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A Coordinate Geometry Problem

Date: 11/26/96 at 21:48:34
From: Simon Lud
Subject: Small problem

Here is my problem?

       \        D
        \      /
         \   /
         /  \
       /     \
     B        C

AC is 3 meters long, BD is 2 meters long, E is 1 meter above the 
ground, and there is a 90 degree angle between AB and BC as well as 
between BC and DC.

How long is BC?

My math teacher was not able to solve the problem.


Date: 12/01/96 at 19:23:46
From: Doctor Anthony
Subject: Re: Small problem

Using coordinate geometry, take the origin at B, and let BC = a.  
Now, in terms of 'a', you can write down the equations of the straight 
lines BD and CA and impose the condition that these meet with a y 
value equal to 1 (we're going to make heavy use of the Pythagorean 

The coordinates of D are [a,sqrt(4-a^2)]

The coordinates of C are [a,0]

The coordinates of A are [0,sqrt(9-a^2)] and B is of course [0,0]

The equation of BD is y = [sqrt(4-a^2)/a]*x

The equation of CA is y = [-sqrt(9-a^2)/a](x-a)

We put x = ay/sqrt(4-a^2) (from the equation for BD) into the equation 
for CA to get:

   y = [-sqrt(9-a^2)/a][ay/sqrt(4-a^2) - a]

At this stage we can divide out 'a' from the top and bottom lines of 
the right hand side, and also now put y=1: 

  1 = -sqrt(9-a^2)/sqrt(4-a^2) + sqrt(9-a^2)

which leads to:

     sqrt(4-a^2) + sqrt(9-a^2) - sqrt(4-a^2)sqrt(9-a^2) = 0

This would require further heavy algebra to simplify, so reaching for 
my handy TI-92 I get the answer a = 1.2311857

And so the distance BC is 1.2312 meters.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Associated Topics:
High School Coordinate Plane Geometry
High School Geometry

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