Math Is Fun Forum

Au101

Member

Registered: 2010-12-01

Posts: 353

Transformations

This time, I was confident:

24. Under a certain enlargement (3,4) → (0,-2) and (-5,2) → (4,-1). Find the image of (-3,-8) under this enlargement.

Now, I'm a little rusty, so I looked up the form of the matrix for the "scaling" transformation and found:

Gives a scaling by k in all directions.

I tried this, but my answer was wrong. Then I thought: 'hey! maybe the scaling isn't the same in both directions,' so I tried instead:

Still wrong. Then I realised: 'oh! the centre may not be (0,0),' so I tried:

Still wrong. Then I thought: 'well, what if the x and y coordinates of the centre of enlargement are different?' So that was when I had my last try:

Running the numbers gives me:

But:

Does not give the right answer.

So my question for the good folks at mathisfun is: who's wrong - me, or the book?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,643

Re: Transformations

hi Au101,

The wording doesn't say there's a matrix for this.

To get a 2 by 2 matrix for any transformation the origin has to be an invariant point, because M times origin = origin for all M.

So I thought maybe the centre of enlargement isn't the origin.

I plotted the points and joined the start point to the end point for the two given points and their image points.  Where those lines cross would be the centre of enlargement.

Looks like the scale factor is -1/2

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Au101

Member

Registered: 2010-12-01

Posts: 353

Re: Transformations

I see! So, in general, how should I proceed with a question like this? Matrix algebra seemed the best way, simply because it's the way I'm familiar with!

Last edited by Au101 (2015-01-07 11:42:01)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,643

Re: Transformations

You have to remember that the origin will always have to be an invariant point.  This means you can only find a matrix for:

Rotations around (0,0)
Reflections in lines of the form y = mx
Enlargements centred on (0,0)
Shears where an axis is invariant.
Stretches from either axis.

There are others but I'm not sure what to call them.

My first thought for the question was to plot the points.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Au101

Member

Registered: 2010-12-01

Posts: 353

Re: Transformations

What do you get as the final answer? Because I get (3,4), while the book has (7,16).

Au101

Member

Registered: 2010-12-01

Posts: 353

Re: Transformations

I do agree with you about the centre of enlargement and the scale factor, though, thanks!

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: Transformations

(3,4) is correct. (7,16) seems to be the point that maps to (-3,-8).

---

“Here lies the reader who will never open this book. He is forever dead.
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

Au101

Member

Registered: 2010-12-01

Posts: 353

Re: Transformations

You're right, thanks for the help

Au101

Member

Registered: 2010-12-01

Posts: 353

Re: Transformations

Wahey, let's have some fun with part 2 before bed

It's been a long time, I'm not entirely sure how the vector:

Would look on a diagram, but my gut reaction is that it would like the graph of y = x.

So what I did was:

Which I guess would look like the graph of x = 0?

Next I did:

Which I imagine looks like the graph of y = 2x.

Am I on the right track. I hate drawing diagrams

Bob

Administrator

Registered: 2010-06-20

Posts: 10,643

Re: Transformations

These are both called shears.  Sorry, but I think you'll have to draw a graph to see what happens.  Rather than (x,y) try drawing a polygon and transforming the points.  When I did this with classes I would start with an F shape for transformations because it has no symmetry and so you can tell when it gets reflected / rotated etc.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Olinguito

Member

Registered: 2014-08-12

Posts: 649

Re: Transformations

24. Under a certain enlargement (3,4) → (0,-2) and (-5,2) → (4,-1). Find the image of (-3,-8) under this enlargement.

The quickest way to solve this is probably to plot a graph, as Bob Bundy did in post #2. However, if you need to solve it algebraically, you can do this.

Let the centre of enlargement be (a,b). We apply the translation (x,y) → (x−a,y−b) to make the centre of enlargement coincide with the origin. Under the new coordinates the enlargement becomes

[list=*]
[*]

[/*]
[/list]

Thus if k is the enlargement factor we have the following equations:

[list=*]
[*]

[/*]
[/list]

Solve these equations for a, b, k. The image of (−3,−8) under the enlargement is ((−3−a)k+a,(−8−b)k+b).

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Bassaricyon neblina

Au101

Member

Registered: 2010-12-01

Posts: 353

Re: Transformations

Thanks for all the help so far.

Okay, so - sorry, I've only just got a chance to work on this - I tried applying S1 to an F shape and then to a simple right-angled triangle and, well, all it did was distort my shapes! My nice right-angled triangle has been moved and the angles have changed dramatically I'm not really sure how to proceed and I'm also not really sure what the vector:

looks like, either :S

Bob

Administrator

Registered: 2010-06-20

Posts: 10,643

Re: Transformations

hi Au101

This is what I would do:

You can see from this that points on the x axis do not move.  We call this an invariant line.

Points 1 unit above the invariant line move one unit left.

Points 2 units above move two units.

Points 3 units above move three units.

Points 4 units above move four units left.

This transformation is called a shear.  To fully describe it you must also give the invariant line and the 'shear factor' ; in this case, minus one as distance from invariant line x shear factor = distance moved.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Au101

Member

Registered: 2010-12-01

Posts: 353

Re: Transformations

Oh yes Yes, I ended up with that shape, but I didn't know what to do with it! So what would the final answer look like? How do I

Last edited by Au101 (2015-01-09 04:37:50)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,643

Re: Transformations

Well the new y is the same as the old.

The new x will be a point that has moved left from the old position by a distance y

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Au101

Member

Registered: 2010-12-01

Posts: 353

Re: Transformations

So could I do something like this, maybe?

(Sorry. Edit:)

Last edited by Au101 (2015-01-09 08:19:57)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,643

Re: Transformations

You have put (x,y) on the line that goes through (1,1).  I would choose a point that isn't anything special.  Then you need make clear how much the point moves and in what direction, so I would put an arrow from old point to new point that shows that the movement is parallel to the x axis and equal to y in size.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Au101

Member

Registered: 2010-12-01

Posts: 353

Re: Transformations

Okay, thanks bob bundy, I think we're good now!