Arbitrarily choose x = -5 and 
substitute into

3x + 4y = 1

by replacing x by (-5)

3(-5) + 4y = 1
  -15 + 4y = 1

Add 15 to both sides:

       4y = 1+15
       4y = 16

Divide both sides by 4

       y = 4

So (-5,4) is a point on the line

Arbitrarily choose x = 7 and 
substitute into

3x + 4y = 1

by replacing x by (7)

3(7) + 4y = 1
   21 + 4y = 1

Add -21 to both sides:

       4y = 1-21
       4y = -20

Divide both sides by 4

       y = -5

So (7,-5) is a point on the line

So plot those two points:

 

Now draw a line through them:

 

Now plot the point (2,5)



Now construct a perpendicular line from (2,5) to the line



It appears that this perpendicular touches the line at the point
(-1,1).  Let's stop and make sure that point really is on the line:

Substituting (x,y) = (-1,1)

     3x + 4y = 1
3(-1) + 4(1) = 1
      -3 + 4 = 1
           1 = 1

Yes (-1,1) is on the line.  So all we need to do is to use
the distance formula:
     ___________________
d = Ö(x2-x1)² + (y2-y1)²

where (x1,y1) = (2,5) and (x2,y2) = (-1,1)
     ________________
d = Ö(-1-2)² + (1-5)²
     _____________
d = Ö(-3)² + (-4)²
     ______
d = Ö9 + 16
     __
d = Ö25

d = 5

So the distance is 5 units.

Edwin