Question

15.
Write an equation for the line that is parallel to the given line and that passes through the given point.

y = –5x + 3; (–6, 3)

Answer 1

To write the equation of a line that is parallel to the given line \( y = -5x + 3 \) and passes through the point \((-6, 3)\), we need to follow these steps:

1. **Identify the slope of the given line.**
The equation \( y = -5x + 3 \) is in slope-intercept form, \( y = mx + b \), where \( m \) is the slope. Here, the slope \( m \) is \(-5\).

2. **Use the same slope for the parallel line.**
Since parallel lines have the same slope, our new line will also have a slope of \(-5\).

3. **Use the point-slope form of a line equation.**
The point-slope form of a line is given by:
\[
y - y_1 = m(x - x_1)
\]
where \( m \) is the slope and \((x_1, y_1)\) is the point the line passes through. Here, \((x_1, y_1) = (-6, 3)\) and \( m = -5 \).

4. **Substitute the values into the point-slope form:**
\[
y - 3 = -5(x - (-6))
\]
\[
y - 3 = -5(x + 6)
\]

5. **Distribute and simplify:**
\[
y - 3 = -5x - 30
\]
\[
y = -5x - 30 + 3
\]
\[
y = -5x - 27
\]

Thus, the equation of the line that is parallel to the given line and passes through the point \((-6, 3)\) is:
\[
\boxed{y = -5x - 27}
\]