Why is the multiplication of the two gradients of two perpendicualr lines always equal to -1?

Question

This question has been bugging me for quite some time now and my maths teachers can't seem to come up with any answers. Most books just state that's how it is without any further explanation. Sure the formulae can be proven by plotting two perpendicular lines and measuring their gradient, but is there any theory (like trigometry, calculus, etc) to show where this formulae was derived from?

Answer 1

angle x between two lines with slopes m1 and m2 is given by

tan(x) = (m1 - m2)/(1+m1*m2)

when the lines are perpendicular x=pi/2 so that tan(x) = infinity

and hence (1+m1*m2) = 0 or m1*m2 = -1

Answer 2

I have an answer for you if you can follow basic trigonometry. Here are the steps

1) Look at the angles the lines make with the positive x-axis. Notice that if the first line makes an angle of alpha degrees, the other has to make an angle of 90+alpha degrees. For this you may need to draw a picture to convince yourself, but it is no big deal.

2) Write the slopes of these two angles using trigonometric functions. As the slope of a line is the tan of the angle it makes with the positive x-axis these two slopes are
sin(alpha)/cos(alpha) and sin(90+alpha)/cos(90+alpha).
This requires the most basic definitions from trigonometry.

3) Notice that sin(90+alpha) is the same as the cos(alpha) and cos(90+alpha) is the same as -sin(alpha). This is maybe the second or the third thing you learn in a trigonometry class.

4) Multiply the two slopes, you get
(sin(alpha)/cos(alpha))(sin(90+alpha)/cos(90+alpha))
= (sin(alpha)/cos(alpha))(cos(alpha)/-sin(alpha))= -1
This is just algebra and most basics of fraction multiplication.

By the way if by gradient you don't mean the slope of a line but the slope of the normal to a line, then the proof above will still work because if two lines are perpendicular, then so are the normals to those lines. Also, if you do not like trigonometry you can of course use analytic geometry to come up with anice proof but it will be longer and you will work with considerably longer formulas.

Answer 3

r and s are two perpendicular straight lines.

You can think of this: tan (alpha- beta) = [tan alpha - tan beta]/[1+tan alpha tan beta]

You have to consider that the angle that rOx is alpha, and the angle sOx is beta, the angle rOs is alpha - beta, because alpha is the external angle to the triangle r, s Ox, Ox is the x axis, O the origin (0,0).

So, if you consider the formula that I put before, you will notice that tan (alpha-beta) is a real number for all cases, but no when alpha-beta is pi/2.

1 + tan alpha tan beta = 0 in this case, and tan alpha is the gradient of r, tan beta is the gradient of s.

I dont know if this is a very good explanation, some teachers disagree with it, but this is the only one that I know.

Ana

Answer 4

Hmmm - let's see the gradient of the lines would give the slope.

Figure the first line is

y = mx (arbtrarily choose origin as one point)

rise over run is m, so tan(angle) = m = sin/cos

Then it could be defined by a vector, v = ( cos(angle), sin(angle) )

The dot product with the perpendicular line would be zero

Say the perp has vector u = (ux, uy)

then ux*cos(angle) + uy*sin(angle) = 0

then ux/uy = -sin/cos = -m

so then if uy = -(1/m)*ux

The gradient of the first line is m, the second is -1/m.

Take the product.

Answer 5

The answer is simple ... because the two lines are perpendicular ... you really shouldn't read into it too much.

BTW, there exist perpendicular lines whose slopes do not evaluate to -1 when multiplied.

Hint : the X-axis and Y-axis are one of the infinite number of line pairs to which this rule does not apply.