The given line passes through the points (0, -3) and (2, 3).What is the equation, in point-slope form of the line that isparallel to the given line and passes through the point-1, - 1)?y+1=3(x+1)y+1=-=(x + 1)-532v+1={(x+1)y+1 =3(x+1)Mark this and returnSave and ExitS

Answer:[tex]y+1=3(x+1)[/tex]Step-by-step explanation:Ok so we are looking for line parallel to the line containing points (0,-3) and (2,3).Parallel lines have the same slope.So let's find the slope of the line containing the points (0,-3) and (2,3).You can use the formula [tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex].However, I just like to line up the points vertically and subtract them vertically, then put 2nd difference over 1st difference. Like this: (0 , -3)-(2 ,  3)------------2     -6So the slope is -6/-2 or just 3.So the slope of the line we are looking for has slope 3 (or m=3) and your line should contain the point (-1,-1).The point slope form of a line is:[tex]y-y_1=m(x-x_1)[/tex] where [tex]m[/tex] is the slope and [tex](x_1,y_1)[/tex] is a point you know on the line.So we just plug into that equation now.  That gives us:[tex]y-(-1)=3(x-(-1))[/tex]Simplify a bit:[tex]y+1=3(x+1)[/tex]