}{-2}=-2\)
\(\therefore \quad \text { Slope of } \mathrm{FD}=\frac{1}{2} \quad \ldots[\because \mathrm{FD} \perp \mathrm{BC}]\)
Since FD passes through \((5,1)\) and has slope \(\frac{1}{2}\), equation of \(\mathrm{FD}\) is
\(\begin{aligned}
& y-1=\frac{1}{2}(x-5) \\
\therefore \quad & 2(y-1)=x-5 \\
\therefore \quad & 2 y-2=x-5 \\
\therefore \quad & x-2 y-3=0
\end{aligned}\)
Since both the points \(\mathrm{A}\) and \(\mathrm{C}\) have same \(y\) co-ordinates i.e. 3 ,
the given points lie on the line \(y=3\).
Since the equation \(\mathrm{FE}\) passes through \(\mathrm{E}(1,3)\), the equation of \(\mathrm{FE}\) is \(x=1\).
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value of \(x\) in (i), we get
\(\begin{array}{ll}
& 1-2 y-3=0 \\
\therefore \quad & y=-1 \\
\therefore \quad & \text { Co-ordinates of circumcentre } \mathrm{F} \equiv(1,-1) .
\end{array}\)