Question 164587
1. Each block is 1 unit. Distance is 4 units.
{{{drawing( 300, 300, -5, 5, -5, 5,grid( 1 ),green(line( -100,-3,100,-3)),
green(line(-100,1,100,1)))}}}

2.Distance is 6 units.
{{{drawing( 300, 300, -5, 5, -5, 5,grid( 1 ),green(line( -2,-100,-2,100)),
green(line(4,-100,4,100)))}}}


3.This distance takes a little bit more work.
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{{{drawing( 300, 300, -5, 5, -5, 5,grid( 1 ),green(line( -5,-8,5,12)),
green(line(-5,-13,5,7)))}}}
A line perpendicular to the parallel lines will have a slope that satisfies this relationship,
{{{m[1]m[2]=-1}}}
{{{2*m[2]=-1}}}
{{{m[2]=-1/2}}}
The equation of the line that has that slope and goes through point (0,2) (that is, intersects y=2x+2 at (0,2))is
{{{y-y[p]=m(x-x[p])}}}
{{{y-2=-(1/2)x}}}
{{{y=-(1/2)x+2}}}
This new line intersects line y=2x-3 at the following x,
{{{2x-3=-(1/2)x+2}}}
{{{2.5x=5}}}
{{{x=2}}}
with corresponding y,
{{{y=2x-3}}}
{{{y=4-3=1}}}
Intersects at (2,1)
{{{drawing( 300, 300, -5, 5, -5, 5,grid( 1 ),circle(0,2,.2),circle(2,1,0.2),blue(line(2,1,0,2)),green(line( -5,-8,5,12)),
green(line(-5,-13,5,7)))}}}
The distance between the lines equals the distance between (0,2) and (2,1).
{{{D^2=(x[1]-x[2])^2+(y[1]-y[2])^2}}}
{{{D^2=(0-2)^2+(2-1)^2}}}
{{{D^2=4+1}}}
{{{D=sqrt(5)}}}