1. ## Perpendicular lines

Prove that two lines are perpendicular if its product's slopes are $-1$

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

2. Originally Posted by Krizalid
Prove that two lines are perpendicular if its product's slopes are $-1$
What about $x=1 \mbox{ and }y=1$.

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 , ... , just joking).

3. 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.

4. Hello, Krizalid!

I have a rather klunky proof . . .

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

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)
|
*          |          *
*         |         *
*       |       *
* * *
|

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

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

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

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

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

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