Question

Identify an equation in point-slope form for the line perpendicular to y = -4x - 1
that passes through (-2,7)

Answer 1

To find the equation of a line perpendicular to \( y = -4x - 1 \) that passes through the point \((-2, 7)\), we need to follow these steps:

1. **Determine the slope of the given line.**
The equation \( y = -4x - 1 \) is in slope-intercept form \( y = mx + b \), where \( m \) is the slope. Here, the slope \( m \) is \(-4\).

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 slope. Therefore, if the slope of the original line is \(-4\), the slope \( m_{\text{perpendicular}} \) is:
\[
m_{\text{perpendicular}} = -\frac{1}{-4} = \frac{1}{4}
\]

3. **Use the point-slope form of the equation of a line.**
The point-slope form of a line is given by the equation:
\[
y - y_1 = m(x - x_1)
\]
where \( (x_1, y_1) \) is a point on the line and \( m \) is the slope. Here, we will use the point \((-2, 7)\) and the slope \(\frac{1}{4}\):

4. **Substitute the values into the point-slope form.**
\[
y - 7 = \frac{1}{4}(x - (-2))
\]
Simplifying this:
\[
y - 7 = \frac{1}{4}(x + 2)
\]

This equation in point-slope form is:
\[
y - 7 = \frac{1}{4}(x + 2)
\]

This is the required equation of the line that is perpendicular to \( y = -4x - 1 \) and passes through the point \((-2, 7)\).