To find the equation of the line that is perpendicular to a given line and passes through a specific point, follow these steps:
1. Identify the Slope of the Given Line:
The given line equation is [tex]\(2x + 12y = -1\)[/tex]. We need to convert this into the slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope.
[tex]\[
2x + 12y = -1
\][/tex]
Solve for [tex]\(y\)[/tex]:
[tex]\[
12y = -2x - 1 \\
y = \left(-\frac{2}{12}\right)x - \frac{1}{12} \\
y = \left(-\frac{1}{6}\right)x - \frac{1}{12}
\][/tex]
So, the slope [tex]\(m\)[/tex] of the given line is [tex]\(-\frac{1}{6}\)[/tex].
2. Determine the Slope of the Perpendicular Line:
The slope of a line that is perpendicular to another is the negative reciprocal of the original slope.
[tex]\[
\text{Given slope} = -\frac{1}{6} \\
\text{Perpendicular slope} = -\left(-\frac{1}{6}\right)^{-1} \\
= 6
\][/tex]
3. Use the Point-Slope Form:
We know the perpendicular line passes through the point [tex]\((0, 9)\)[/tex]. Using the point-slope form of a line equation, [tex]\(y - y_1 = m(x - x_1)\)[/tex], where [tex]\((x_1, y_1)\)[/tex] is the point it passes through and [tex]\(m\)[/tex] is the slope.
[tex]\[
y - 9 = 6(x - 0) \\
y - 9 = 6x \\
y = 6x + 9
\][/tex]
Thus, the equation of the line that is perpendicular to the given line and passes through the point [tex]\((0, 9)\)[/tex] is:
[tex]\[
y = 6x + 9
\][/tex]
So, the correct option is:
[tex]\[
\boxed{y = 6x + 9}
\][/tex]
