Respuesta :
The Solution:
Given the equation of a line:
[tex]6x-y-4=0[/tex]We are required to find the equation of a line that is perpendicular to the given line and passes through the point (-12,-1).
Step 1:
Determine the slope of 6x-y-4=0.
[tex]\begin{gathered} \text{ Express y in terms of x.} \\ 6x-4=y \\ y=6x-4 \end{gathered}[/tex]The coefficient x, after making y the subject of the formula, is the slope of the given line. That is:
[tex]\begin{gathered} \text{ The slope of }6x-y-4=0\text{ is 6} \\ \\ Slope=m_1=6 \end{gathered}[/tex]Step 2:
Find the slope of the required line:
Since the two lines are perpendicular, the slope is:
[tex]\begin{gathered} m_2=\frac{-1}{m_1}=\frac{-1}{6} \\ \end{gathered}[/tex]Step 3:
Find the equation of the line that passes through point (-12,-1)
By formula,
[tex]y-y_1=m_2(x-x_1)[/tex]In this case,
[tex]\begin{gathered} x_1=-12 \\ y_1=-1 \end{gathered}[/tex]Substituting these values in the formula, we get the required equation is:
[tex]\begin{gathered} y--1=\frac{-1}{6}(x--12) \\ \\ y+1=-\frac{1}{6}(x+12) \end{gathered}[/tex]Cross multiplying, we get
[tex]\begin{gathered} 6y+6=-x-12 \\ \\ 6y+x+6+12=0 \\ \\ 6y+x+18=0 \end{gathered}[/tex]Therefore, the correct answer is:
[tex]6y+x+18=0[/tex]
