Understanding Matrix Transformations on the x-axis

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Homework Help Overview

The discussion revolves around understanding the properties of matrix transformations, specifically how a point P is transformed to P' using a given matrix D. The original poster questions why the line segment connecting P and P' is bisected by the x-axis and maintains a constant angle relative to it, particularly when P is not located on the x-axis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the points P and P' after transformation, questioning the geometric implications of the midpoint being on the x-axis. There are attempts to derive the midpoint and slope of the line segment PP' and to understand the significance of these properties.

Discussion Status

There is an ongoing exploration of the relationship between the slope of the line segment and its angle with the x-axis. Some participants express confusion about the definitions and implications of the terms used, while others attempt to clarify the geometric relationships involved. The discussion reflects a mix of understanding and uncertainty, with no explicit consensus reached.

Contextual Notes

Participants note that the original question may involve assumptions about the behavior of the transformation and the properties of the x-axis, particularly when P is not on the x-axis. There is a recognition that the problem may be constrained by the definitions and properties of the matrix transformation being discussed.

Natasha1
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My question is:

If P' is the image of P under a matrix D = (1, -4, 0, -1) as follows (top left, top right, bottom left, bottom right). If P is not on the x-axis, why is PP' bisected by the x-axis and is at a constant angle to the x-axis, for any choice of P? :confused:

I can visually see what's happening but how can I show it? Please help
 
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Your notation is exceedingly unclear.
 
What is P'P?

I'm completely lost myself.

I see your matrices and all, but that's about it.
 
Can anyone crack this one??:rolleyes:
 
Natasha1 said:
My question is:

If P' is the image of P under a matrix D = (1, -4, 0, -1) as follows (top left, top right, bottom left, bottom right). If P is not on the x-axis, why is PP' bisected by the x-axis and is at a constant angle to the x-axis, for any choice of P? :confused:

I can visually see what's happening but how can I show it? Please help
I started to do this yesterday but the Tex apparently wasn't working.

The best I can say is do it! Let P be (x,y) (with y non-zero) and multiply it by D: P'= DP so that you have P' in terms of x and y.

What is the midpoint of the line segment PP'? what is the slope of that line?
 
HallsofIvy said:
I started to do this yesterday but the Tex apparently wasn't working.

The best I can say is do it! Let P be (x,y) (with y non-zero) and multiply it by D: P'= DP so that you have P' in terms of x and y.

What is the midpoint of the line segment PP'? what is the slope of that line?


Well I get that the midpoint of the line segment PP' is the x-axis and the line is y = 1/4 x which means the slope is tan-1 (0.25) so theta = 14.04 degrees? Is this correct?
 
Natasha1 said:
Well I get that the midpoint of the line segment PP' is the x-axis and the line is y = 1/4 x which means the slope is tan-1 (0.25) so theta = 14.04 degrees? Is this correct?

The midpoint of PP' is on the x-axis. What does that tell you about "why is PP' bisected by the x-axis"?

Technically, the slope is 1/4, slope is not the angle. In any case, the slope does not depend on the point P?? What does that tell you about "and is at a constant angle to the x-axis, for any choice of P"?
 
HallsofIvy said:
The midpoint of PP' is on the x-axis. What does that tell you about "why is PP' bisected by the x-axis"?

Technically, the slope is 1/4, slope is not the angle. In any case, the slope does not depend on the point P?? What does that tell you about "and is at a constant angle to the x-axis, for any choice of P"?

I can't answer the why in the "why is PP' bisected by the x-axis"? I can only say that the x-axis will be the midpoint of any P chosen except if it's on the x-axis itself. But that of course is a consequence, can't answer why :-(

Honestly I can't answer this question, I'm going to give it a miss I think. I can visualise see it by I can't explain why. The lines PP' will always be parallel but a part from that what can I say?
 
"Bisect" means "divide into equal parts". If the midpoint of PP' is always on the x-axis then the x-axis bisects the PP'!

And if the lines PP' are always parallel, no matter what P is, then they cross the x-axis at a constant angle, don't they?

That's all this problem is asking you to say!
 
  • #10
HallsofIvy said:
"Bisect" means "divide into equal parts". If the midpoint of PP' is always on the x-axis then the x-axis bisects the PP'!

And if the lines PP' are always parallel, no matter what P is, then they cross the x-axis at a constant angle, don't they?

That's all this problem is asking you to say!

Thanks Sir that will do :wink:
 

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