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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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"Plane line"?.. –  DonAntonio Nov 2 '12 at 3:28
    
common someone explains why the thumbs down, I tryed to solve it before asking, did some research, used math ml thing, made it as clear as possible the way I understand it. Plane line is because the problem says " La droite planaire" which I translated to plane line... –  Dave Nov 2 '12 at 14:12
    
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... –  DonAntonio Nov 2 '12 at 14:43

1 Answer 1

up vote 1 down vote accepted

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!

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