Question 1132638
The line {{{y=-(12/5) x+ 2}}} is exactly {{{3}}} units away from two other lines parallel to it. 

{{{y=-(12/5) x+ 2}}}...both sides multiply by {{{5}}}

{{{5y=-12 x+ 10}}}

{{{12x + 5y - 10 = 0 }}}


The distance from a point ({{{x}}},{{{ y}}}) to a line {{{ax + by + c = 0}}} is given by 

 distance = {{{ax + by + c)/sqrt(a^2+b^2) }}}


You have the line 

{{{12x + 5y - 10 = 0 }}}=>{{{a=12}}},{{{b=5}}},{{{c=10}}}


You want the distance from y-intercept point ({{{0}}}, {{{y}}}) to be {{{3}}}. 


{{{(12*0 + 5*y - 10)/sqrt(12^2+5^2)  = 3 }}}

{{{(5*y - 10)/sqrt(169)  = 3 }}}

{{{(5*y - 10)/13 = 3 }}} ... multiply by 13 

{{{5*y - 10 = 39 }}}

{{{5*y  = 39+10 }}}

{{{5*y  = 49}}}

{{{y  = 49/5}}}


This has two solutions:
 
{{{ 5y - 10 = 39}}} 
{{{ y = 49/5 }}}

And 
{{{ 5y - 10 = -39 }}}
{{{ y = -29/5 }}}


=> two other lines parallel to given line are:

{{{y=-(12/5) x+ 49/5}}} and {{{y=-(12/5) x-29/5}}} 


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(0,2,3),
 graph( 600, 600, -10, 10, -10, 10,-(12/5) x+ 49/5, -(12/5) x+ 2,-(12/5) x- 29/5)) }}}

In the graphic, the radius of the circle is {{{3}}}, showing the distance to the other two lines is {{{3}}}.


The difference of these two y-intercept values is 
{{{ 49/5 - (-29)/5 = 78/5 = 15.6 }}}=>{{{15}}}{{{6/10}}}=>{{{15}}}{{{3/5}}}

answer:  D){{{15}}}{{{ 3/5}}}