Hello, symmetry!
An interesting problem . . . I had to "invent" my own approach.
I'm certain others have shorter, more elegant proofs.
Prove that if two nonvertical lines have slopes whose product is 1,
then the lines are perpendicular.
$\displaystyle \text{Let }m_1 = \frac{q}{p}\:\text{ and }\:m_2 = \frac{p}{q}$
Then: .$\displaystyle m_1\cdot m_2 \:=\:\left(\frac{1}{p}\right)\left(\frac{p}{q}\right) \:=\:1$
. . We have two slopes whose product is $\displaystyle 1$.
We know that: $\displaystyle slope \:=\:\frac{rise}{run}$
So $\displaystyle m_1 = \frac{q}{p}$ looks like this: Code:
*
* q
* 
*      *
p
And $\displaystyle m_2 = \frac{p}{q}$ looks like this; Code:
q
*   *

* 
 p
* 

*
Sketch them on the same graph. Code:
 F
 * (p,q)
 a * 
 * 
O *   +   + 
(0,0) 
 * 
 
 b * 
 
 * (q,p)
 G
Now consider the triangle $\displaystyle FOG$.
Let $\displaystyle a \:= \:OF \:=\:\sqrt{(p0)^2 + (q0)^2} \:=\:\sqrt{p^2+q^2} $
Let $\displaystyle b \:=\:OG\:=\:\sqrt{(q0)^2+(\text{}p0)^2} \:=\:\sqrt{p^2+q^2}$
Let $\displaystyle c \:=\:GF \:=\:\sqrt{(pq)^2 + (q+p)^2} \:=\:\sqrt{2p^2 + 2q^2}$
We note that: .$\displaystyle a^2 + b^2 \:=\:\left(\sqrt{p^2+q^2}\right)^2 + \left(\sqrt{p^2 + q^2}\right)^2 \:=\:2p^2+2q^2$
. . . and that: .$\displaystyle c^2\:=\:\left(\sqrt{2p^2+2q^2}\right)^2\:=\:2p^2+2 q^2$
Since $\displaystyle a^2 + b^2\:=\:c^2$, then $\displaystyle \Delta FOG$ is a right triangle with $\displaystyle \angle FOG = 90^o$.