$y = 1, A(1,-2)$ This is what I have so far.
since I only have y, how do I figure out x for my normal vector ? I guessed .... [0,1]
$\bigg |$ $ ([x1,y1] - [x0,y0] )$ $ \cdot [0,1] \over \sqrt{1} $ $\bigg |$
$\bigg |$ $ ([1,-2] - [x0,y0] )$ $ \cdot [0,1] \over \sqrt{1} $ $\bigg |$
Not sure where to go from there.
Thanks