Slope of the perpendicular line on the other line

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Discussion Overview

The discussion centers on the mathematical relationship between the slopes of two perpendicular lines, exploring the definition and derivation of the condition that the product of their slopes equals -1. Participants examine the implications of this relationship in the context of line equations and geometric interpretations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions why the slope of a perpendicular line is defined such that the product of the slopes equals -1, seeking clarification on the underlying reasoning.
  • Another participant suggests visualizing the relationship by drawing lines to understand the conditions under which two lines are perpendicular.
  • A different participant reiterates the need for a formal proof or algebraic statement to support the claim about perpendicular slopes, expressing skepticism about the explanation provided.
  • One participant offers a geometric interpretation involving angles and the tangent function, attempting to derive the relationship between the slopes through trigonometric identities.
  • Another participant responds to the skepticism by indicating that the derivation can be approached independently, referencing a previous contribution for guidance.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of a formal proof for the relationship between perpendicular slopes. While some provide geometric and algebraic reasoning, others remain unconvinced and seek more rigorous justification.

Contextual Notes

There is an emphasis on the geometric interpretation of slopes and angles, but the discussion does not resolve the need for a formal proof or consensus on the explanation provided.

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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