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$y = 1, A(1,-2)$ 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

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    I almost don't downvote and even less for calling things this or that way, if that's what you were implying. Anyway, There's an answer to your question donw here already...2012-11-02

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What do you mean you "only have $\,y"\,$? You have the straight, horizontal line $\,y=1\,$ and the point $\,A(1,-2)\,$ . The distance of these to things is, of course, the absolute value of the difference $\,|-2-1|=3\,$ since clearly the point $\,(1,1)\,$ on the given line is on the same vertical line as $\,A\,$ and is thus on the perpendicular line to $\,y=1\,$ passing through $\,A\,$ ... Draw a diagram!