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A given line has the equation [tex]2x + 12y = -1[/tex].

What is the equation, in slope-intercept form, of the line that is perpendicular to the given line and passes through the point [tex](0, 9)[/tex]?

A. [tex]y = -6x + 9[/tex]

B. [tex]y = -\frac{1}{6}x + 9[/tex]

C. [tex]y = \frac{1}{6}x + 9[/tex]

D. [tex]y = 6x + 9[/tex]


Sagot :

To solve the problem of finding the equation of the line perpendicular to the given line [tex]\(2x + 12y = -1\)[/tex] that passes through the point [tex]\((0, 9)\)[/tex], we can follow these steps:

### Step 1: Convert the given line equation to slope-intercept form.
We begin with the given line equation:
[tex]\[ 2x + 12y = -1 \][/tex]

First, we need to solve for [tex]\(y\)[/tex] to get the equation in the form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\(b\)[/tex] is the y-intercept.

[tex]\[ 12y = -2x - 1 \][/tex]
[tex]\[ y = \frac{-2}{12}x - \frac{1}{12} \][/tex]
[tex]\[ y = -\frac{1}{6}x - \frac{1}{12} \][/tex]

So, the slope [tex]\(m\)[/tex] of the given line is [tex]\(-\frac{1}{6}\)[/tex].

### Step 2: Determine the slope of the perpendicular line.
The slope of a line that is perpendicular to another line is the negative reciprocal of the original line's slope. Therefore, if the slope of the given line is [tex]\(-\frac{1}{6}\)[/tex], the slope of the perpendicular line is:
[tex]\[ m_{\text{perpendicular}} = -\left( -\frac{1}{6} \right)^{-1} = 6 \][/tex]

### Step 3: Use the point-slope form to find the equation of the perpendicular line.
We know the slope [tex]\(m_{\text{perpendicular}} = 6\)[/tex] and the line passes through the point [tex]\((0, 9)\)[/tex]. We can use the point-slope form of the equation of a line, which is [tex]\(y = mx + b\)[/tex], to find the equation of the perpendicular line.

Since the line passes through [tex]\((0, 9)\)[/tex], the y-intercept [tex]\(b\)[/tex] is simply 9.

Therefore, the equation of the line in slope-intercept form is:
[tex]\[ y = 6x + 9 \][/tex]

### Conclusion:
The equation of the line that is perpendicular to the given line [tex]\(2x + 12y = -1\)[/tex] and passes through the point [tex]\((0, 9)\)[/tex] is:
[tex]\[ y = 6x + 9 \][/tex]

So, the correct answer is:
[tex]\[ \boxed{y = 6x + 9} \][/tex]