# Free Body Diagram for a Body on a Smooth Inclined Plane

If instead of decomposing into perpendicular and parallel force components, you use the 'regular' $x$ and $y$ axis as you did to get Eq. (2), then you will have a non-zero acceleration in both $x$ and $y$ directions. So, what you really should end up with for Eq. (2) would be $$ma_y=mg-N\cos\theta\tag2$$
because the forces are **not** balanced. Therein lies the reason why we decompose into parallel and perpendicular components -- because the block will only slide parallel to the incline, so we can nicely conclude that $\Sigma F_\perp=0$. However, in the usual $xy$ coordinate system, there will be an acceleration in both $x$ and $y$ directions. So you can't just assume one of them is zero.