Perpendicular lines

**Krizalid** · August 20th 2007, 06:08 PM

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

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

**ThePerfectHacker** · August 20th 2007, 06:43 PM

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).

**rualin** · August 21st 2007, 01:09 AM

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.

**Soroban** · August 21st 2007, 04:28 AM

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

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