Expert: Josh - 12/19/2008

Question
Hi,
Find the equation of line that is parallel to another line whose equation is known and the distance between these two lines is known?
No point is given through which line is passing.

Answer
Jerry

If the known equation is given by y=mx+b, then the equation of the new line parallel to this must also have the same slope "m".

That is, it will take the form of y=mx+k, for some constant k.

So, your task is to find the value of k given that m is known.

The distance between two parallel lines is related to the y-intercept (the point where it cuts the vertical axis is y=k).

Using trigonometry, if we write "q" for the angle of inclination, as measured from the positive x-axis in an anti-clockwise direction, q=tan^(-1)(m), where tan^(-1) is the inverse tangent function.

Draw a diagram to understand this:

.............D
|
|
A............E
|
|
|
B---H--------------
|
.............F
|
|
C............G

For illustration purpose, think of AD as the line whose equation is known. Here, the angle q is subtended by DAE. The line has slope m=tan(q)=|DE|/|AE|.....[1], where |DE| is the distance from point D to point E, similarly, |AE| is the distance from point A to point E.

The vertical separation between line AD and the other parallel line through CF is given by |AB|+|BC|. By definition the point A=(0,b), where b is the y-intercept in the original equation for the straight line AD. Likewise, |BC|=k. The total length |AC|=|b-k|.....[2]

AH is perpendicular to both line AD and line CF. The angle BAH = q because DAH and BAE are both 90 degrees. The geometry leads to this final result:

D=|AC|*cos(q) .....[3]

where D is the distance of separation between the two parallel lines AD and CF, as measured along the normal AH.

Remember that the angle q is related to the gradient "m" in [1].

To consolidate this discussion, let us consider this example.
A straight line is described by the equation y=x+4. Suppose a parallel line further below it has equation y=px+k and is separated by a distance of D=10. Since it is parallel, p=m=1.

Using [1], tan(q)=m=1. This implies q=45 degrees.
From [2], the vertical distance between the y-intercepts is |b-k|=|4-k|. Using [3], 10=|4-k|*cos(45). It follows that |4-k|=10*sqrt(2).  But the question here tells us the second line is below the first one, meaning that k<4. So, the solution is obtained by considering the case 4-k=10*sqrt(2). This yields k=4-10*sqrt(2).

Summary: For this sample question,
equation 1:   y=x+4
equation 2:   y=x+k, where k=4-10*sqrt(2) is approximately -10.1421.
perpendicular line |AH|:  y=-x+4 intersects with equation 2 at (after some algebra) x=(b-k)/2=7.071, y=x+k=-3.071. Using the formula for Euclidean distance [d=sqrt((x1-x0)^2+(y1-y0)^2)], we can verify that the distance of separation between the lines, i.e., between (0,4) and (7.071,-3.071) is indeed 10.

I would really like to see you have a go at the question and show me some working the next time. Good luck!