The distance between two parallel lines is the length of the perpendicular segment connecting the two lines.
We find the slope of the perpendicular:
\(m_{perpendicular}=-\frac{1}{m}=-\frac{1}{2}\)
Pick a point on \(y=2x+3\) (let's go with \((0, 3)\)) and find where it intersects \(y=2x-3\)
The perpendicular line will be \(y=-\frac{1}{2}x+3\)
We have a system of equations:
\(y=-\frac{1}{2}x+3\)
\(y=2x-3\)
Solve:
\(2x-3=-\frac{1}{2}x+3\)
⇒\(\frac{5}{2}x-3=3\)
⇒\(\frac{5}{2}x=6\)
⇒\(5x=12\)
⇒\(x=\frac{12}{5}\)
Plug into \(y=2x-3\):
\(y=2(\frac{12}{5})-3=\frac{24}{5}-\frac{15}{5}=\frac{9}{5}\)
So our second point is \((\frac{12}{5}, \frac{9}{5})\)
The distance between the points is:
\(d=\sqrt{(\frac{12}{5}-0)^{2}+(\frac{9}{5}-3)^{2}}\)
⇒\(d=\sqrt{(\frac{12}{5})^{2}+(-\frac{6}{5})^{2}}\)
⇒\(d=\sqrt{\frac{144}{25}+\frac{36}{25}}\)
⇒\(d=\sqrt{\frac{180}{25}}\)
⇒\(d=\frac{\sqrt{180}}{5}\)
⇒\(d=\frac{6\sqrt{5}}{5}\)
So the distance between the two lines is \(\frac{6\sqrt{5}}{5}\)
