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M1.m2 = -1 ?

  1. Aug 30, 2011 #1
    M1.m2 = -1 ??

    Why is the product of the gradients of 2 straight line perpendicular to each other is always equal to -1 ?? Is there any theory to explain this ??
     
  2. jcsd
  3. Aug 30, 2011 #2

    WannabeNewton

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    Re: M1.m2 = -1 ??

    The slopes of perpendicular lines are negative reciprocals of each other. So if [itex]y_{1}[/itex] and [itex]y_{2}[/itex] are perpendicular lines (in standard form) then [itex]\frac{\mathrm{d} y_{1}}{\mathrm{d} x} = m[/itex] and [itex]\frac{\mathrm{d} y_{2}}{\mathrm{d} x} = -m^{-1}[/itex] so their products will simply be -1.
     
  4. Aug 31, 2011 #3

    uart

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    Re: M1.m2 = -1 ??

    I think the easiest proof is a geometric one using similar triangles.

    - The attachment shows a line, gradient m1, and a perpendicular.

    - The two angles marked with dots are equal. (Complementary to common angle).

    - By similar triangles d/1 = 1/m1

    - By definition (rise/run) the gradient of the perpendicular = -d/1 = -1/m1
     

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    Last edited: Aug 31, 2011
  5. Aug 31, 2011 #4
    Re: M1.m2 = -1 ??

    There are many ways to prove this. One and the simpler way is illustrated to you by WannaBeNewton in above post .

    Other ways are by trigonometry , use of trigonometrical coordinates , Use of other theorems etc.

    I am showing you another simpler way by similarity of triangles ( I hope you know it.)

    Let one line with side of course x1-x and y1-y be intersected by other line with side of course x3-x2 and y3-y2 be intersected at angle 90o

    Now you complete their intersection by joining their ends with dotted lines (note one end will be common ie intersection point). You'll notice it will form two right angled triangles .
    Now you prove those two triangles similar by their same shape .

    By this ,
    All corresponding sides of 2 triangles will be proportional , ie
    y1-y/x1-x = x3-x2/y3-y2

    or1

    m1=-1/m2
    (Realize one line inclination is > 90 degree hence slope is negative.)

    m1 x m2 = -1

    QED .
     
  6. Aug 31, 2011 #5
    Re: M1.m2 = -1 ??

    I think he meant: explain the cause of the result

    attachment.php?attachmentid=38422&stc=1&d=1314772955.png

    we have 2 perpendicular lines l and l1. the slope of l = -y/x and the slope of l1 = y1/x1

    triangle (O,x1,y1) is similar to triangle (M,x1,x) (two angles are the same)
    This triangle is also similar to triangle (O, y, x)

    therefore y/x = x1/y1 and if you substitute that in the slope of l1, you'll see it's equal to x/y, so the product of the two slopes = -1
     

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  7. Aug 31, 2011 #6
    Re: M1.m2 = -1 ??

    Think in 3D.

    Put differently;<m,1> vector is forming 90 degrees with <n,1> vector 2D gradients since original graphs retain perpendicularity.

    thus m*n+1=0*sqrt(n**2+1)sqrt(m**2+1)

    You can further solve to find
    m*n = -1

    m=-1/n
     
    Last edited: Aug 31, 2011
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