Question

6.
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)

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 can follow these steps:

1. **Find the slope of the given line**:

We need to rewrite the equation of the line in slope-intercept form (y = mx + b), where m is the slope.

Start with the equation:
\[
2x + 9y = -30
\]

Isolate \(y\):
\[
9y = -2x - 30
\]
\[
y = -\frac{2}{9}x - \frac{30}{9}
\]
\[
y = -\frac{2}{9}x - \frac{10}{3}
\]

The slope of the given line \( m_1 = -\frac{2}{9} \).

2. **Determine the slope of the perpendicular line**:

The slope of the line that is perpendicular to another line is the negative reciprocal of the original slope. Thus,
\[
m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{2}{9}} = \frac{9}{2}
\]

3. **Use the point-slope formula**:

Now that we have the slope \( m_2 = \frac{9}{2} \) and the point \((-8, 0)\), we can use the point-slope form of a line:
\[
y - y_1 = m(x - x_1)
\]

Plugging in our values:
\[
y - 0 = \frac{9}{2}(x - (-8))
\]
\[
y = \frac{9}{2}(x + 8)
\]

4. **Simplify the equation**:

Distributing the slope:
\[
y = \frac{9}{2}x + \frac{9 \cdot 8}{2}
\]
\[
y = \frac{9}{2}x + 36
\]

5. **Convert to standard form (if needed)**:

If we want to express it in standard form \(Ax + By = C\), we can multiply through by 2 to eliminate the fraction:
\[
2y = 9x + 72
\]
Rearranging gives us:
\[
-9x + 2y = 72
\]
or equivalently,
\[
9x - 2y = -72
\]

Thus, the equation of the line that is perpendicular to the given line and passes through the point \((-8, 0)\) is:
\[
\boxed{9x - 2y = -72}
\]