Find a vector that is perpendicular to $u = (9,2)$ Attempt:
We know perpendicular vectors have dot product
$u \cdot v = 0$
therefore 
$[9,2] \cdot [x,y]$ = 0
$9x + 2y = 0$
what would I do now?
thanks!
 A: You've got your equation. Now solve!
$9x +2y=0$
$4.5x+y=0$
$y = -4.5x$
Now we have
$[x,y]=[x,-4.5x]$
Where $x$ is a free parameter. In order to not be confusing, we can use a different variable:
$[1k,-4.5k]$
or
$k[1,-4.5]$
We can multiply back that 2 to make it look pretty:
$k[2,-9]$
What's up with that free parameter? It helps to think about it geometrically. What's a vector that's perpendicular to the first? Well, there are an infinite number of them, all of different lengths - plus you have two directions to go in (positive and negative $k$).
Edit: $k=0$ is fine. 
A: I am very happy for your attempt... :) 
You have $9x+2y=0$ 
which would imply $9x=-2y$ 
which would imply (assuming $y\neq 0$) $$\frac{x}{y}=\frac{-2}{9}$$
least possible thing you could do is just let $x=-2$ and $y=9$
A: Why not simply draw a picture? For any vector $(a,b)$ the vector $(-b,a)$ is perpendicular to $(a,b)$.  There's definitely no need to perform any calculations.
A: Every point $(x,y)$ that satisfies the equation $9x+2y = 0$ is perpendicular . 
Example $x=2\ ,y=-9$ . In terms of direction there is only one vector that is perpendicular that to $u = [9,2]$ . You can generate different magnitudes for them by just multiplying a $k $ with it . 
