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Suppose there are two parallel lines: $w_1x_1+w_2x_2=c_1$ (Line 1) and $w_1x_1+w_2x_2=c_2$ (Line 2). What is the distance between them (the shortest distance between any two points)?

I know the answer is $d=\frac{|c_1-c_2|}{||w||}$ where $||w||=\sqrt{w_1^2+w_2^2}$.

The method I was going to calculate is as follows:

1) find any point on Line 1 $(x_1^0,x_2^0)$ such that $w_1x_1^0+w_2x_2^0=c_1$

2) calculate the perpendicular line (Line 3) to Line 1 and passing through $(x_1^0,x_2^0)$

3) find the point $(x_1^1,x_2^1)$ where Line 3 intersects Line 2

4) calculate the distance between $(x_1^0,x_2^0)$ and $(x_1^1,x_2^1)$

However I couldn't figure out the algebra of this method. Can someone show me the steps of the above calculation? Or is there any simpler way to calculate this? Thanks.

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  • $\begingroup$ I would say sipler way to calculate distance between two parallel lines you wrote in the introduction of the query. $\endgroup$ – georg Aug 11 '19 at 19:12
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    $\begingroup$ Line 3 given by $w_2x_1-w_1x_2=0$ is through the origin and perpendicular to Line 1 and Line 2; can you find the distance from the origin to the intersection of Line 1 (and Line 2) with Line 3 ? $\endgroup$ – J. W. Tanner Aug 11 '19 at 19:15
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    $\begingroup$ Which step are you having trouble with? Perpendicular is of form $w_2x_1-w_1x_2=c$,.where $c=w_2x_1^0-w_1x_2^0$. $\endgroup$ – herb steinberg Aug 11 '19 at 19:16
  • $\begingroup$ Since you're doing linear algebra, the simplest method is to realize that $w_1x_1+w_2x_2$ is simply the inner product (aka dot product) of $(w_1,w_2)$ with $(x_1,x_2)$ and that the vector $(w_1,w_2)$ is orthogonal to both lines. Also, the inner product of $(w_1,w_2)$ with any point's position vector gives you the (scaled) distance from the line $w_1x_2+w_2x_2=0,$ scaled by $\|w\|.$ $\endgroup$ – David K Aug 11 '19 at 19:59
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Another possible way: Let $(a_1, a_2)$ be a point on $w_1x_1+w_2x_2-c_1=0$. Then we want to minimalize $$f(x)=(x_1-a_1)^2+(x_2-a_2)^2$$ With the constraint $$g(x)=w_1x_1+w_2x_2-c_2=0$$ Using Lagrange multipliers: $$\nabla f + 2\lambda\nabla g = 0$$ The derivatives are $$\nabla f=[2(x_1-a_1), 2(x_2-a_2)]$$ $$\nabla g = [w_1, w_2]$$ Which means that $$2(x_1-a_1)+2\lambda w_1=0$$ $$2(x_2-a_2)+2\lambda w_2=0$$ So the system of equations we need to solve is $$x_1=-\lambda w_1+a_1$$ $$x_2=-\lambda w_2+a_2$$ $$w_1x_1+w_2x_2=c_2$$ Which has the following solution: $$x_1=\frac{-a_2w_1w_2+a_1w_2^2+c_2w_1}{w_1^2+w_2^2}$$ $$x_2=\frac{-a_1w_1w_2+a_2w_1^2+c_2w_2}{w_1^2+w_2^2}$$ $$\lambda=\frac{a_1w_1+a_2w_2-c_2}{w_1^2+w_2^2}$$ So the minimal distance squared is \begin{align} d^2 &=\left(\frac{-a_2w_1w_2+a_1w_2^2+c_2w_1}{w_1^2+w_2^2}-a_1\right)^2+\left(\frac{-a_1w_1w_2+a_2w_1^2+c_2w_2}{w_1^2+w_2^2}-a_2\right)^2\\ &=\left(\frac{-a_2w_1w_2+a_1w_2^2+c_2w_1}{w_1^2+w_2^2}-\frac{a_1w_1^2+a_1w_2^2}{w_1^2+w_2^2}\right)^2+\left(\frac{-a_1w_1w_2+a_2w_1^2+c_2w_2}{w_1^2+w_2^2}-\frac{a_2w_1^2+a_2w_2^2}{w_1^2+w_2^2}\right)^2\\ &=\left(\frac{-a_2w_1w_2-a_1w_1^2+c_2w_1}{w_1^2+w_2^2}\right)^2+\left(\frac{-a_1w_1w_2-a_2w_2^2+c_2w_2}{w_1^2+w_2^2}\right)^2\\ &=\left(\frac{-c_1w_1+c_2w_1}{w_1^2+w_2^2}\right)^2+\left(\frac{-c_1w_2+c_2w_2^2}{w_1^2+w_2}\right)^2\\ &=\frac{(-c_1w_1+c_2w_1)^2+(-c_1w_2+c_2w_2)^2}{(w_1^2+w_2^2)^2}\\ &= \frac{(c_2-c_1)^2}{w_1^2+w_2^2} \end{align} As we expected.

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  • $\begingroup$ Thanks for this excellent answer. I changed the selected answer to this one because I was looking for a solution that started closer from first principals. $\endgroup$ – Catiger3331 Aug 15 '19 at 14:31
  • $\begingroup$ @Catiger3331 thank you :) To be honest, this answer is a bit far away from linear algebra and geometry, and more calculation-heavy; But I like this method because of my geometry skills are not so great :) $\endgroup$ – Botond Aug 15 '19 at 14:52
  • $\begingroup$ I like this and this is similar to en.wikipedia.org/wiki/Distance_between_two_straight_lines $\endgroup$ – David Wu Oct 30 '19 at 20:17
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There is an easier way to find the distance using dot product.

Pick the point $A=(x_1,y_1)$ on the fist line and $B=(x_2,y_2)$ on the second line.

The distance between the lines is the length of the projection of $AB$ on the normal vector to the parallel lines

$$d=\frac {|AB.N|}{||N||}=\frac {|c_2-c_1|}{\sqrt {w_1^2+w_2^2}}$$

You may fill in the details of simplifying the dot product and the norm in the above fraction.

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It is easier to figure out the distance from the right triangle formed by one of the lines, the vertical axis and the distance-line itself, as shown in the graph.

The intersections of the vertical axis with the two lines are $-c_1/w_2$ and $-c_2/w_2$, respectively. And the tangent of the two line is $\tan\theta=-w_1/w_2$. According to the right triangle in the graph. the distance $d$ is simply,

$$d=|c_1/w_2-c_2/w_2|\cos\theta=\frac{|(c_1-c_2)/w_2|}{\sqrt{1+w_1^2/w_2^2}}=\frac{|c_1-c_2|}{\sqrt{w_1^2+w_2^2}}$$

where $\cos\theta=1/\sqrt{1+\tan^2\theta}$ is used.

enter image description here

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