Given :-

$x-y=0$is perpendicular bisector of side $AB$.

$\therefore$Equation of side $AB$ is.

$x+y+\lambda =0$

$\because$It passes through $A\left(2,3\right)$

$\therefore \lambda =-5$

$\therefore$Equation of $AB$ is,

$x+y-5=0 ....\left(1\right)$

Solving $x-y=0 \& x+y-5=0,$we get mid-point $D$ of $AB$.

$\therefore$ Mid point is $\left(\frac{5}{2},\frac{5}{2}\right)$

Let co-ordinate of $B$ be $\left(x_1,y_1\right)$

$\therefore \frac{x_1+2}{2}=\frac{5}{2} \& \frac{y_1+3}{2}=\frac{5}{2}$

$\therefore x_1=3, y_1=2$

$\therefore$Co-ordinates of $B$ is $\left(3,2\right)$

Also, $\frac{x}{2}+\frac{y}{2}=1$, i.e., $x+y-2=0$is perpendicular bisector for $AC$ .

$\therefore$Equation of $AC$ is

$x-y+k=0$

It passes through $\left(2,3\right),$we get $k=1$

$\therefore$Equation of $AC$ is

$x-y+1=0$

Solving $x+y-2=0 \& x-y+1=0,$we get mid-point $E$ of $AC$.

$2x=1 \left[\text{By Adding} x+y-2=0 \text{\&} x-y+1=0\right]$

$\therefore x=\frac{1}{2}$

& $y=\frac{3}{2}$[ substitute $x=\frac{1}{2}$in $x-y+1=0$ ]

So, Co-ordinates of $E$ are $\left(\frac{1}{2}, \frac{3}{2}\right)$

Let coordinates of $C$ be $\left(x_2,y_2\right)$

$\therefore \frac{x_2+2}{2}=\frac{1}{2} \& \frac{y_2+3}{2}=\frac{3}{2}$

$\therefore x_2=-1 \& y_2=0$

$\therefore$Co-ordinates of $C$ are $\left(-1,0\right)$

Now, equation of side $BC$ in two point form is

$\frac{y-2}{0-2}=\frac{x-3}{-1-3}$

$\Rightarrow y-2=\frac{x-3}{2}$

$\Rightarrow 2y-4=x-3$

$\Rightarrow x-2y+1=0$

$\therefore$option $1$ is correct.