Question

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

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 first need to determine the slope of the given line.

The equation of the line is in the slope-intercept form \(y = mx + b\), where \(m\) is the slope. From the equation \(y = -19x - 6\), we can see that the slope \(m\) is \(-19\).

For a line to be perpendicular to another line, the slopes of the two lines must multiply to \(-1\). Therefore, if the slope of the given line is \(-19\), the slope of the line we want to find (let's call it \(m'\)) is given by:

\[
m' = -\frac{1}{-19} = \frac{1}{19}
\]

Now, we have the slope of the line we want to find, which is \(\frac{1}{19}\), and it passes through the point \((-3, -1)\).

We can use the point-slope form of the equation of a line, which is:

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

where \((x_1, y_1)\) is the point the line passes through.

Substituting \((-3, -1)\) for \((x_1, y_1)\) and \(\frac{1}{19}\) for \(m\):

\[
y - (-1) = \frac{1}{19}(x - (-3))
\]

This simplifies to:

\[
y + 1 = \frac{1}{19}(x + 3)
\]

Next, we will distribute \(\frac{1}{19}\):

\[
y + 1 = \frac{1}{19}x + \frac{3}{19}
\]

Now, subtract \(1\) from both sides to solve for \(y\):

\[
y = \frac{1}{19}x + \frac{3}{19} - 1
\]

To combine \(\frac{3}{19}\) and \(-1\), we can rewrite \(-1\) as \(-\frac{19}{19}\):

\[
y = \frac{1}{19}x + \frac{3}{19} - \frac{19}{19}
\]

Now, simplifying that gives:

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

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

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