Question

Write the equation of a line that is perpendicular to the given line and that passes through the given point.

2x + 9y = –30; (–8, 0)


y = x + 36


y = x + 36


y = x – 8


y = x + 36

Answer 1

To find the equation of a line that is perpendicular to the given line \(2x + 9y = -30\) and passes through the point \((-8, 0)\), we need to follow these steps:

1. **Find the slope of the given line.**
First, we will convert the equation of the given line into the slope-intercept form \(y = mx + b\).

\[
2x + 9y = -30
\]
Subtract \(2x\) from both sides:

\[
9y = -2x - 30
\]
Now, divide by 9:

\[
y = -\frac{2}{9}x - \frac{30}{9}
\]
This shows that the slope \(m_1\) of the given line is \(-\frac{2}{9}\).

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, the slope \(m_2\) of the line we want to find is:

\[
m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{2}{9}} = \frac{9}{2}
\]

3. **Use the point-slope form to write the equation of the perpendicular line.**
We can use the point-slope form of a linear equation:

\[
y - y_1 = m(x - x_1)
\]

Here, \((x_1, y_1) = (-8, 0)\) and \(m = \frac{9}{2}\):

\[
y - 0 = \frac{9}{2}(x - (-8))
\]

Simplifying this gives:

\[
y = \frac{9}{2}(x + 8)
\]
\[
y = \frac{9}{2}x + \frac{9}{2} \cdot 8
\]
\[
y = \frac{9}{2}x + 36
\]

So, the equation of the line that is perpendicular to the given line \(2x + 9y = -30\) and passes through the point \((-8, 0)\) is:

\[
y = \frac{9}{2}x + 36
\]