The equation of the given line is,
[tex]y=-3x-1\text{ ------(1)}[/tex]We have to find the equation of a line parallel to y=-3x-1 and passing through the point (x1, y1)=(2, -3).
The general equation of a line is,
[tex]y=mx+c\text{ ------(2)}[/tex]Here, m is the slope of the line.
Comparing equations (1) and (2), we find that the slope of the line y=-3x-1 is m=-3.
The slope of two parallel lines are always equal. Hence, the slope of a line parallel to the line y=-3x-1 is also m=-3.
Now, the formula for the point slope form of the equation of a line can be written as,
[tex]y-y1=m(x-x1)[/tex]Substitute m=-3, x1=2 and y1=-3 in the above equation to find the equation of a line paralle to y=-3x-1 and passing through point (2, -3) is,
[tex]\begin{gathered} y-(-3)=-3(x-2) \\ y+3=-3x-3\times(-2) \\ y+3=-3x+6 \\ y=-3x+6-3 \\ y=-3x+3 \end{gathered}[/tex]Therefore, the equation of a line paralle to y=-3x-1 and passing through point (2, -3) is y=-3x+3.
