Question 1035086
Change this description a little bit.  The slopes of both lines must be  {{{5/12}}} for the description to be meaningful.  The two equations of these lines are then  5x+12y-1=0  and  5x+12y+j=0,  and you want to find j.  The two lines are parallel, and the distance between them is 2.



Here is 5x+12y-1=0 graphed:
{{{graph(300,300,-3,4,-3,4,-5x/12+1/12)}}}
The equation can also be stated {{{y=-(5/12)x+1/12}}}.


Choose a convenient point on this line, maybe  the x-intercept,  ( 1/5 ,0).


You want the other line,  {{{5x+12y+j=0}}}, to be 2 unit away from  5x+12y-1=0; and you can pick either above OR below.  Put the equation of the line you wish to find in slope-intercept form, as {{{y=-(5/12)x-j/12}}}.


Summarizing all this,  we want  distance from {{{y=-(5/12)x+1/12}}}  to {{{y=-(5/12)x-j/12}}}  to be equal to 2.  To be much more specific, referring to a given or desired point on the first line graphed,  <b>we want  the distance between  ( 1/5, 0 ) and {{{y=-(5/12)x+j/12}}}  to be equal to 2.</b>



...
Keep reading and thinking on that.
A general point for the equation of the line you want to find is an ordered pair point   ( x, -(5/12)x+j/12 ).
When you have this then use the distance formula to set up the equation necessary to solve for j.