Find an equation of the line that satisfies the given conditions. Through \( (1,-1) \); parallel to the line \( x+2 y=6 \)

To find the equation of the line that passes through the point \( (1, -1) \) and is parallel to the line given by the equation \( x + 2y = 6 \), we can follow these steps: 1. **Determine the slope of the given line**: We need to rewrite the equation \( x + 2y = 6 \) in slope-intercept form \( y = mx + b \), where \( m \) is the slope. Rearranging the equation: \[ 2y = -x + 6 \] \[ y = -\frac{1}{2}x + 3 \] From this, we can see that the slope \( m \) of the given line is \( -\frac{1}{2} \). 2. **Use the slope for the new line**: Since the new line is parallel to the given line, it will have the same slope. Therefore, the slope of the new line is also \( -\frac{1}{2} \). 3. **Use the point-slope form of the equation**: The point-slope form of a line is given by: \[ y - y_1 = m(x - x_1) \] where \( (x_1, y_1) \) is a point on the line and \( m \) is the slope. Here, \( (x_1, y_1) = (1, -1) \) and \( m = -\frac{1}{2} \). Plugging in the values: \[ y - (-1) = -\frac{1}{2}(x - 1) \] Simplifying this: \[ y + 1 = -\frac{1}{2}x + \frac{1}{2} \] \[ y = -\frac{1}{2}x + \frac{1}{2} - 1 \] \[ y = -\frac{1}{2}x - \frac{1}{2} \] 4. **Final equation**: The equation of the line in slope-intercept form is: \[ y = -\frac{1}{2}x - \frac{1}{2} \] Thus, the equation of the line that passes through the point \( (1, -1) \) and is parallel to the line \( x + 2y = 6 \) is: \[ y = -\frac{1}{2}x - \frac{1}{2} \]