1. **State the problem:** Convert the equation $2x - 4y + 11 = 0$ into normal form. 2. **Recall the normal form of a line:** The normal form is given by $$x \cos \alpha + y \sin \alpha = p$$ where $p$ is the perpendicular distance from the origin to the line, and $\alpha$ is the angle the perpendicular makes with the positive $x$-axis. 3. **Rewrite the given equation:** The line equation is $2x - 4y + 11 = 0$. 4. **Find the length of the normal vector:** The coefficients of $x$ and $y$ form the normal vector $\vec{n} = (2, -4)$. Its length is $$\sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}.$$ 5. **Divide the entire equation by the length of the normal vector to normalize:** $$\frac{2x}{2\sqrt{5}} - \frac{4y}{2\sqrt{5}} + \frac{11}{2\sqrt{5}} = 0$$ 6. **Simplify the fractions:** $$\frac{2}{2\sqrt{5}} = \frac{1}{\sqrt{5}}, \quad \frac{4}{2\sqrt{5}} = \frac{2}{\sqrt{5}}$$ So the equation becomes: $$\frac{1}{\sqrt{5}} x - \frac{2}{\sqrt{5}} y + \frac{11}{2\sqrt{5}} = 0$$ 7. **Rewrite to isolate the constant term on the right:** $$\frac{1}{\sqrt{5}} x - \frac{2}{\sqrt{5}} y = - \frac{11}{2\sqrt{5}}$$ 8. **Multiply both sides by $-1$ to make the right side positive:** $$-\frac{1}{\sqrt{5}} x + \frac{2}{\sqrt{5}} y = \frac{11}{2\sqrt{5}}$$ 9. **Identify $\cos \alpha$ and $\sin \alpha$:** $$\cos \alpha = -\frac{1}{\sqrt{5}}, \quad \sin \alpha = \frac{2}{\sqrt{5}}$$ 10. **Identify $p$ (distance from origin):** $$p = \frac{11}{2\sqrt{5}}$$ **Final answer:** The normal form of the line is $$-\frac{1}{\sqrt{5}} x + \frac{2}{\sqrt{5}} y = \frac{11}{2\sqrt{5}}.$$