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Finding a point on a triangle

  1. Sep 11, 2005 #1
    if A(1,2) and B(8,3) find any point P on the x axis such that AP is perpendicular to BP.

    Here's what I know:
    P (P,0)
    perpendicular slopes are opposite reciprocals of each other
    ABP is a right triangle

    any ideas how to start this problem?
     
  2. jcsd
  3. Sep 11, 2005 #2

    TD

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    Homework Helper

    We have A(1,2), B(8,3) and P(p,0).

    Then vector PA = A-P = (1-p,2) and vector PB = B - P = (8-p,3).

    You want those vectors to be perpendicular so their inner product has to be 0.

    [tex]\left( {1 - p,2} \right) \cdot \left( {8 - p,3} \right) = 0 \Leftrightarrow \left( {1 - p} \right)\left( {8 - p} \right) + 6 = 0 \Leftrightarrow p = 7 \,\,\vee \,\, p = 2[/tex]
     
  4. Sep 11, 2005 #3
    Thanks very much, I forgot that perpendicular slopes multiplied to zero. You helped a ton!
     
  5. Sep 11, 2005 #4

    TD

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    No problem, don't forget we're talking about the scalar (or inner) product of vectors though, not just multiplication.
     
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