Question: Write an equation for line L in point-slope form and slope-intercept form.
L is perpendicular to $y = 4x$.
$y = 4x$
$(1,4)$
1. **State the problem:** We need to find the equation of line $L$ that is perpendicular to the line $y = 4x$ and passes through the point $(1,4)$.
2. **Recall the slope of the given line:** The line $y = 4x$ is in slope-intercept form $y = mx + b$ where $m = 4$ is the slope.
3. **Find the slope of the perpendicular line:** The slope of a line perpendicular to another is the negative reciprocal of the original slope.
$$m_{L} = -\frac{1}{4}$$
4. **Write the point-slope form:** The point-slope form of a line is
$$y - y_1 = m(x - x_1)$$
where $(x_1, y_1)$ is a point on the line and $m$ is the slope.
Using point $(1,4)$ and slope $m_L = -\frac{1}{4}$:
$$y - 4 = -\frac{1}{4}(x - 1)$$
5. **Convert to slope-intercept form:** Simplify the point-slope form to get $y = mx + b$.
$$y - 4 = -\frac{1}{4}x + \frac{1}{4}$$
Add $4$ to both sides:
$$y = -\frac{1}{4}x + \frac{1}{4} + 4$$
$$y = -\frac{1}{4}x + \frac{1}{4} + \frac{16}{4}$$
$$y = -\frac{1}{4}x + \frac{17}{4}$$
**Final answers:**
- Point-slope form: $$y - 4 = -\frac{1}{4}(x - 1)$$
- Slope-intercept form: $$y = -\frac{1}{4}x + \frac{17}{4}$$