Question
Find the line that is perpendicular to $ y = - \frac{ 8 }{ 3 } x + 5 $ and passes though the point $ \left(3,~5\right) $.
Answer
The equation of the line perpendicular to the given line that contains point $ A $ is:
$ \color{blue}{ 3x-8y+31=0 }$ ( General form )
$ \color{blue}{ y = \dfrac{ 3 }{ 8 } x + \dfrac{ 31 }{ 8 } } ~~~$ ( Slope y-intercept form )
Explanation
Step 1:The slope of a given line is $ m = -\frac{ 8 }{ 3 } $.
Step 2: The perpendicular slope ($ m_1 $) is negative reciprocal of the slope $ m $.
$$ m_1 = - \frac{1}{m} = -\frac{ 1 }{ -\frac{ 8 }{ 3 } } = \frac{ 3 }{ 8 } $$So the perpendicular line will have a slope of $ m_1 = \frac{ 3 }{ 8 } $
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 }{ 8 } $ , $ x_0 = 3 $ and $ y_0 = 5 $. After substitution we have:
$$ \begin{aligned} y - y_0 =& ~ m_1 (x - x_0) \\ y - 5 =& ~ \frac{ 3 }{ 8 } ( x - 3) \\y -5 =& ~ \frac{ 3 }{ 8 } x -\frac{ 9 }{ 8 } \\y =& ~ \frac{ 3 }{ 8 } x -\frac{ 9 }{ 8 } + 5 \\y =& ~ \frac{ 3 }{ 8 } x + \frac{ 31 }{ 8 }\\ \end{aligned} $$This page was created using
Parallel and perpendicular lines calculator