A Coordinate Geometry ProblemDate: 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) 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/