The equation of the line parallel to the given line that contains point $ A $ is:
$ \color{blue}{ 2x+3y-11=0 }$ ( General form )
$ \color{blue}{ y = - \frac{ 2 }{ 3 } x + \frac{ 11 }{ 3 } } ~~~$ ( Slope y-intercept form )
Step 1: The first thing that we need to do is to find the slope of a given line.
The explicit equation of a given line is: $ y = - \frac{ 2 }{ 3 } x - \frac{ 1 }{ 9 } $
So the slope is $ m = -\frac{ 2 }{ 3 } $.
Step 2: Parallel lines have the same slope, so the slope of the unknown line ($ m_1 $) will also be $ -\frac{ 2 }{ 3 } $. So the parallel line will have a slope of $ m_1 = -\frac{ 2 }{ 3 } $
Step 3: Now we have a point and the slope so we can use point-slope form, which is:
$$ y - y_0 = m_1 (x - x_0) $$In this example we have: $ m_1 = -\frac{ 2 }{ 3 } $ , $ x_0 = 4 $ and $ y_0 = 1 $. After substitution we have:
$$ \begin{aligned} y - y_0 =& ~ m_1 (x - x_0) \\ y - 1 =& ~ -\frac{ 2 }{ 3 } ( x - 4) \\y -1 =& ~ -\frac{ 2 }{ 3 } x + \frac{ 8 }{ 3 } \\y =& ~ -\frac{ 2 }{ 3 } x + \frac{ 8 }{ 3 } + 1 \\y =& ~ - \frac{ 2 }{ 3 } x + \frac{ 11 }{ 3 }\\ \end{aligned} $$