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Math Help - Coordinate geomtry

  1. #1
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    Coordinate geomtry

    Hi, I have a problem that is giving me trouble solving. The problem reads The vertices of a \triangle ABC are at the points A (a, 0), B (0, b), C (c, d). If \angleB = 90 degrees. find a relaionship between a, b, c and d. The book answere is b(d-b)=ac. Can someone help me to fugure out this proble please thank you.
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  2. #2
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    It is simple. You gotta tally lengths:
    |AB|=\sqrt{a^2 + b^2}
    |BC|=\sqrt{c^2+ (d-b)^2}
    |AC|= \sqrt{(c-1)^2 + d^2}

    From Pitagoras' theorem:
    |AB|^2 + |BC|^2 = |AC|^2

    After simple transformation you should get:
    b^2 - bd + ca=0
    And finally the answer you wrote.
    Last edited by Oloria; January 12th 2010 at 11:17 AM.
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  3. #3
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    Quote Originally Posted by Oloria View Post
    It is simple. You gotta tally lengths:
    |AB|=\sqrt{a^2 + b^2}
    |BC|=\sqrt{x^2+ (d-b)^2}
    |AC|= \sqrt{(c-1)^2 + d^2}

    From Pitagoras' theorem:
    |AB|^2 = |BC|^2 = |AC|^2

    After simple transformation you should get:
    b^2 - bd + ca=0
    And finally the answer you wrote.
    Thank you very much. When i add the results I am having trouble comming up with the answer. Are you suppose to just add the results?
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  4. #4
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    Show me your solution, please. Then I will point you to error.
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  5. #5
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    Quote Originally Posted by Oloria View Post
    Show me your solution, please. Then I will point you to error.
    Hi, this is what I got so far
     a^2 + b^2 +c^2-(d-b)^2=(c-1)^2+d^2

     <br />
a^2-b^2-2b(-d)-(-d)^2+b^2+c^2+2c(-1)+(-1)^2+c^2+d^2

     <br />
a^2+2c^2+2bd-2c+1<br />
    I have a feeling that I am on the wrong track.
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  6. #6
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    Quote Originally Posted by scrible View Post
    Hi, this is what I got so far
     a^2 + b^2 +c^2-(d-b)^2=(c-1)^2+d^2
    On the left side you subtract (d-b)^2 and you should sum it. Correct:
     a^2 + b^2 +c^2+(d-b)^2=(c-1)^2+d^2

    I have a feeling that I am on the wrong track.
    Wrong feeling. I did it to the end and my result is the same as in the book.
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