Re: Find point given 2 points

From: Lynn Kurtz (kurtzDELETE-THIS_at_asu.edu)
Date: 06/25/04 

Date: Thu, 24 Jun 2004 19:03:45 -0700

On Thu, 24 Jun 2004 15:35:23 -0500, Michael Hill
<hillmw@ram.lmtas.lmco.com> wrote:

>Two questions about finding the 3rd point:
>
>I have a line consisting of the coordinates (4,6) and (12,2).
>The midpoint of the line can be calculated as (8,4) using the
>midpoint formula.
>A right triangle is formed at the midpoint (8,4).
>
>1) Using 2 points (8,4) and point (12,2) and the lengths of the legs
>I'd like to find the 3rd point.
>
>Assume the following data:
>The small leg of the triangle is formed from (8,4) to (12,2) (call
>this leg b).
>The large leg of the triangle has a length of 10 (call this leg a).

Leg a lies along the vector < 4, -2 >. Leg b lies along the direction
of the perpendicular vector < 1, 2>. Make a unit vector out of this by
dividing by its length: u = (1/sqrt(5))< 1, 2>. The third point is 10
units in this direction from (8, 4) so its position is:

< 8, 4 > + 10/sqrt(5)< 1, 2 > = < 8 + 2 sqrt(5), 4 + 4 sqrt(5) >

>2) Using 2 points (8,4) and point (12,2) and the angles of the
>triangle I'd like to find the 3rd point.
>
>Assume the following data:
>The small leg of the triangle is formed from (8,4) to (12,2) (call
>this leg b). This is a 30,60,90 triangle.
>
>Can't figure out the last step

Apparently this is a different question from 1. In this case b is the
distance from (8,4) to (12, 2) which is b = 5 sqrt(2). So side a would
be b*sqrt(3) = 5 sqrt(6). Now just repeat part 1 using this instead of
10.

--Lynn

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