1. **Problem Statement:** Find the shortest distance between two parallel lines given in the slope-intercept form $y=mx+c$. 2. **Formula and Explanation:** The shortest distance $d$ between two parallel lines $y=mx+c_1$ and $y=mx+c_2$ is given by the formula: $$d=\frac{|c_2-c_1|}{\sqrt{1+m^2}}$$ This formula comes from the perpendicular distance between a point and a line, adapted for two lines with the same slope $m$. 3. **Derivation:** - Both lines have slope $m$, so they are parallel. - The vertical intercepts are $c_1$ and $c_2$. - The distance between them is the length of the perpendicular segment connecting the two lines. - The denominator $\sqrt{1+m^2}$ normalizes the difference in intercepts to the perpendicular direction. 4. **Example:** If the lines are $y=2x+3$ and $y=2x-1$, then $$d=\frac{|(-1)-3|}{\sqrt{1+2^2}}=\frac{4}{\sqrt{5}}=\frac{4\sqrt{5}}{5}$$ 5. **Summary:** To find the shortest distance between two parallel lines in the form $y=mx+c$, subtract their intercepts, take the absolute value, and divide by $\sqrt{1+m^2}$.