Transformations - centre of rotation?

OP

Original post by notnek

You could try sketching (without a compass) two perpendicular bisectors which will give you an approximate location of the centre of rotation. Then you can use your tracing paper to find it's accurate location.

Thanks for the reply. I've tried that (not sure if I'm doing it right) but I still can't get the centre of rotation. I swear I've tried every single point on this grid, but obviously not lol. Any advice?

Reply 4

Original post by TheWantedGuy

Thanks for the reply. I've tried that (not sure if I'm doing it right) but I still can't get the centre of rotation. I swear I've tried every single point on this grid, but obviously not lol. Any advice?

Check what you did along with this:

1. Pick a point P on the shape pre-transformation and locate the respective point post-transformation P'.

2. Draw line PP' .

3. Locate the midpoint M of P and P'.

4. Draw a perpendicular bisector (intersecting PP' at a right angle at M).

5. Repeat steps 1-4 for a second point Q.

6. Extend the perpendicular bisectors (if necessary) so that they intersect.

7. (Since perpendicular bisectors intersect the center of a circle, and since the circle containing P and P' and the circle containing Q and Q' are both centered at the center of rotation), the intersection of of the two perpendicular bisectors is the center of rotation.

(edited 11 years ago)

Reply 5

Notnek

21

Original post by TheWantedGuy

Thanks for the reply. I've tried that (not sure if I'm doing it right) but I still can't get the centre of rotation. I swear I've tried every single point on this grid, but obviously not lol. Any advice?

It's hard for me to help you with this via an internet forum since you seem to have tried everything. It's possible that you are performing some of the methods incorrectly but I can't really check that.

I'll narrow down your search by telling you that the centre lies in the bottom left quadrant (i.e. the x and y coordinates are both negative).

Spoiler

An algebraic way would be to find the equations of two perpendicular bisectors and then solve them simultaneously to find the intersection. But I assume you're a GCSE student which probably means that this method is too advanced.

(edited 11 years ago)

Reply 6

Original post by TheWantedGuy

...

Perhaps try the tracing paper method systematically. Mark the points as you've tried them, so you know you're checking everything.

Reply 7

BabyMaths

There is a simple algebraic method. From (-2,1) to (3,-2) we have differences of 5 and 3. Find two numbers satisfying a+b=5 and a-b=3. These are 4 and 1.
(-2,1) +(1,-4)=(-1,-3)
(3,-2)+(-4,1)=(-1,-3).

..but I'd stick with the tracing paper.

Reply 8

OP

Original post by aznkid66

...

Thanks for the suggestion. But every time I try to draw those 'curves' (circle bit on the line) in order to do the perpendicular biesctor, the width of the compass messes up and changes. I cannot keep it the same, assumingly because the width of it is so small due to the small line :/

Original post by notnek

I'll narrow down your search by telling you that the centre lies in the bottom left quadrant (i.e. the x and y coordinates are both negative).

Spoiler

Thank you very much. I got it using the tracing paper, eventually!

Original post by BabyMaths

There is a simple algebraic method. From (-2,1) to (3,-2) we have differences of 5 and 3. Find two numbers satisfying a+b=5 and a-b=3. These are 4 and 1.
(-2,1) +(1,-4)=(-1,-3)
(3,-2)+(-4,1)=(-1,-3).

..but I'd stick with the tracing paper.

Thanks for the reply. I guess the tracing paper method is a lot easier, but in case I cannot seem to nail the point (such as in this instance) in the exam, I would like to know the algebraic method, if you wouldn't mind explaining a little bit more. So, if the two points are (-5,3) and (2,-4), there is a difference of 7 and 7. I get lost after that bit... a+b=7, a-b=7?

Reply 9

Original post by TheWantedGuy

the width of the compass messes up and changes. I cannot keep it the same, assumingly because the width of it is so small due to the small line :/

Firstly, get a better compass. If the separation of the two ends changes without you wanting them to, then it's of little use to you.

Second. How are you constructing the perpendicular bisector? What I'd do is, having chosen the two points. Select a width on the compass well over half the distance between them. Draw an arc from each point. These arcs will cross in two places. Draw a line through those two points, and you have the perpendicular bisector.

Reply 10

Original post by TheWantedGuy

Thanks for the suggestion. But every time I try to draw those 'curves' (circle bit on the line) in order to do the perpendicular biesctor, the width of the compass messes up and changes. I cannot keep it the same, assumingly because the width of it is so small due to the small line :/

You don't need to draw arcs with the same radius ^^; As long as they're centered around the points, they'll intersect on the same perpendicular bisector.

Original post by TheWantedGuy

Thanks for the reply. I guess the tracing paper method is a lot easier, but in case I cannot seem to nail the point (such as in this instance) in the exam, I would like to know the algebraic method, if you wouldn't mind explaining a little bit more. So, if the two points are (-5,3) and (2,-4), there is a difference of 7 and 7. I get lost after that bit... a+b=7, a-b=7?

If two corresponding points have the same distance d, then a=d, b=0 or a=0, b=d, so the center of rotation is at (x+d,y) or (x,y+d).

(edited 11 years ago)

Reply 11

Original post by aznkid66

You don't need to draw arcs with the same radius ^^; As long as they're centered around the points, they'll intersect on the same perpendicular bisector.

Huh! Perpendicular, but not a bisector. Radii must be the same.

Reply 12

Original post by ghostwalker

Huh! Perpendicular, but not a bisector. Radii must be the same.

D'oh!

That shows how little I know about geometry and using compasses...

Reply 13

OP

Original post by ghostwalker

Firstly, get a better compass. If the separation of the two ends changes without you wanting them to, then it's of little use to you.

Second. How are you constructing the perpendicular bisector? What I'd do is, having chosen the two points. Select a width on the compass well over half the distance between them. Draw an arc from each point. These arcs will cross in two places. Draw a line through those two points, and you have the perpendicular bisector.

Thank you for the help.

That's how I do the perpendicular bisectors, yeah. But, with regard the compass, it happens on all the compasses I use. I thought it was normal that if the distance between compass point and pencil was very small (e.g 0.5cm) then it was impossible to draw a circle without the point moving. Or does that only happen to me?

(edited 11 years ago)

Reply 14

Original post by TheWantedGuy

it was normal that if the distance between compass point and pencil was very small (e.g 0.5cm) then it was impossible to draw a circle without the point moving. Or does that only happen to me?

Yeah, 0.5 cm is very small and it would be almost impossible to draw circles with that radius, with almost any compass.

However, there is no reason to be drawing such small circles with this question. How big are your circles here? And what's the grid size?

(edited 11 years ago)

Reply 15

BabyMaths

Original post by TheWantedGuy

Thanks for the reply. I guess the tracing paper method is a lot easier, but in case I cannot seem to nail the point (such as in this instance) in the exam, I would like to know the algebraic method, if you wouldn't mind explaining a little bit more. So, if the two points are (-5,3) and (2,-4), there is a difference of 7 and 7. I get lost after that bit... a+b=7, a-b=7?

Perfect example to wreck my method. :tongue:

Can you see that this is a special case. In this case the center is directly below one of the points and directly to the left of the other point.

For another example try (0,3) and (3,2) [90 clockwise].

Differences are 3 and 1.

a+b=3
a-b=1

Add the equations 2a=4, a=2, b=1

From one point go 1 right and 2 down.
From the other go 2 right and 1 up.
Both ways bring you to the center: (1,1).

(edited 11 years ago)

Reply 16

Original post by BabyMaths

Perfect example to wreck my method. :tongue:

Can you see that this is a special case. In this case the center is directly below one of the points and directly to the left of the other point.

For another example try (0,3) and (3,2) [90 clockwise].

Differences are 3 and 1.

a+b=3
a-b=1

Add the equations 2a=4, a=1, b=1

From one point go 1 right and 2 down.
From the other go 2 left and 1 up.
Both ways bring you to the center: (1,1).

Hm? From a single pair of points, where will always be multiple centers of rotation (any that point that lies on the perpendicular bisector, so 2 that are the same distance/angle from the points), right?

The first example still works, for a=7,b=0. Left 7, 0 up from the point on the right gives you 90 clockwise.
The second example has another center at (2,4) [90 counter-clockwise].

(edited 11 years ago)

Reply 17

BabyMaths

Original post by aznkid66

Hm? From a single pair of points, where will always be multiple centers of rotation (any that point that lies on the perpendicular bisector, so 2 that are the same distance/angle from the points), right?

Yes but if you know it's a 90 degree clockwise rotation the center is unique.

Reply 18

OP

Original post by BabyMaths

Yes but if you know it's a 90 degree clockwise rotation the center is unique.

I think I can do it now, thanks to all for the help.

Just a final question about transformations: how do I know if the original shape has been transformed or rotated, because in both transformations, the position + orientation of the shape changes.

(edited 11 years ago)

Reply 19