# Finding perpendicular slope

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

1. ### john hiteGuest

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?

thanx,
jh

john hite, Jan 30, 2005

2. ### Kevin AnthoneyGuest

Find a tangent vector to each curve, and calculate the scalar product.

Kevin Anthoney, Jan 30, 2005

3. ### john hiteGuest

No curves involved, just lines with cartesian coordinates.

jh

john hite, Jan 30, 2005
4. ### Travis WillseGuest

John,

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
C=(0,0).

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.
QED.

Cheers,
Travis

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 hiteGuest

Thank you all for posting.

jh

john hite, Jan 31, 2005
7. ### Gerry WildenbergGuest

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. 