Question

Write an equation of the line that passes through (−3,−1)
and is perpendicular to the line y=−19x−6
.(1 point)

Answer 1

To find the equation of the line that passes through the point \((-3, -1)\) and is perpendicular to the line given by the equation \(y = -19x - 6\), we need to determine the slope of the given line and then use the negative reciprocal of that slope for the perpendicular line.

1. **Identify the slope of the given line**: The equation \(y = -19x - 6\) is in slope-intercept form \(y = mx + b\), where \(m\) is the slope. Here, the slope (\(m_1\)) is \(-19\).

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. Therefore, if \(m_1 = -19\), then the slope of the perpendicular line (\(m_2\)) is:
\[
m_2 = -\frac{1}{m_1} = -\frac{1}{-19} = \frac{1}{19}
\]

3. **Use the point-slope form of the line equation**: 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. In our case, \((x_1, y_1) = (-3, -1)\) and \(m = \frac{1}{19}\).

Plugging in these values:
\[
y - (-1) = \frac{1}{19}(x - (-3))
\]
Simplifying this, we have:
\[
y + 1 = \frac{1}{19}(x + 3)
\]

4. **Rearranging to slope-intercept form**: Now, distribute \(\frac{1}{19}\):
\[
y + 1 = \frac{1}{19}x + \frac{3}{19}
\]

Then, subtract \(1\) (which is \( \frac{19}{19} \)) from both sides to solve for \(y\):
\[
y = \frac{1}{19}x + \frac{3}{19} - \frac{19}{19}
\]
\[
y = \frac{1}{19}x - \frac{16}{19}
\]

Thus, the equation of the line that passes through the point \((-3, -1)\) and is perpendicular to the line \(y = -19x - 6\) is:
\[
y = \frac{1}{19}x - \frac{16}{19}
\]