Question 146251
We can see that the equation {{{y=(2/3)x+1}}} has a slope {{{m=2/3}}} and a y-intercept {{{b=1}}}.



Since parallel lines have equal slopes, this means that we know that the slope of the unknown parallel line is {{{m=2/3}}}.

Now let's use the point slope formula to find the equation of the parallel line by plugging in the slope {{{m=2/3}}}  and the coordinates of the given point *[Tex \LARGE \left\(0,-3\right\)].



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



{{{y--3=(2/3)(x-0)}}} Plug in {{{m=2/3}}}, {{{x[1]=0}}}, and {{{y[1]=-3}}}



{{{y+3=(2/3)(x-0)}}} Rewrite {{{y--3}}} as {{{y+3}}}



{{{y+3=(2/3)x+(2/3)(-0)}}} Distribute



{{{y+3=(2/3)x+0}}} Multiply



{{{y=(2/3)x+0-3}}} Subtract 3 from both sides. 



{{{y=(2/3)x-3}}} Combine like terms. 



So the equation of the line parallel to {{{y=(2/3)x+1}}} that goes through the point *[Tex \LARGE \left\(0,-3\right\)] is {{{y=(2/3)x-3}}}.



Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,(2/3)x+1,(2/3)x-3),
circle(0,-3,0.08),
circle(0,-3,0.10),
circle(0,-3,0.12))}}}Graph of the original equation {{{y=(2/3)x+1}}} (red) and the parallel line {{{y=(2/3)x-3}}} (green) through the point *[Tex \LARGE \left\(0,-3\right\)].