we have that
5x-12y-5=0 ----> 12y=5x-5----> y=(5/12)x-(5/12)------> line 1
and
5x-12y+21=0--> 12y=5x+21----> y=(5/12)x+(21/12)------> line 2
we know that
Line 1 and Line 2 are parallel-------> m1=m2 (the slopes are the same)
the shortest distance between the line 1 and line 2, will be located on the perpendicular line
step 1
select any point on line 1
for x=0
y=(5/12)*0-(5/12)------> y=-5/12----> y=-0.417
Let
A (0,-0.417)
step 2
find the equation of the perpendicular line to line 1
m1=5/12
the slope of the perpendicular line is m=-1/m1-----> m=-12/5
with point A (0,-0.417) and m=-12/5
y-y1=m*(x-x1)-----> y-(-0.417)=(-12/5)*(x-0)----> y=(-12/5)x-(0.417)
step 3
find the intersection point of the perpendicular line and line 2
so
resolve the system
y=(-12/5)x-(5/12)
y=(5/12)x+(21/12)
using a graph tool
see the attached figure
the solution is the point B (-0.769,1.429)
step 4
find the distance between point A and point B
A (0,-0.417)
B (-0.769,1.429)
d=√[(y2-y1)²+(x2-x1)²]-----> d=√[(1.429+0.417)²+(-0.769)²]
d=√4------> d=2 units
the answer is
the shortest distance is 2 units
