Another example

Here's another example but where neither line passes through the origin. The lines are y = -2/3x + 1 and y = -2/3x - 4.

[Graphics:../HTMLFiles/index_21.gif]

Again, we can use any line perpendicular to these. Such a line has slope 3/2.  We'll use the one passing through the point (0, 1), since we already know it is on one of the parallel lines (namely, y = -2/3x + 1). An equation for this perpendicular line is  y = 3/2x + 1. (Another good choice would be y = 3/2x - 4.)

[Graphics:../HTMLFiles/index_27.gif]

We now find the point of intersection with the other line by solving the system

y = -2/3x - 4  FormBox[RowBox[{y, =, , RowBox[{3/2x, +, 1.}]}], TraditionalForm]

You can check that the solution is the point (x, y) = (-30/13, -32/13). The distance between the lines is therefore the distance between the points (0, 1) and   (-30/13, -32/13), which is

((-30/13 - 0)^2 + (-32/13 - 1)^2)^(1/2) = ((30/13)^2 + (45/13)^2)^(1/2)
= ((15/13)^2 (2^2 + 3^2))^(1/2)
= 15/1313^(1/2) .

Incidentally, you can perform a geometric-numerical check of your answer. If you are using graph paper and have a compass, draw a circle centered at the point (0, 1) whose radius is the distance between the lines. Using a calculator, you can see that FormBox[RowBox[{15/1313^(1/2), , ≈, , 4.16025}], TraditionalForm]. You should be able to see that this corresponds to what you see in the resulting picture. (Do you see any obvious lengths that agree with a value of about FormBox[4.16025, TraditionalForm]?)

[Graphics:../HTMLFiles/index_39.gif]

Created by Mathematica  (December 3, 2004)