Question 180706

{{{2x+y=-3}}} Start with the given equation.



{{{y=-3-2x}}} Subtract {{{2x}}} from both sides.



{{{y=-2x-3}}} Rearrange the terms.



We can see that the equation {{{y=-2x-3}}} has a slope {{{m=-2}}} and a y-intercept {{{b=-3}}}.



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

Now let's use the point slope formula to find the equation of the parallel line by plugging in the slope {{{m=-2}}}  and the coordinates of the given point *[Tex \LARGE \left\(-4,-5\right\)].



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



{{{y--5=-2(x--4)}}} Plug in {{{m=-2}}}, {{{x[1]=-4}}}, and {{{y[1]=-5}}}



{{{y--5=-2(x+4)}}} Rewrite {{{x--4}}} as {{{x+4}}}



{{{y+5=-2(x+4)}}} Rewrite {{{y--5}}} as {{{y+5}}}
 


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Answer:


So the equation of the line, in point-slope form, parallel to {{{2x+y=-3}}} that goes through the point *[Tex \LARGE \left\(-4,-5\right\)] is {{{y+5=-2(x+4)}}}.



In short, the equation that you're looking for is {{{y+5=-2(x+4)}}}



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,(-3-2x)/1,-2(x+4)-5),
circle(-4,-5,0.08),
circle(-4,-5,0.10),
circle(-4,-5,0.12)
)}}}Graph of the original equation {{{2x+y=-3}}} (red) and the parallel line {{{y+5=-2(x+4)}}} (green) through the point *[Tex \LARGE \left\(-4,-5\right\)].