Question
Find the line that is perpendicular to $ y = \frac{ 1 }{ 4 } x - 2 $ and passes though the point $ \left(4,~-1\right) $.
Answer
The equation of the line perpendicular to the given line that contains point $ A $ is:
$ \color{blue}{ 4x+y-15=0 }$ ( General form )
$ \color{blue}{ y = - 4 x + 15 } ~~~$ ( Slope y-intercept form )
Explanation
Step 1:The slope of a given line is $ m = \frac{ 1 }{ 4 } $.
Step 2: The perpendicular slope ($ m_1 $) is negative reciprocal of the slope $ m $.
$$ m_1 = - \frac{1}{m} = -\frac{ 1 }{ \frac{ 1 }{ 4 } } = -4 $$So the perpendicular line will have a slope of $ m_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 = -4 $ , $ x_0 = 4 $ and $ y_0 = -1 $. After substitution we have:
$$ \begin{aligned} y - y_0 =& ~ m_1 (x - x_0) \\ y - \left( -1\right) =& ~ -4 ( x - 4) \\y + 1 =& ~ -4 x + 16 \\y =& ~ -4 x + 16 -1 \\y =& ~ - 4 x + 15\\ \end{aligned} $$This page was created using
Parallel and perpendicular lines calculator