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Slope of the perpendicular line on the other line

  1. Apr 9, 2014 #1
    Why is the slope of perpendicular line on the the other defined this way:
    m = slope of a particular line
    m'= slope of the perpendicular line on that particular line

    m*m' = -1 OR m' = -1/m

  2. jcsd
  3. Apr 9, 2014 #2

    Simon Bridge

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    If you have a line y=mx+c, and another line y'=m'x+c', what does m' have to be for y' to be perpendicular to y.
    Draw a few lines and see.

    i.e. if m=m', then the two lines are parallel.
  4. Apr 9, 2014 #3
    You mean there is no kind of proof or something like algebraic statement which states that the perpendicular line slope would be concluded via it?
  5. Apr 9, 2014 #4


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    One way to look at it is this: if y= mx+ b is the equation of a line, then the slope, m, is the tangent of the angle, [itex]\theta[/itex], the line makes with the x-axis. If two lines are perpendicular then they form a right triangle with the x-axis as hypotenuse. The angle one of the lines makes with the x- axis, say [itex]\theta[/itex], will be acute, the other, [itex]\phi[/itex], will be obtuse.

    Looking over this I see I have used the wrong words. I meant that one will be less that or equal to 45 degrees, the other larger than or equal to 45 degrees.

    The angles inside that right triangle will be [itex]\theta[/itex] and [itex]\pi- \phi[/itex] and we must have [itex]\theta+ (\pi- \phi)= \pi/2[/itex] so that [itex]\phi- \theta= \pi/2[/itex].

    Now use [itex]tan(a+ b)= \frac{tan(a)+ tan(b)}{1- tan(a)tan(b)}[/itex] with [itex]a= \phi[/itex] and [itex]b= -\theta[/itex]:
    [tex]tan(\phi- \theta)= \frac{tan(\phi)- tan(\theta)}{1+ tan(\phi)tan(\theta)}= tan\left(\frac{\pi}{2}\right)[/tex]

    But [itex]tan(\pi/2)[/itex] is undefined! We must have the fraction on the left undefined which means the denominator must be 0: [itex]1+ tan(\theta)tan(\phi)= 0[/itex] so that [itex]tan(\theta)tan(\phi)= -1[/itex] and that last is just "[itex]mm'= -1[/itex]".
    Last edited by a moderator: Apr 9, 2014
  6. Apr 9, 2014 #5

    Simon Bridge

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    NO - I mean that you can start from what I wrote and work it out for yourself.
    I was trying to set your feet on the right path.

    HallsofIvy showed you one approach.
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