Correct option is (D) (1,-1)
(-4, 2); -4 < 2 \(\Rightarrow\) abscissa < ordinate
(2, 4); 2 < 4 \(\Rightarrow\) abscissa < ordinate
(-1, 1); -1 < 1 \(\Rightarrow\) abscissa < ordinate
(1, -1); 1 > -1 and 1 - (-1) = 2
i.e., abscissa is more than its ordinate by 2.