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I'm trying to find the average distance between a point and a point on a line segment. I mean what would be the result if you take all the points on a line segment and find the average distance between those points and the given line. I guess it is equal to the distance between the given point and the midpoint of the line. But I can't figure out exactly. It's equal to limit of number of there points going to infinity and the average distance of those as shown in the picture enter image description here

Reinstein
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    How do you define average distance? – Allawonder Sep 12 '19 at 17:27
  • You first have to decide what you mean by “average distance.” You’ll get a different value if you average over distance moved along the line segment then you do if you average over the angle that the segment between the fixed and moving point makes with the line. – amd Sep 13 '19 at 01:10
  • @Reinstein Since you are new here, please visit https://math.stackexchange.com/tour – Robert Z Oct 18 '19 at 12:26

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If you let the origin of coordinates be the left endpoint of the segment, and let the segment lie on the $x$-axis, and you let the $x$-coordinate of the other endpoint be $k,$ and the fixed point be $(a,b),$ then the distance from a point on the segment to the fixed point is given by $$D(x)=\sqrt{b^2+(x-a)^2},$$ where we choose $x>0$ (so that $k>0$ too). Thus, if we require the average distance $D_a$ to satisfy the condition $$\int_0^kD(x)\mathrm dx=D_a\int_0^k\mathrm dx,$$ then the average distance is given by the $$\frac1k\int_0^k\sqrt{b^2+(x-a)^2}\mathrm dx,$$ which may be explicitly calculated by making the substitution $x-a=b\tan y.$

Allawonder
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  • I got that integral too. but i wanna know if it's equal to the distance between the given point and the midpoint of line – Reinstein Sep 13 '19 at 08:01
  • @Reinstein To see what it's equal to, you'd have to evaluate the integral and see if it gives $D(k/2).$ – Allawonder Sep 13 '19 at 08:05
  • I integrate that with an online calculator, but the answer too complicated – Reinstein Sep 13 '19 at 10:38
  • @Reinstein But surely you can tell whether it's equal to $D(k/2).$ The form should not be so complicated as not to be able to see whether the one can be transformed into the other. – Allawonder Sep 13 '19 at 15:50