Distance between lines

[medicalphysicsguy]

In Strang 1991 Calculus, Chap 12 is a deceptively simple question: how far apart are the parallel lines \(\displaystyle x=y\) and \(\displaystyle x=y+1\)?

I can intuit my way though this problem. The line from (0,0) to (1,0) would be the hypotenuse of an isoceles right triangle, one leg of which would be on the line \(\displaystyle x=y+1\) and the other of which would be the plumline distance. So it's \(\displaystyle \frac{\sqrt{2}}{2}\).

But, I can't algebra or calc my way through this problem at all. Any suggestions for how to think about it?

Thanks, mpg

 

[pappus]

In Strang 1991 Calculus, Chap 12 is a deceptively simple question: how far apart are the parallel lines \(\displaystyle x=y\) and \(\displaystyle x=y+1\)?

I can intuit my way though this problem. The line from (0,0) to (1,0) would be the hypotenuse of an isoceles right triangle, one leg of which would be on the line \(\displaystyle x=y+1\) and the other of which would be the plumline distance. So it's \(\displaystyle \frac{\sqrt{2}}{2}\).

But, I can't algebra or calc my way through this problem at all. Any suggestions for how to think about it?

Thanks, mpg

1. Do you know basic vector geometry? If so your problem can be reduced to the question "What is the perpendicular distance of P(0, -1) to the line l: x - y = 0"
Re-write the equation of l into normal form then use the equation:

\(\displaystyle d(P, l) = |\overrightarrow{n_0} \cdot \vec p - e|\)

replacing the vector \(\displaystyle \vec x\) by the stationary vector of P.

2. If this sounds to you like Namreg (= German reverse) I'll translate the calculation into Algebra:

The line l has the equation: \(\displaystyle l: Ax + By + C = 0\)

Then the distance of P(p, q) to the line l is calculated by:

\(\displaystyle \displaystyle{d(P,l)=\left| \frac{Ap + Bq + C}{\sqrt{A^2+B^2}}\right|}\)

 

[srmichael]

In Strang 1991 Calculus, Chap 12 is a deceptively simple question: how far apart are the parallel lines \(\displaystyle x=y\) and \(\displaystyle x=y+1\)?

I can intuit my way though this problem. The line from (0,0) to (1,0) would be the hypotenuse of an isoceles right triangle, one leg of which would be on the line \(\displaystyle x=y+1\) and the other of which would be the plumline distance. So it's \(\displaystyle \frac{\sqrt{2}}{2}\).

But, I can't algebra or calc my way through this problem at all. Any suggestions for how to think about it?

Thanks, mpg

So if we find a line that is perpendicular to y=x and then find out the point where that line intersects y=x+1, then we have two points and we can use the distance formula to find the distance between the two lines.

1) The line perpendicular to y=x is y=-x (the y-intercept is irrelevant)

2) We can now find where y=-x and y=x+1 intersect for any point by setting them equal to each other:

-x = x+1
-2x = 1
x = -1/2

So for any point (x,y) on the line y=x, the point where the line y=-x intersects y=x+1 would be half a coordinate less than x, or x-1/2.
Thus, if the line y=-x intersects the line y=x at (x,y), it intersects the line y=x+1 at (x-1/2, y). Now replacing the y coordinates in each of these points with the corresponding equations you get (x, x) and (x-1/2, x+1/2).

3) Use these two points in the distance formula and you get:

\(\displaystyle \displaystyle d=\sqrt{[x-(x-\frac{1}{2})]^2+[x-(x+\frac{1}{2})]^2}\)

\(\displaystyle \displaystyle d=\sqrt{(\frac{1}{2})^2+(\frac{-1}{2})^2}\)

\(\displaystyle \displaystyle d=\sqrt{\frac{1}{4}+\frac{1}{4}}\)

\(\displaystyle \displaystyle d=\sqrt{\frac{1}{2}}\)

\(\displaystyle \displaystyle d=\frac{\sqrt{2}}{2}\)

 

[medicalphysicsguy]

Thanks everyone. Pappus, I think we already had the analytic geom part of the course but I can't follow your first description. So I'll have to go try find a way to expand my understanding of vectors a bit.

 

[HallsofIvy]

Choose any point at all on one of the lines. For example,setting x= 0 in x= y gives y= 0. (0, 0) is a point on that line. The line x= y has slope 1 so any line perpendicular to it will have slope -1. The line y= -x has slope -1 and passes through the point (0, 0).

Find the point where y= -x intersects the line x= y+ 1. The distance between that point and (0, 0) is the distance between the two lines.