Finding perpendicular slope

Discussion in 'General Math' started by john hite, Jan 30, 2005.

  1. john hite

    john hite Guest

    Intuitively it seem that given a line the slope of a perpendicular would
    be -1/m. Is this correct and do you know of a proof?

    john hite, Jan 30, 2005
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  2. Find a tangent vector to each curve, and calculate the scalar product.
    Kevin Anthoney, Jan 30, 2005
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  3. john hite

    john hite Guest

    No curves involved, just lines with cartesian coordinates.

    john hite, Jan 30, 2005
  4. John,

    How about this?

    Assume that the line of interest is of the form y = mx (i.e., it passes
    through the origin, which is fine, since perpendicular-ness doesn't
    depend on the position of a line, only its slope). Also, let's first
    consider m != 0 (so that the calculated slope of the perpendicular line
    isn't infinity). Now, consider the line y* = (-1/m)x

    Consider the points A = (-1,1/m) on y*, B = (1,m) on y, and the origin

    Then, the distance a from B to C satisfies a^2 = 1+m^2; the distance b
    from A to C satsifies b^2 = 1+(1/m)^2. Also, the distance c from A to B
    satisfies 2^2+(m-(1/m))^2 = m^2+2+(1/m)^2.

    By the law of cosines, where m<C means "the measure of angle C,"

    c^2 = a^2+b^2-2ab*cos(m<C)

    Substituting gives a^2+b^2 = c^2, or equivalently,

    cos(m<C) = 0
    m<C = pi/2

    Thus, the measure of angle C (the angle between y, y*) is a right angle.

    Travis Willse, Jan 31, 2005
  5. Here's a simple way to see it: Consider a point (x,y) in the
    first quadrant and the right triangle with vertices (0,0), (x,0),
    (x,y). The slope of the hypotenuse is y/x. Now rotate this
    triangle 90 degrees counterclockwise. The hypotenuse now
    terminates at (-y,x). The slope of the rotated hypotenuse, which
    makes a 90 degree angle with the first, is x/(-y), and we're done.
    The World Wide Wade, Jan 31, 2005
  6. john hite

    john hite Guest

    Thank you all for posting.


    john hite, Jan 31, 2005
  7. Yes it's correct -- if you found this yourself, your intuition is excellent.

    A proof using trig identities is easy. Note that the trig identities follow
    from purely geometric arguments that don't use any concept of "slope".

    However once you have the definitions of sin, cos and tan and the two
    identities: sin(a+b)=sin(a)*cos(b)+cos(a)*sin(b)
    and cos(a+b)=cos(a)*cos(b)-sin(a)*sin(b), then you can define slope of a
    line as m=tan(angle the line makes with the x-axis)

    Then use the fact that:

    sin(a+b) sin(a)*cos(b)+cos(a)*sin(b)
    tan(a+b)= ---------- = --------------------------------
    cos(a+b) cos(a)*cos(b)-sin(a)*sin(b)

    Now let L1 and L2 be perpendicular lines and let L1 make an angle of a with
    the x-axis. Finally observe that the perpendicular line L2 will make an
    angle with the x-axis which is 90 degrees greater (or less) than angle a.
    Assume 90 degrees greater (less is about the same).

    So the slope of L2 will be:

    sin(a+90) sin(a)*cos(90)+cos(a)*sin(90)
    tan(a+90)= ---------- = --------------------------------
    cos(a+b) cos(a)*cos(90)-sin(a)*sin(90)

    which simplifies to -1/tan(a).
    Gerry Wildenberg, Feb 11, 2005
  8. But that is a lot of needless work. You might want to look at the
    proof I gave.
    The World Wide Wade, Feb 11, 2005
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