Question
Find the line that is perpendicular to $ y = - 4 x + 9 $ and passes though the point $ \left(4,~6\right) $.
Answer
The equation of the line perpendicular to the given line that contains point $ A $ is:
$ \color{blue}{ x-4y+20=0 }$ ( General form )
$ \color{blue}{ y = \dfrac{ 1 }{ 4 } x + 5 } ~~~$ ( Slope y-intercept form )
Explanation
Step 1:The slope of a given line is $ m = -4 $.
Step 2: The perpendicular slope ($ m_1 $) is negative reciprocal of the slope $ m $.
$$ m_1 = - \frac{1}{m} = -\frac{ 1 }{ -4 } = \frac{ 1 }{ 4 } $$So the perpendicular line will have a slope of $ m_1 = \frac{ 1 }{ 4 } $
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{ 1 }{ 4 } $ , $ x_0 = 4 $ and $ y_0 = 6 $. After substitution we have:
$$ \begin{aligned} y - y_0 =& ~ m_1 (x - x_0) \\ y - 6 =& ~ \frac{ 1 }{ 4 } ( x - 4) \\y -6 =& ~ \frac{ 1 }{ 4 } x -1 \\y =& ~ \frac{ 1 }{ 4 } x -1 + 6 \\y =& ~ \frac{ 1 }{ 4 } x + 5\\ \end{aligned} $$This page was created using
Parallel and perpendicular lines calculator