If $A\left(1,1\right)$  and $P\left(x,y\right)$  such that slope of $\overleftrightarrow{AP}$  is $1$  and $AP=\sqrt{2}$ , then  $\left(x,y\right)$  is

Slope of $\overleftrightarrow{AP}$  is $=1$

$\frac{y_2-y_1}{x_2-x_1}=1$

$\frac{y-1}{x-1}=1$

$y-1=x-1$

$x=y$

$AP=\sqrt{2}$

$\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}=\sqrt{2}$

${\left(x-1\right)}^2+{\left(x-1\right)}^2=2$

$2{\left(x-1\right)}^2=2$

$x-1=\sqrt{1}$

$x=2$

$\therefore$ $y=2$

$(x,y)=\left(2,2\right)$