Slope of the perpendicular line on the other line

AI Thread Summary
The slope of a line and its perpendicular counterpart are defined by the relationship m * m' = -1, where m is the slope of the original line and m' is the slope of the perpendicular line. This relationship arises from the geometric properties of angles formed by the lines with the x-axis, specifically when they create a right triangle. The angles of these lines must satisfy the condition that their tangent values multiply to -1, leading to the conclusion that the slopes are negative reciprocals. The discussion emphasizes that this can be derived through trigonometric identities and geometric reasoning. Understanding this relationship is crucial for determining the conditions under which two lines are perpendicular.
Alg0r1thm
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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

Thanks
 
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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.
 
Simon Bridge said:
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.

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?
 
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, \theta, 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 \theta, will be acute, the other, \phi, 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 \theta and \pi- \phi and we must have \theta+ (\pi- \phi)= \pi/2 so that \phi- \theta= \pi/2.

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

But tan(\pi/2) is undefined! We must have the fraction on the left undefined which means the denominator must be 0: 1+ tan(\theta)tan(\phi)= 0 so that tan(\theta)tan(\phi)= -1 and that last is just "mm'= -1".
 
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Alg0r1thm said:
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?
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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