Slope of the perpendicular line on the other line

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

Simon Bridge
Homework Helper
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.

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?

HallsofIvy
Homework Helper
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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Simon Bridge