Find the line that is parallel to $ 3x-5y-30=0 $ and passes though the point $ \left(5,~-3\right) $.
The equation of the line parallel to the given line that contains point $ A $ is:
$ \color{blue}{ 3x-5y-30=0 }$ ( General form )
$ \color{blue}{ y = \dfrac{ 3 }{ 5 } x - 6 } ~~~$ ( 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{ 3 }{ 5 } x - 6 $
So the slope is $ m = \frac{ 3 }{ 5 } $.
Step 2: Parallel lines have the same slope, so the slope of the unknown line ($ m_1 $) will also be $ \frac{ 3 }{ 5 } $. So the parallel line will have a slope of $ m_1 = \frac{ 3 }{ 5 } $
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{ 3 }{ 5 } $ , $ x_0 = 5 $ and $ y_0 = -3 $. After substitution we have:
$$ \begin{aligned} y - y_0 =& ~ m_1 (x - x_0) \\ y - \left( -3\right) =& ~ \frac{ 3 }{ 5 } ( x - 5) \\y + 3 =& ~ \frac{ 3 }{ 5 } x -3 \\y =& ~ \frac{ 3 }{ 5 } x -3 -3 \\y =& ~ \frac{ 3 }{ 5 } x - 6\\ \end{aligned} $$