1. **State the problem:** We are given two lines, $L_1$ and $L_2$, where $L_1$ has equation $y=3x-2$ and $L_2$ passes through the point $(0,7)$. The lines are perpendicular. We need to find the equation of $L_2$ in the form $y=mx+c$. 2. **Recall the rule for perpendicular slopes:** If two lines are perpendicular, the slope of one line is the negative reciprocal of the slope of the other. That is, if $m_1$ is the slope of $L_1$, then the slope of $L_2$, $m_2$, satisfies: $$m_2 = -\frac{1}{m_1}$$ 3. **Identify the slope of $L_1$:** From $y=3x-2$, the slope $m_1=3$. 4. **Calculate the slope of $L_2$:** $$m_2 = -\frac{1}{3}$$ 5. **Use the point-slope form to find $c$:** Since $L_2$ passes through $(0,7)$, the $y$-intercept $c=7$. 6. **Write the equation of $L_2$:** $$y = -\frac{1}{3}x + 7$$ **Final answer:** $y = -\frac{1}{3}x + 7$