To determine the equation of the line using the given conditions – a slope of [tex]\(-\frac{1}{6}\)[/tex] and the point [tex]\((4, -2)\)[/tex] – we will proceed in two parts.
### Part 1: Point-Slope Form
The point-slope form of a linear equation is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Here, [tex]\( m \)[/tex] is the slope, and [tex]\( (x_1, y_1) \)[/tex] is the point through which the line passes.
Given:
- Slope [tex]\( m = -\frac{1}{6} \)[/tex]
- Point [tex]\( (x_1, y_1) = (4, -2) \)[/tex]
Plug these values into the point-slope form formula:
[tex]\[ y - (-2) = -\frac{1}{6}(x - 4) \][/tex]
Simplify the left side:
[tex]\[ y + 2 = -\frac{1}{6}(x - 4) \][/tex]
So, the equation of the line in point-slope form is:
[tex]\[ y + 2 = -\frac{1}{6}(x - 4) \][/tex]
### Part 2: Slope-Intercept Form
The slope-intercept form of a linear equation is given by:
[tex]\[ y = mx + b \][/tex]
We start with the point-slope form equation obtained:
[tex]\[ y + 2 = -\frac{1}{6}(x - 4) \][/tex]
Expand the right side:
[tex]\[ y + 2 = -\frac{1}{6}x + \frac{4}{6} \][/tex]
Simplify the constant term on the right side:
[tex]\[ y + 2 = -\frac{1}{6}x + \frac{2}{3} \][/tex]
Next, subtract 2 from both sides to solve for [tex]\( y \)[/tex]:
[tex]\[ y = -\frac{1}{6}x + \frac{2}{3} - 2 \][/tex]
Convert -2 to fractions to combine with [tex]\(\frac{2}{3}\)[/tex]:
[tex]\[ y = -\frac{1}{6}x + \frac{2}{3} - \frac{6}{3} \][/tex]
Subtract [tex]\(\frac{6}{3}\)[/tex] from [tex]\(\frac{2}{3}\)[/tex]:
[tex]\[ y = -\frac{1}{6}x - \frac{4}{3} \][/tex]
Thus, the equation of the line in slope-intercept form is:
[tex]\[ y = -\frac{1}{6}x - \frac{4}{3} \][/tex]
In summary:
- Point-slope form: [tex]\( y + 2 = -\frac{1}{6}(x - 4) \)[/tex]
- Slope-intercept form: [tex]\( y = -\frac{1}{6}x - \frac{4}{3} \)[/tex]
