Question
Find the line that is parallel to $ y = \frac{ 3 }{ 10 } x + 8 $ and passes though the point $ \left(-7,~-2\right) $.
Answer
The equation of the line parallel to the given line that contains point $ A $ is:
$ \color{blue}{ 3x-10y+1=0 }$ ( General form )
$ \color{blue}{ y = \dfrac{ 3 }{ 10 } x + \dfrac{ 1 }{ 10 } } ~~~$ ( Slope y-intercept form )
Explanation
Step 1:The slope of a given line is $ m = \frac{ 3 }{ 10 } $.
Step 2: Parallel lines have the same slope, so the slope of the unknown line ($ m_1 $) will also be $ \frac{ 3 }{ 10 } $. So the parallel line will have a slope of $ m_1 = \frac{ 3 }{ 10 } $
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 }{ 10 } $ , $ x_0 = -7 $ and $ y_0 = -2 $. After substitution we have:
$$ \begin{aligned} y - y_0 =& ~ m_1 (x - x_0) \\ y - \left( -2\right) =& ~ \frac{ 3 }{ 10 } ( x - \left( -7\right)) \\y + 2 =& ~ \frac{ 3 }{ 10 } x + \frac{ 21 }{ 10 } \\y =& ~ \frac{ 3 }{ 10 } x + \frac{ 21 }{ 10 } -2 \\y =& ~ \frac{ 3 }{ 10 } x + \frac{ 1 }{ 10 }\\ \end{aligned} $$This page was created using
Parallel and perpendicular lines calculator