Question 16749
y=1/2x is the graph of a straight line with y intercept 0 and slope m=1/2
y=1/2x + 5 is the graph of a straight line with y intercept 5 and slope m = 1/2.


The second line is exactly the same as the first line, except that it is moved "up" 5 units.  


If you rewrite these equations in standard form there is a formula for the distance from a point on a given line to another given line.  Let's rewrite these equations in standard form as 
y = 1/2x,  which is 2y= x  or x-2y=0
y=1/2x + 5, which is 2y = x + 10 or x-2y +10= 0


The formula (if I remember correctly from about 15 years ago!!)is : 
Distance from a point to a given line:  {{{ d = (abs(Ax+By + C))/(sqrt(A^2 + B^2)) }}}, where Ax+By+C = 0 is the equation of the line, and (x,y) is a given point (in this case, any point you choose to select from the other line--let's say the point (0,0) on the first line, since this is an easy point to see and to calculate).


So, let (0,0) be a point on the first line, and find the perpendicular distance from (0,0) to the line x-2y+10 = 0.


In the formula, A= 1, B=-2, C= 10, x=0, and y=0:

{{{ d = (abs(Ax+By + C))/(sqrt(A^2 + B^2)) }}}
{{{ d = (abs(1*0+(-2)*0 + 10))/(sqrt(1^2 + (-2)^2)) }}}
{{{d= 10/sqrt(5) = (10*sqrt(5))/(sqrt(5)*sqrt(5)) }}}
{{{d = (10*sqrt(5))/5}}}
{{{d = 2*sqrt(5)}}}


Someone may have an easier way, but this is what I recall from a LONG time ago!


R^2 at SCC