I suspect you didn't spend much time reading the np.delete docs. You didn't specify an axis.

In [96]: arr = np.array([[1,2],[3,4],[5,6],[1,3]])

Removing the first row is easy:

In [97]: np.delete(arr, 0, axis=0)
Out[97]: 
array([[3, 4],
       [5, 6],
       [1, 3]])

Without axis,delete flattens the array first:

In [98]: np.delete(arr, 0, )
Out[98]: array([2, 3, 4, 5, 6, 1, 3])

It's also easy to remove/select rows by indexing, or boolean masking

In [99]: arr[[1,2,3],:]
Out[99]: 
array([[3, 4],
       [5, 6],
       [1, 3]])

That's ultimately what delete does; it translates the "remove" list into a "keep" list.

The other part of your problem is identifing which rows you want to remove. That is, converting values into row indices. It's a good idea to keep the two tasks separate.

Exploring the use of isin

In [100]: np.isin(arr, [[1,2]])
Out[100]: 
array([[ True,  True],
       [False, False],
       [False, False],
       [ True, False]])

This is a 2d boolean array. Applying all:

In [101]: np.isin(arr, [[1,2]]).all(axis=0)
Out[101]: array([False, False])

oops, wrong axis:

In [102]: np.isin(arr, [[1,2]]).all(axis=1)
Out[102]: array([ True, False, False, False])

we can use that boolean array index the rows:

In [103]: arr[np.isin(arr, [[1,2]]).all(axis=1),:]
Out[103]: array([[1, 2]])

and for removing a row:

In [104]: arr[~np.isin(arr, [[1,2]]).all(axis=1),:]
Out[104]: 
array([[3, 4],
       [5, 6],
       [1, 3]])

Yes, we could use where to convert that to an index, but it isn't needed.

I encourage you to examine and understand each piece in the process, both identifying the values, and removing them. Don't treat any as a 'block box' that you don't need to understand.

So, for this particular case, I suggest using a structured view of your array, this allows you to treat it like an "array of objects" (of structs):

>>> import numpy as np
>>> arr = np.array([[1,2],[3,4],[5,6],[1,3]])
>>> structured = arr.ravel().view("i8, i8")
>>> structured
array([(1, 2), (3, 4), (5, 6), (1, 3)],
      dtype=[('f0', '<i8'), ('f1', '<i8')])

Then, you can simply do:

>>> np.isin(structured, np.array([(1,2)], dtype="i8, i8"))
array([ True, False, False, False])
>>> arr[np.isin(structured, np.array([(1,2)], dtype="i8, i8"))]
array([[1, 2]])
>>> arr[~np.isin(structured, np.array([(1,2)], dtype="i8, i8"))]
array([[3, 4],
       [5, 6],
       [1, 3]])

Or equivalently using np.delete specifying the correct axis:

>>> np.delete(arr, np.isin(structured, np.array([(1,2)], dtype="i8, i8")), axis=0)
array([[3, 4],
       [5, 6],
       [1, 3]])

Important to note, you can add more "items" to delete trivially:

>>> np.delete(arr, np.isin(structured, np.array([(1,2), (5,6)], dtype="i8, i8")), axis=0)
array([[3, 4],
       [1, 3]])

Just make sure you use a list with tuple's when you create the structured array using the np.array constructor, so that it understands how you want them interpreted correctly.