1. **State the problem:** We have a line $L_1$ with equation $y = 2x + 5$. We want to find the equation of line $L_2$ which is perpendicular to $L_1$ and passes through the point $(-5, -4)$. 2. **Recall the formula and rules:** The slope of $L_1$ is $m_1 = 2$. For two lines to be perpendicular, their slopes satisfy $m_1 \times m_2 = -1$. So, the slope of $L_2$ is $m_2 = -\frac{1}{m_1} = -\frac{1}{2}$. 3. **Use point-slope form:** The equation of a line with slope $m$ passing through point $(x_1, y_1)$ is: $$y - y_1 = m(x - x_1)$$ Substitute $m = -\frac{1}{2}$ and point $(-5, -4)$: $$y - (-4) = -\frac{1}{2}(x - (-5))$$ which simplifies to $$y + 4 = -\frac{1}{2}(x + 5)$$ 4. **Simplify the equation:** $$y + 4 = -\frac{1}{2}x - \frac{5}{2}$$ Subtract 4 from both sides: $$y = -\frac{1}{2}x - \frac{5}{2} - 4$$ Express 4 as $\frac{8}{2}$ to combine: $$y = -\frac{1}{2}x - \frac{5}{2} - \frac{8}{2}$$ $$y = -\frac{1}{2}x - \frac{13}{2}$$ 5. **Final answer:** The equation of line $L_2$ in the form $y = mx + c$ is: $$y = -\frac{1}{2}x - \frac{13}{2}$$