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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?

A
\
\        D
\      /
\   /
E
/  \
/     \
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.

cu
```

```
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
theorem):

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)

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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