Question 339008
I'll do the first one to get you started. Please only post one problem at a time. Thanks.




{{{3x+6y=-12}}} Start with the given equation.



{{{6y=-12-3x}}} Subtract {{{3x}}} from both sides.



{{{6y=-3x-12}}} Rearrange the terms.



{{{y=(-3x-12)/(6)}}} Divide both sides by {{{6}}} to isolate y.



{{{y=((-3)/(6))x+(-12)/(6)}}} Break up the fraction.



{{{y=-(1/2)x-2}}} Reduce.



We can see that the equation {{{y=-(1/2)x-2}}} has a slope {{{m=-1/2}}} and a y-intercept {{{b=-2}}}.



Since parallel lines have equal slopes, this means that we know that the slope of the unknown parallel line is {{{m=-1/2}}}.

Now let's use the point slope formula to find the equation of the parallel line by plugging in the slope {{{m=-1/2}}}  and the coordinates of the given point *[Tex \LARGE \left\(4,-2\right\)].



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



{{{y--2=(-1/2)(x-4)}}} Plug in {{{m=-1/2}}}, {{{x[1]=4}}}, and {{{y[1]=-2}}}



{{{y+2=(-1/2)(x-4)}}} Rewrite {{{y--2}}} as {{{y+2}}}



{{{y+2=(-1/2)x+(-1/2)(-4)}}} Distribute



{{{y+2=(-1/2)x+2}}} Multiply



{{{y=(-1/2)x+2-2}}} Subtract 2 from both sides. 



{{{y=(-1/2)x+0}}} Combine like terms. 



{{{y=(-1/2)x}}} Remove the trailing zero



So the equation of the line parallel to {{{3x+6y=-12}}} that goes through the point *[Tex \LARGE \left\(4,-2\right\)] is {{{y=(-1/2)x}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,-(1/2)x-2,(-1/2)x),
circle(4,-2,0.08),
circle(4,-2,0.10),
circle(4,-2,0.12))}}} Graph of the original equation {{{y=-(1/2)x-2}}} (red) and the parallel line {{{y=(-1/2)x}}} (green) through the point *[Tex \LARGE \left\(4,-2\right\)]. 



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Jim