Understanding Matrix Transformations on the x-axis

[Natasha1]

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?

I can visually see what's happening but how can I show it? Please help

 

[0rthodontist]

Your notation is exceedingly unclear.

 

[JasonRox]

What is P'P?

I'm completely lost myself.

I see your matrices and all, but that's about it.

 

[Natasha1]

Can anyone crack this one??

 

[HallsofIvy]

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?

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?

 

[Natasha1]

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?

 

[HallsofIvy]

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"?

 

[Natasha1]

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?

 

[HallsofIvy]

"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!

 

[Natasha1]

"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