Question 166654

{{{3y-12x=7}}} Start with the given equation.



{{{3y=7+12x}}} Add {{{12x}}} to both sides.



{{{3y=12x+7}}} Rearrange the terms.



{{{y=(12x+7)/(3)}}} Divide both sides by {{{3}}} to isolate y.



{{{y=((12)/(3))x+(7)/(3)}}} Break up the fraction.



{{{y=4x+7/3}}} Reduce.




We can see that the equation {{{y=4x+7/3}}} has a slope {{{m=4}}} and a y-intercept {{{b=7/3}}}.



Since parallel lines have equal slopes, this means that we know that the slope of the unknown parallel line is {{{m=4}}}.



Now let's use the point slope formula to find the equation of the parallel line by plugging in the slope {{{m=4}}}  and the coordinates of the given point *[Tex \LARGE \left\(0,6\right\)].



{{{y-y[1]=m(x-x[1])}}} Start with the point slope formula



{{{y-6=4(x-0)}}} Plug in {{{m=4}}}, {{{x[1]=0}}}, and {{{y[1]=6}}}



{{{y-6=4x+4(0)}}} Distribute



{{{y-6=4x+0}}} Multiply



{{{y=4x+0+6}}} Add 6 to both sides. 



{{{y=4x+6}}} Combine like terms. 



So the equation of the line parallel to {{{3y-12x=7}}} that goes through the point *[Tex \LARGE \left\(0,6\right\)] is {{{y=4x+6}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,4*x+7/3,4x+6),
circle(0,6,0.08),
circle(0,6,0.10),
circle(0,6,0.12))}}}Graph of the original equation {{{y=4x+7/3}}} (red) and the parallel line {{{y=4x+6}}} (green) through the point *[Tex \LARGE \left\(0,6\right\)].