The point P (1, 1) is translated parallel to $2x=y$ in the first quadrant through a unit distance. The coordinates of the new position of P are

 $\tan \theta =2$ 

$\therefore \sin \theta =\frac{2}{\sqrt{5}},\cos \theta =\frac{1}{\sqrt{5}}$

$Letnewco-ordinateis(x_1,y_1)$

$\therefore \frac{x_1-1}{\cos \theta }=\frac{y_1-1}{\sin \theta }=\pm 1$

$\therefore x_1=1\pm \cos \theta ,y_1=1\pm \sin \theta$

$x_1=1\pm \frac{1}{\sqrt{5}},y_1=1\pm \frac{2}{\sqrt{5}}$

$Hence,newpositionofPare\left(1\pm \frac{1}{\sqrt{5}},1\pm \frac{2}{\sqrt{5}}\right)$