Question 978141
Line {{{blue(y=2x-3)}}} is parallel to line {{{red(y=2x+7)}}} ,
which is perpendicular to {{{green(y=-0.5x+7)}}} ,
so {{{blue(y=2x-3)}}} and {{{green(y=-0.5x+7)}}} are perpendicular to each other.
So the distance between {{{blue(y=2x-3)}}} and {{{red(y=2x+7)}}}
is the distance, along {{{green(y=-0.5x+7)}}} ,
between the points where {{{green(y=-0.5x+7)}}} intersects the other two lines.
Lines {{{green(y=-0.5x+7)}}} and {{{red(y=2x+7)}}} obviously intersect at point {{{Q(0,7)}}} , the y-intercept for both lines.
We can find the point {{{P}}} where {{{green(y=-0.5x+7)}}} and {{{blue(y=2x-3)}}} intersect
by solving {{{system(green(y=-0.5x+7),blue(y=2x-3))}}}
I wanted to graph anyway, and in the graph it looked like it was {{{P(4,5)}}} .
Substituting {{{x=4}}} to see that {{{y=5}}} for both lines confirmed that solution:
{{{green(y=-0.5*4+7=-2+7=5)}}} and {{{blue(y=2*4-3=8-3=5))}}}


{{{drawing(300,400,-2,6,-4,8,grid(0),
red(line(-3,1,1,9)),
blue(line(-2,-7,6,9)),
green(line(-2,8,6,4)),
line(0.4,7.133,0.133,7.267),line(0.267,6.867,0.4,7.133),
line(3.866,5.4,4.133,5.267),line(3.866,5.4,3.733,5.133),
triangle(0,5,0,7,4,5),
rectangle(0,5,0.3,5.3),
locate(4.2,5.3,P(4,5)),locate(-1.5,7.1,Q(0,7)),
locate(1,-1,blue(y=2x-3)),
locate(3.8,4,green(y=-0.5x+7)),
locate(-2,3,red(y=2x+7))
 )}}} 
The distance from {{{Q{0,7)}}} and {{{P(4,5)}}}
is the distance between {{{red(y=2x+7)}}} and {{{blue(y=2x-3)}}} .
The distance between two points can be calculated according to a formula:
{{{PQ=sqrt((x[P]-x[Q])^2+(y[P]-y[Q])^2)}}} ,
but there is no need (usually) to memorize formulas,
if you understand their meaning.
That distance is the hypotenuse of a right triangle,
like the one in my drawing,
whose legs, parallel to the x-axis and y-axis,
are the horizontal distance and the vertical distance between the points.
In this case,
{{{distance=sqrt(4^2+2^2)=sqrt(16+4)=sqrt(20)=2sqrt(5)=about4.47}}}(rounded).