Re: Find point given 2 points

From: Virgil (ITSnetNOTcom/virgil_at_COMCAST.com)
Date: 06/24/04 

Date: Thu, 24 Jun 2004 16:52:08 -0600

In article <40DB3B0B.DB9C2204@ram.lmtas.lmco.com>,
 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).
>
> 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

Given that one leg goes from (8,4) to (12,3) and that the other leg is
at right angles to it along a line through (8,4) then the point you are
looking for will be will be on a line through (8,4) with a slope of

 m = - (8-12)/(4-3) = 4
so the equation is
y - 4 = 4*(x - 8) or y = 4*x - 28.

The distance from (x,y) on that line to (8,4) is
dist = sqrt( (x-8)^2 + (y-4)^2 )
     = sqrt( (x-8)^2 + (4*(x-8))^2 )
     = sqrt(1 + 4^2) * |x-8|
     = sqrt(17) * |x-8|
>
> In solving 1) I can calculate the length of the hypotenuse (call this
> leg c) as:
> 10^2 + 4^2 = c^2
> 100 + 16 = c^2
> 10.77 = c
>
> Since I know the length of all the legs and 2 points I should be able
> to apply a formula to it.

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