Question

Write an equation of the line that passes through negative 3, -1 and is perpendicular to the line y = -1/9x - 6

Answer 1

To find the equation of a line that passes through the point \((-3, -1)\) and is perpendicular to the line given by the equation \(y = -\frac{1}{9}x - 6\), we need to follow these steps:

1. **Determine the slope of the given line:**
The slope of the line \(y = -\frac{1}{9}x - 6\) is \(-\frac{1}{9}\).

2. **Find the slope of the perpendicular line:**
The slope of a line that is perpendicular to another is the negative reciprocal of the original slope. Thus, the slope \(m\) of the line we want to find is:
\[
m = -\frac{1}{(-\frac{1}{9})} = 9
\]

3. **Use the point-slope form to write the equation of the line:**
The point-slope form of the equation of a line is given by:
\[
y - y_1 = m(x - x_1)
\]
where \((x_1, y_1)\) is a point on the line and \(m\) is the slope. Using the point \((-3, -1)\) and the slope \(9\):
\[
y - (-1) = 9(x - (-3))
\]
Simplifying this, we get:
\[
y + 1 = 9(x + 3)
\]

4. **Distribute and rearrange to the slope-intercept form:**
Distributing \(9\):
\[
y + 1 = 9x + 27
\]
Then, subtract \(1\) from both sides:
\[
y = 9x + 26
\]

Thus, the equation of the line that passes through \((-3, -1)\) and is perpendicular to the line \(y = -\frac{1}{9}x - 6\) is:
\[
\boxed{y = 9x + 26}
\]