Answer :
The given point is (-2,1) and the given line is x-3y=5.
The slope of the given line can be determined as,
[tex]m=-\frac{1}{-3}=\frac{1}{3}[/tex]The equation of the line which is parallel to the given line will have the same slope.
Thus, the equation of the new line passing through (-2,1) can be determined as,
[tex]\begin{gathered} \frac{y-1}{x-(-2)}=\frac{1}{3} \\ 3(y-1)=x+2 \\ 3y-3=x+2 \\ 3y=x+5 \\ y=\frac{x}{3}+\frac{5}{3} \end{gathered}[/tex]Thus, the equation of the line in slope intercept form is y=x/3+5/3.
The equation of the line in the point slope form can be determined as,
[tex]\begin{gathered} (y-1)=\frac{1}{3}(x-(-2) \\ (y-1)=\frac{1}{3}(x+2) \end{gathered}[/tex]Thus, the equation of the line in the point slope form is (y-1)=(x+2)/3
