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I have two parallel lines and I know the coordinates for one of them where it interesects with a horizontal line on the top and a horizontal line on the bottom and the distance between them.

What I am trying to fathom is how to work out the coordinates where the second parallel line intersects with said horizontal lines.

Here is a picture of what I am whittering on about (I want to work out the coordinates labelled H and I circled in blue. All the other coordinates I know are lettered A to G and I show some sample values for the coordinates and the distance between the two lines):

enter image description here

I figured I could take the normal vector and use that to work out two coordinates on the (red) parallel line but that would not be where they intersect with the green (horizontal) lines. Is that a good place to start?

I've been drawing triangles like crazy but I am ashamed to say I do not think I know enough math for this. Is it possible? Do I need another measurement perhaps?

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up vote 0 down vote accepted

Drop a perpendicular from $E$ to $CD$; call the intersection point $J$. Drop another perpendicular from $H$ to $EF$; call that intersection point $K$. Then $EFJ$ and $HEK$ are similar right triangles; between the similarity relationship and the Pythagorean theorem, you should have enough information to determine the length of $EH$, which will give you the coordinates you're looking for.

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Thanks for the answer - since I do not know the coordinate H how can I drop a perpendicular? – kmp May 15 '12 at 14:22
Forget I said that - I see what you mean now. – kmp May 15 '12 at 14:26

You know it's slope and a point it passes through so it is enough information. Consider mid-point between E and F; it's coordinates are (55,100). On the other hand, G is at (100,100) so it is shifted by 45 in the direction of positive x-axis. Since two lines are parallel, this observation applies to other points as well. So you have H = (95,0) and I = (105,200).

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Alternatively, drop a perpendicular from $E$ to $CD$ call the intersection point $R$. Drop (raise) another perpendicular from $F$ to $HI$, call the intersection point $L$. Now, triangles $REF$ and $FIL$ are similar, right angled triangles, and we can use Pythagoras and similarity, to find the length of the hypotenuse of $FIL$.

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In fact, your drawing is inaccurate. The orange line between $EF$ and $HI$ that is labeled with a magnitude of $45$ is not at an angle with $EF$ and $HI$, it should be parallel with $EH$ and $FI$. So the midpoint approach works fine here. We can find the midpoint of $EF$ to be $(55, 100)$, and since we know the $x$-coordinate of $HI$ at the same height of $100$ to be $100$, we know the horizontal distance to be $100-55=45$, which agrees with the distance provided. This distance will be the same between $EH$ and $FI$ by the definition of parallel. So this gives us $H = (50+45, 0) = (95, 0)$ and $I = (60+45, 200) = (105, 200)$.

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