Step-by-step Solution:

If a parallelogram is formed by the set of two parallel lines \( \ell_1, \ell_2 \) and \( \ell_3, \ell_4 \), then the area of the parallelogram is given by: \[ \text{Area} = \frac{d_1 d_2}{\sin \theta}, \] where \( d_1 \) is the distance between the lines \( \ell_1 \) and \( \ell_2 \), \( d_2 \) is the distance between the lines \( \ell_3 \) and \( \ell_4 \), and \( \theta \) is the angle between the lines. Here, \[ d_1 = \frac{1}{\sqrt{4^2 + 1^2}} = \frac{1}{\sqrt{17}}, \quad d_2 = \frac{1}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}}. \] The angle between the lines is given by: \[ \tan \theta = \left| \frac{4 - (-1)}{1 - 4} \right| = \frac{5}{3}, \] and \[ \sin \theta = \frac{5}{\sqrt{34}}. \] Thus, the area of the parallelogram is: \[ \text{Area} = \frac{\frac{1}{\sqrt{17}} \times \frac{1}{\sqrt{2}}}{\frac{5}{\sqrt{34}}} = \frac{1}{5}. \]