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Math Help - Perpendicular lines

  1. #1
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    Perpendicular lines

    Prove that two lines are perpendicular if its product's slopes are -1

    Without trig., vectors, just using basic analytic geometry tools.
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  2. #2
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    Quote Originally Posted by Krizalid View Post
    Prove that two lines are perpendicular if its product's slopes are -1
    What about x=1 \mbox{ and }y=1.

    Without trig., vectors, just using basic analytic geometry tools.
    I can prove it by trigonometry. Are you sure you do not want that?
    (Are we allowed to use conformal mapping , ... , just joking).
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  3. #3
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    Algebra, Analytic Geometry: Slopes of Perpendicular Lines

    It's full of typos and wrong angles, ugly pictures, and it's too long to be elegant yet the idea seems correct. But then again, I know nothing about proving.
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  4. #4
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    Hello, Krizalid!

    I have a rather klunky proof . . .


    Prove that two lines are perpendicular if the product of their slopes is -1

    Without trig, vectors . . . just using basic analytic geometry tools.

    The first problem is creating a right angle.

    I used that fact that an angle inscribed in a semicircle is a right angle.
    Code:
                      |
                    * * *     P
              *       |       o(x,y)
            *         |   o    o*
           *          o         o*
                  o   |          o
          *   o       |           o*
        B o-----------+------------o A
    (-r,0)*           |            *(r,0)
                      |
           *          |          *
            *         |         *
              *       |       *
                    * * *
                      |

    P(x,y) is a point on the semicircle: . y \:=\:\sqrt{r^2-x^2} .[1]

    The slope of PA is: . m_{PA} \:=\:\frac{y}{x-r} .[2]

    The slope of PB is: . m_{PB} \:=\:\frac{y}{x+r} .[3]


    Substitute [1] into [2]: . m_{PA} \:=\:\frac{\sqrt{r^2-x^2}}{x-r} \:=\:\frac{\sqrt{(r-x)(r+x)}}{-(r-x)} \:=\:-\sqrt{\frac{r+x}{r-x}}

    Substitute [1] into [3]: . m_{PB} \:=\:\frac{\sqrt{r^2-x^2}}{x+r} \:=\:\frac{\sqrt{(r-x)(r+x)}}{r+x} \:=\:\sqrt{\frac{r-x}{r+x}}


    Therefore: . \left(m_{PA}\right)\left(m_{PB}\right) \;=\;-\sqrt{\frac{r+x}{r-x}}\cdot\sqrt{\frac{r-x}{r+x}} \;=\;-1

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