- #1

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m = slope of a particular line

m'= slope of the perpendicular line on that particular line

m*m' = -1 OR m' = -1/m

Thanks

- Thread starter Alg0r1thm
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- #1

- 11

- 0

m = slope of a particular line

m'= slope of the perpendicular line on that particular line

m*m' = -1 OR m' = -1/m

Thanks

- #2

Simon Bridge

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Draw a few lines and see.

i.e. if m=m', then the two lines are parallel.

- #3

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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?

Draw a few lines and see.

i.e. if m=m', then the two lines are parallel.

- #4

HallsofIvy

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

Now use [itex]tan(a+ b)= \frac{tan(a)+ tan(b)}{1- tan(a)tan(b)}[/itex] with [itex]a= \phi[/itex] and [itex]b= -\theta[/itex]:

[tex]tan(\phi- \theta)= \frac{tan(\phi)- tan(\theta)}{1+ tan(\phi)tan(\theta)}= tan\left(\frac{\pi}{2}\right)[/tex]

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

The angles inside that right triangle will be [itex]\theta[/itex] and [itex]\pi- \phi[/itex] and we must have [itex]\theta+ (\pi- \phi)= \pi/2[/itex] so that [itex]\phi- \theta= \pi/2[/itex].

Now use [itex]tan(a+ b)= \frac{tan(a)+ tan(b)}{1- tan(a)tan(b)}[/itex] with [itex]a= \phi[/itex] and [itex]b= -\theta[/itex]:

[tex]tan(\phi- \theta)= \frac{tan(\phi)- tan(\theta)}{1+ tan(\phi)tan(\theta)}= tan\left(\frac{\pi}{2}\right)[/tex]

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

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- #5

Simon Bridge

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NO - I mean that you can start from what I wrote and work it out for yourself.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?

I was trying to set your feet on the right path.

HallsofIvy showed you one approach.

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