To determine which equation represents a line that passes through the point [tex]\((5, 1)\)[/tex] with a slope of [tex]\(\frac{1}{2}\)[/tex], we can start by using the point-slope form of a linear equation, which is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Here, [tex]\((x_1, y_1)\)[/tex] is a point on the line and [tex]\(m\)[/tex] is the slope. Given:
- Point: [tex]\((5, 1)\)[/tex]
- Slope: [tex]\(\frac{1}{2}\)[/tex]
We can plug these values into the point-slope form equation:
[tex]\[ y - 1 = \frac{1}{2}(x - 5) \][/tex]
Now we compare this with the given equations:
1. [tex]\( y - 5 = \frac{1}{2}(x - 1) \)[/tex]
- Rewriting this in slope-intercept form:
[tex]\[ y - 5 = \frac{1}{2}(x - 1) \][/tex]
[tex]\[ y - 5 = \frac{1}{2}x - \frac{1}{2} \][/tex]
[tex]\[ y = \frac{1}{2}x - \frac{1}{2} + 5 \][/tex]
[tex]\[ y = \frac{1}{2}x + \frac{9}{2} \][/tex]
- This does not match the derived equation [tex]\( y - 1 = \frac{1}{2}(x - 5) \)[/tex].
2. [tex]\( y - \frac{1}{2} = 5(x - 1) \)[/tex]
- Rewriting this in slope-intercept form:
[tex]\[ y - \frac{1}{2} = 5(x - 1) \][/tex]
[tex]\[ y - \frac{1}{2} = 5x - 5 \][/tex]
[tex]\[ y = 5x - 5 + \frac{1}{2} \][/tex]
[tex]\[ y = 5x - \frac{9}{2} \][/tex]
- This does not match the derived equation [tex]\( y - 1 = \frac{1}{2}(x - 5) \)[/tex].
3. [tex]\( y - 1 = \frac{1}{2}(x - 5) \)[/tex]
- This is exactly the same as our derived equation.
- Therefore, this is the correct equation.
4. [tex]\( y - 1 = 5\left(x - \frac{1}{2}\right) \)[/tex]
- Rewriting this in slope-intercept form:
[tex]\[ y - 1 = 5\left(x - \frac{1}{2}\right) \][/tex]
[tex]\[ y - 1 = 5x - \frac{5}{2} \][/tex]
[tex]\[ y = 5x - \frac{5}{2} + 1 \][/tex]
[tex]\[ y = 5x - \frac{3}{2} \][/tex]
- This does not match the derived equation [tex]\( y - 1 = \frac{1}{2}(x - 5) \)[/tex].
Therefore, the correct equation that represents a line passing through [tex]\((5, 1)\)[/tex] with a slope of [tex]\(\frac{1}{2}\)[/tex] is:
[tex]\[ \boxed{y - 1 = \frac{1}{2}(x - 5)} \][/tex]
