Question: Find the equation of the
line that is perpendicular to
y = 1/6 x + 3 and contains the
point (-3,23).
y = [?]x + [ ]
1. **State the problem:** We need to find the equation of a line that is perpendicular to the line $$y = \frac{1}{6}x + 3$$ and passes through the point $$(-3, 23)$$.
2. **Recall the slope of the given line:** The slope $$m_1$$ of the given line is $$\frac{1}{6}$$.
3. **Find the slope of the perpendicular line:** The slope $$m_2$$ of a line perpendicular to another with slope $$m_1$$ is the negative reciprocal:
$$
m_2 = -\frac{1}{m_1} = -\frac{1}{\frac{1}{6}} = -6
$$
4. **Use point-slope form:** The equation of a line with slope $$m_2$$ passing through point $$(x_1, y_1) = (-3, 23)$$ is:
$$
y - y_1 = m_2 (x - x_1)
$$
Substitute values:
$$
y - 23 = -6 (x - (-3)) = -6 (x + 3)
$$
5. **Simplify the equation:**
$$
y - 23 = -6x - 18
$$
Add 23 to both sides:
$$
y = -6x - 18 + 23
$$
$$
y = -6x + 5
$$
6. **Final answer:** The equation of the line perpendicular to $$y = \frac{1}{6}x + 3$$ and passing through $$(-3, 23)$$ is:
$$
y = -6x + 5
$$