The given equation is $y-2=0$

This can be written as

$0.x+1.y=2$

$ \sqrt{0^2+1^2}=1$

Hence dividing by 1 on both sides we get

$0.x+1.y=2$

i.e., $ x \cos 90^{\circ}+ y \sin 90^{\circ} = 2$

The above equation can be compared with the normal form of the equation.

$x \cos \alpha + y \sin \alpha = p$

Here $p=2$ and the angle $ \alpha = 90^{\circ}$.

Hence the perpendicular distance is 2 units and the angle $ \alpha = 90^{\circ}$