The image of origin in the line \(\mathrm{x}+4 \mathrm{y}=1 \mathrm{is}\) (a) \([(2 / 17),-(8 / 17)]\) (b) \([-(2 / 17),-(8 / 17)]\) (c) \([-(2 / 17),(8 / 17)]\) (d) \([(2 / 17),(8 / 17)]\)

To find the distance between the origin (0, 0) and the line \(x + 4y = 1\), we use the point to line distance formula: \[d = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}\] where the line is given in the form Ax + By + C = 0 and (x, y) is the point. For our line, A = 1, B = 4, C = -1, x = 0, and y = 0. Plugging in these values, we get: \[d = \frac{|(1)(0) + (4)(0) + (-1)|}{\sqrt{(1)^2 + (4)^2}} = \frac{1}{\sqrt{17}}\]

To find this point, we can imagine moving from the origin in the opposite direction of the normal vector of the line by a distance equal to the distance between the origin and the line. The normal vector of the line is given by the coefficients of x and y in the equation of the line: \( \begin{pmatrix} 1 \\ 4 \end{pmatrix}\). We can calculate the normalized normal vector by dividing each component of the normal vector by the length of the normal vector: \[\begin{pmatrix} \frac{1}{\sqrt{17}} \\ \frac{4}{\sqrt{17}} \end{pmatrix}\] Now, we need to find the point which is twice the calculated distance away from the origin in the direction opposite of the normalized normal vector: \[\begin{pmatrix} 0 \\ 0 \end{pmatrix} - 2 \cdot \frac{1}{\sqrt{17}} \cdot \begin{pmatrix} \frac{1}{\sqrt{17}} \\ \frac{4}{\sqrt{17}} \end{pmatrix} = \begin{pmatrix} - \frac{2}{17} \\ - \frac{8}{17} \end{pmatrix}\] Comparing this point to the options given in the exercise, we see that it matches the point of option (c). So the image of the origin in the line \(x + 4y = 1\) is \[\boxed{[-(2/17), (8/17)]}\] or option (c).