If a line L is perpendicular to the line 5x−y=1, and the area of the triangle formed by the line L and the coordinate axes is 5, then the distance of line L from the line x+5y=0 is 7√13, 7√5, 5√13, 5√7

The line 5x - y = 1 has a slope of 5. A line perpendicular to this line will have a slope of -1/5. Therefore, the equation of the line L can be written in the form y = (-1/5)x + c, where c is the y-intercept.

The x-intercept is found by setting y = 0: 0 = (-1/5)x + c => x = 5c
The y-intercept is c.

The area of the triangle formed by the line L and the coordinate axes is given by (1/2) * base * height = (1/2) * 5c * c = 5

Solving for c: (5c^2)/2 = 5 => c^2 = 2 => c = ±√2

Therefore, the equation of line L is either y = (-1/5)x + √2 or y = (-1/5)x - √2.
We can rewrite these equations in the form x + 5y = 5√2 or x + 5y = -5√2.

The distance between two parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is given by |C1 - C2| / √(A^2 + B^2).

In our case, the lines are x + 5y - 5√2 = 0 and x + 5y = 0. The distance is:
|(-5√2) - 0| / √(1^2 + 5^2) = 5√2 / √26 = 5√2 / √26 = 5√52 / 26 = 5 * 2√13 / 26 = 5√13 / 13

However, if we use the other equation, x + 5y + 5√2 = 0, the distance becomes:
|5√2 - 0| / √26 = 5√13/13

There must be a mistake in the problem statement or the provided options. Let's reconsider the area calculation. The area of the triangle formed by the line and the axes is given as 5. Let the intercepts be a and b. Then (1/2)ab = 5, so ab = 10. The equation of the line is x/a + y/b = 1. Since the line is perpendicular to 5x - y = 1, its slope is -1/5. Thus, -b/a = -1/5, which implies b = a/5. Substituting into ab = 10, we get a(a/5) = 10, so a^2 = 50, and a = 5√2. Then b = √2. The equation of the line is x/(5√2) + y/√2 = 1, which simplifies to x + 5y = 5√2.

The distance from this line to x + 5y = 0 is |5√2| / √(1^2 + 5^2) = 5√2 / √26 = 5√52/26 = 10√13/26 = 5√13/13. This doesn't match any of the options.

There appears to be an error in either the question or the provided options.