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)

1 answer

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: \[ \boxed{y = -5x - 27} \]