This is another way to find a distance between two parallel lines. This derivation was suggested to me by Andre and I highly recommend him and his answers to any student, who wants to learn math ans physics. This derivation requires the knowledge of trigonometry
and some simple trigonometric identities, so this may be suitable for more advanced students.

Once again, we have two lines.

y=mx+b

_{1}(1)--equation for the first line.y=mx+b

_{2}(2)--equation for the second line.Now recall that the slope of the line is the tangent of an angle this line forms with the x-axis. Indeed, m=(y

_{2}-y_{1})/(x_{2}-x_{1}), where x_{1}, x_{2}, y_{1}, y_{2 }are the x- and y-coordinates of any two distinct points on the line. If one draws the picture, it will be immediately obvious that m is the tangent of the angle between the line and the x-axis.The difference b

_{2}-b_{1}gives the relative displacement along the y-axis of two lines. Since b_{2}is the y-intercept of the second line and b_{1}is the y-intercept of the first line, the vertical displacement of one line with respect to another is given by |b_{2}-b_{1}|. The perpendicular line segment between two lines and this displacement form the same angle between each other as the lines and the x-axis have between each other. So its tangent is m. Now recall the identity:tan

^{2}(x)+1=1/cos^{2}(x) (3)From (3) it follows that cos(x)=1/√(1+tan

^{2}(x))In our case the cosine of the angle α between the vertical line, which length is |b

_{2}-b_{1}|, and the perpendicular line, which length d we are trying to determine, is 1/√(1+m^{2}). The distance d is them given by:**d=|b**_{2}-b_{1}|cos(α)=|b_{2}-b_{1}|/√(1+m^{2}) (4)

_{2}-b

_{1}|cos(α)=|b

_{2}-b

_{1}|/√(1+m

^{2}) (4)

This once again proves the formula derived earlier using coordinate approach.