Perpendicular lines

• Aug 20th 2007, 06:08 PM
Krizalid
Perpendicular lines
Prove that two lines are perpendicular if its product's slopes are $\displaystyle -1$

Without trig., vectors, just using basic analytic geometry tools.
• Aug 20th 2007, 06:43 PM
ThePerfectHacker
Quote:

Originally Posted by Krizalid
Prove that two lines are perpendicular if its product's slopes are $\displaystyle -1$

What about $\displaystyle x=1 \mbox{ and }y=1$. :D

Quote:

Without trig., vectors, just using basic analytic geometry tools.
I can prove it by trigonometry. Are you sure you do not want that?
(Are we allowed to use conformal mapping :D, ... , just joking).
• Aug 21st 2007, 01:09 AM
rualin
Algebra, Analytic Geometry: Slopes of Perpendicular Lines

It's full of typos and wrong angles, ugly pictures, and it's too long to be elegant yet the idea seems correct. But then again, I know nothing about proving.
• Aug 21st 2007, 04:28 AM
Soroban
Hello, Krizalid!

I have a rather klunky proof . . .

Quote:

Prove that two lines are perpendicular if the product of their slopes is $\displaystyle -1$

Without trig, vectors . . . just using basic analytic geometry tools.

The first problem is creating a right angle.

I used that fact that an angle inscribed in a semicircle is a right angle.
Code:

                  |                 * * *    P           *      |      o(x,y)         *        |  o    o*       *          o        o*               o  |          o       *  o      |          o*     B o-----------+------------o A (-r,0)*          |            *(r,0)                   |       *          |          *         *        |        *           *      |      *                 * * *                   |

$\displaystyle P(x,y)$ is a point on the semicircle: .$\displaystyle y \:=\:\sqrt{r^2-x^2}$ .[1]

The slope of $\displaystyle PA$ is: .$\displaystyle m_{PA} \:=\:\frac{y}{x-r}$ .[2]

The slope of$\displaystyle PB$ is: .$\displaystyle m_{PB} \:=\:\frac{y}{x+r}$ .[3]

Substitute [1] into [2]: .$\displaystyle m_{PA} \:=\:\frac{\sqrt{r^2-x^2}}{x-r} \:=\:\frac{\sqrt{(r-x)(r+x)}}{-(r-x)} \:=\:-\sqrt{\frac{r+x}{r-x}}$

Substitute [1] into [3]: .$\displaystyle m_{PB} \:=\:\frac{\sqrt{r^2-x^2}}{x+r} \:=\:\frac{\sqrt{(r-x)(r+x)}}{r+x} \:=\:\sqrt{\frac{r-x}{r+x}}$

Therefore: .$\displaystyle \left(m_{PA}\right)\left(m_{PB}\right) \;=\;-\sqrt{\frac{r+x}{r-x}}\cdot\sqrt{\frac{r-x}{r+x}} \;=\;-1$