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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 find 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], follow these steps:

1. Convert the given line to slope-intercept form ( [tex]\( y = mx + b \)[/tex] ):

Start with the given equation:
[tex]\[ 2x + 12y = -1 \][/tex]

Solve for [tex]\( y \)[/tex]:
[tex]\[ 12y = -2x - 1 \][/tex]

Divide by 12:
[tex]\[ y = -\frac{2}{12}x - \frac{1}{12} \][/tex]

Simplify the expression:
[tex]\[ y = -\frac{1}{6}x - \frac{1}{12} \][/tex]

So, the slope [tex]\( m_1 \)[/tex] of the given line is [tex]\( -\frac{1}{6} \)[/tex].

2. Find the slope of the line perpendicular to the given line:

The slope of a line perpendicular to another is the negative reciprocal of the original line's slope. Therefore:
[tex]\[ m_2 = -\frac{1}{-\frac{1}{6}} = 6 \][/tex]

3. Use the point-slope form of the line equation:

The point-slope form is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\((x_1, y_1)\)[/tex] is the point the line passes through. In this case, the point is [tex]\((0, 9)\)[/tex]:

Substitute [tex]\( m_2 = 6 \)[/tex], [tex]\( x_1 = 0 \)[/tex], and [tex]\( y_1 = 9 \)[/tex] into the equation:
[tex]\[ y - 9 = 6(x - 0) \][/tex]

Simplify the equation:
[tex]\[ y - 9 = 6x \][/tex]

Solving for [tex]\( y \)[/tex]:
[tex]\[ y = 6x + 9 \][/tex]

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

Hence, the correct answer is:
[tex]\[ y = 6x + 9 \][/tex]
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