Answer
The equation of the line which passing through $\left( -12,\ -1 \right)$ and perpendicular to the line with equation $6x-y-4=0$ in the general form is $6x+y+18=0$.
Work Step by Step
The point on the line is given as $\left( -12,\ -1 \right)$ and the equation of the line parallel to the required line is $6x-y-4=0$.
It is known that if two lines are perpendicular, then the product of their slopes is -1.
The slope of the given line equation is 6.
Let the slope of the required line equation be m. Therefore,
$\begin{align}
& 6m=-1 \\
& m=-\frac{1}{6}
\end{align}$
Therefore, the slope of the required line equation will be $-\frac{1}{6}$.
Point-Slope form:
The equation of the line by point-slope form will be:
$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$
Now, substitute the given data in the above equation to get the desired equation of the line:
$\begin{align}
& y-\left( -1 \right)=-\frac{1}{6}\left\{ x-\left( -12 \right) \right\} \\
& -6\left( y+1 \right)=\left( x+12 \right) \\
& -6y-6=x+12 \\
& 6x+y+18=0.
\end{align}$
