The perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x1 ,y1) is given by d = | Ax1 + by1 + c|/ root (A^2 + B^2) ..... so put (0,0) .

Given equation of line is 3x + 4y + 1 = 0.

Finding the distance between the origin and the line:

Let's consider a point P(x1, y1) on the given line such that the line segment OP is perpendicular to the given line.

The equation of the given line is 3x + 4y + 1 = 0, which can be written as y = (-3/4)x - 1/4.

The slope of the line is -3/4, so the slope of the line perpendicular to it is 4/3.

Now, let's assume that point P(x1, y1) lies on the given line. So, we have:

y1 = (-3/4)x1 - 1/4

The equation of the line perpendicular to the given line passing through point P is:

y - y1 = (4/3)(x - x1)

Simplifying this equation, we get:

y = (4/3)x + (4/3)x1 - y1

Now, we need to find the point where this line intersects the x-axis (i.e., y = 0).

Putting y = 0 in the above equation, we get:

x = -(3/4)(4/3)x1 + (3/4)y1

x = -x1 + (3/4)y1

Now, we have two points: O(0, 0) and Q(-x1 + (3/4)y1, 0).

The distance between O and Q is:

D = sqrt[(x1 - (3/4)y1)^2 + 0^2]

We need to minimize this distance, which means we need to minimize the expression (x1 - (3/4)y1)^2.

Using the equation of the given line, we can write y1 = (-3/4)x1 - 1/4.

Substituting this in the above expression, we get:

D^2 = (x1 - (3/4)(-3/4)x1 - 3/16)^2

D^2 = (25/16)x1^2 + (9/16)x1 + 9/256

This is a quadratic equation in x1. To minimize D^2, we need to find the vertex of this parabola.

The x-coordinate of the vertex is given by:

x1 = -b/2a = -(9/16)/(2(25/16)) = -9/25

Substituting this value of x1 in the expression for D^2, we get:

D^2 = 9/256 - (9/25)(3/16) + (25/16)(9/625)

D^2 = 9/625

D = sqrt(9/625) = 3/25

Therefore, the perpendicular distance of the origin from the given line is 3/25, which is option B.