Question 1126593
{{{y=mx+b}}}

 an equation of the line that passes through the  point

 ({{{x}}},{{{   y}}})=({{{8}}},{{{ -3}}})

 
 and is parallel to the given line {{{4y - 3x = 20}}}

first recall that parallel lines have same slope

so, write {{{4y - 3x = 20}}} in slope-intercept form {{{y=mx+b}}}

 {{{4y  =3x+ 20}}}

{{{y  =3x/4+ 20/4}}} 

{{{y  =(3/4)x+ 5}}} => as you can see a slope {{{m=3/4}}}


so far, the equation of the parallel line is

{{{y=(3/4)x+b}}} ...now use given point ({{{x}}},{{{   y}}})=({{{8}}},{{{ -3}}}) to find {{{b}}}

{{{-3=(3/4)8+b}}}

{{{-3=(3/cross(4))cross(8)2+b}}}

{{{-3=3*2+b}}}

{{{-3=6+b}}}

{{{b=-3-6}}}

{{{b=-9}}}

and, your equation is:

{{{y=(3/4)x-9}}}


{{{drawing( 600, 600, -15, 15, -15, 15,
circle(8,-3,.12), locate(8,-3,p(8,-3)),locate(5,-5,y=(3/4)x-9),locate(3,6,y=(3/4)x+ 5),
 graph( 600, 600, -15, 15, -15, 15, (3/4)x-9,(3/4)x+ 5)) }}}