Question 252628


{{{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=-2*x+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\(5,4\right\)].



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



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



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



{{{y-4=-2x+10}}} Multiply



{{{y=-2x+10+4}}} Add 4 to both sides. 



{{{y=-2x+14}}} Combine like terms. 



So the equation of the line parallel to {{{2x+y=3}}} that goes through the point *[Tex \LARGE \left\(5,4\right\)] is {{{y=-2x+14}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,-2*x+3,-2x+14),
circle(5,4,0.08),
circle(5,4,0.10),
circle(5,4,0.12))}}}


Graph of the original equation {{{y=-2*x+3}}} (red) and the parallel line {{{y=-2x+14}}} (green) through the point *[Tex \LARGE \left\(5,4\right\)].