# Slope of the perpendicular line on the other line

1. Apr 9, 2014

### Alg0r1thm

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

2. Apr 9, 2014

### Simon Bridge

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.

3. Apr 9, 2014

### Alg0r1thm

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?

4. Apr 9, 2014

### HallsofIvy

Staff Emeritus
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 "$mm'= -1$".

Last edited: Apr 9, 2014
5. Apr 9, 2014

### Simon Bridge

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.