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Old 01-12-2014, 09:39 AM
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Default Wacko qusetion

Point A to Point B = 26 inches. Point C is a fixed point in the distance.
If I move point A 1/8" to the left. How much do I have to move point B to maintain a straight line to point C?
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Old 01-12-2014, 09:55 AM
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Old 01-12-2014, 10:47 AM
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If A, B, and C are in a straight line to begin with, the amount of movement of point C would equal 1/8" x the ratio of the distances between AB and BC (AB/BC=???? x 1/8")

Bob
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Old 01-12-2014, 10:54 AM
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If A to B was a straight line in the first place, and C is in that straight line, you don't move anything. You still have a straight line. Unless I'm missing something or you are trying to make C equadistant between A and B. Sunday mornings are tough for math class!
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Old 01-12-2014, 11:04 AM
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How much do I have to move point B to maintain a straight line to point C?
This question is worded very poorly. It is not clear if the straight line is from point B to point C, but I assume so.

This is a trick question. You can always draw a straight line between any two points, so you do not need to move point B. Furthermore you can move point B anyplace in the universe and still draw a straight line to point C.

If the question is to get a straight line from point A to point B with point C fixed, after point A was moved, then the previous answer is correct, if you move point B to the right.
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Old 01-12-2014, 11:28 AM
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If A B C is a triangle, you don't have to move B at all. It is already in line with C. If A B C is a straight line, you essentially create a triangle when you move A so you would solve for the height of the triangle which would be the distance from point B to line AC. That would depend on the distance from A to C. Really need more information regarding the preliminary relationship between A B and C. If they are a straight line, it would help to know either the distance AC or BC to have a real solution. It almost sounds like we are adjusting iron sights..
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Old 01-12-2014, 11:52 AM
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Since C is a fixed point and the distance from AC and BC is constant. Then AB is also wants to remain constant along an arc with C being the center point. To keep AB constant you need to move it the same distance along the line of the formed arc. In the end, it doesn't matter, you can always draw a straight line if you're sober enough.
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Old 01-12-2014, 11:55 AM
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If A B C is a triangle, you don't have to move B at all. It is already in line with C. If A B C is a straight line, you essentially create a triangle when you move A so you would solve for the height of the triangle which would be the distance from point B to line AC. That would depend on the distance from A to C. Really need more information regarding the preliminary relationship between A B and C. If they are a straight line, it would help to know either the distance AC or BC to have a real solution. It almost sounds like we are adjusting iron sights..
Agree on the triangle theory, if that's what he is asking. Not clear. Trying to solve dead-center on a circle between AB?
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Old 01-12-2014, 11:56 AM
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"Wacko qusetion"

Yes, it is a Waco QUSETION.
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Old 01-12-2014, 01:35 PM
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My bad. A B C are in a straight line.
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Old 01-12-2014, 02:03 PM
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Question

You would have to see it in order to know what is in it? Just like the OB Care. My answer is leave C along. Sounds like C is on the way to B or C is on the way to A. It doesn't make any difference. Are we all bored today?

Here I sit trying to give away a 110" X 36" one piece mirror? Along with answering a high school math sophomore test question...
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Old 01-12-2014, 04:04 PM
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Old 01-12-2014, 04:08 PM
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OK, so all three points sit on a hyperbolic paraboloid, right?
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Old 01-12-2014, 04:54 PM
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There's always Catia software if you are burning for an answer. Measures all points in space. When you're finished, you can design an airplane or the Disney Concert Hall! I'll just stick to a tape measure and calculator

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Old 01-12-2014, 05:00 PM
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There's always Catia software if you are burning for an answer. Measures all points in space. When you're finished, you can design an airplane or the Disney Concert Hall! I'll just stick to a tape measure and calculator
I've been trying to teach Chas non-Euclidean Geometry for years, but, man, he is a slowwww learner. I got him Mathematica for Christmas -- I'm hoping that will help.
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Old 01-12-2014, 05:03 PM
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My bad. A B C are in a straight line.
Sounds like you need to go straight to your room B, since you, C, changed the criteria and became an A.
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Old 01-12-2014, 05:09 PM
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My son is math and Econ at UCLA. When I bring him a tough one (for me), I get the "One scoop chocolate, One scoop of vanilla...Don't waste my time!"
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Old 01-12-2014, 05:53 PM
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Sounds like you need to go straight to your room B, since you, C, changed the criteria and became an A.
C has no bearing on the question, best I can tell and based on the original question. C could be on the moon. AB could move from here to infinity and still be a straight line. My 2c.
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Old 01-12-2014, 05:57 PM
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AB could move from here to infinity and still be a straight line. My 2c.
Ooooh, you're close.... The real answer will depend on whether you take the view that parallel lines meet at infinity.
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Old 01-12-2014, 06:24 PM
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Did not pick up on parallel lines anywhere. Only one line. You got dat double vision goin' again Patrick?
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