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