Question

Write an equation of the line that passes through (−3,−1)
(
−
3
,
−
1
)
and is perpendicular to the line y=−19x−6
𝑦
=
−
1
9
𝑥
−
6
.



Write a perpendicular line: Find Equation of Line perpendicular to a line through given point

(1 point)

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 \(y = -19x - 6\), we need to follow these steps:

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

2. **Find 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. So, the perpendicular slope will be:
\[
m_{\text{perpendicular}} = -\frac{1}{-19} = \frac{1}{19}
\]

3. **Use the point-slope form of the line equation**: The point-slope form of a line's equation that has a slope \(m\) and passes through a point \((x_0, y_0)\) is:
\[
y - y_0 = m(x - x_0)
\]
In our case, \(m = \frac{1}{19}\) and the point is \((-3, -1)\). Plugging in these values:
\[
y - (-1) = \frac{1}{19}(x - (-3))
\]
Simplifying this, we have:
\[
y + 1 = \frac{1}{19}(x + 3)
\]

4. **Distributing and rearranging to slope-intercept form**:
\[
y + 1 = \frac{1}{19}x + \frac{3}{19}
\]
\[
y = \frac{1}{19}x + \frac{3}{19} - 1
\]
\[
y = \frac{1}{19}x + \frac{3}{19} - \frac{19}{19}
\]
\[
y = \frac{1}{19}x - \frac{16}{19}
\]

So, the equation of the line that passes through \((-3, -1)\) and is perpendicular to \(y = -19x - 6\) is:

\[
y = \frac{1}{19}x - \frac{16}{19}
\]