Math Is Fun Forum

Karuna

Guest

Linear transformations

Find the image of the triangle ABC with the vertices A(2,1),B(4,3),C(3,1) under a stretch, scale factor 2. with invariant line y=x

Bob

Administrator

Registered: 2010-06-20

Posts: 10,602

Re: Linear transformations

hi Karuna,

Welcome to the forum.

In a stretch the movement is perpendicular to the invarient line. For a scale factor of 2, points move away to twice the distance.

Hope that helps.

Bob

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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

rk_nithi

Novice

Registered: Yesterday

Posts: 1

Re: Linear transformations

hi Karuna,

Welcome to the forum.

In a stretch the movement is perpendicular to the invarient line. For a scale factor of 2, points move away to twice the distance.

https://i.imgur.com/QqaD6h5.gif

Hope that helps.

Bob

Thanks for the feedback. Answer is correct. i would like to understand how to prove this with mathematical steps or equation? also can you tell me which software you are using to plot the graph?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,602

Re: Linear transformations

hi rk_nithi

Welcome to you as a new member!

Easy bit first. This is a vector graphics program called Geometer's Sketchpad.  There is also a free program called Geogebra which works similarly.  Geo is more versatile but takes longer to learn.  Sketchpad costs money but I got it years ago (before Geo) and it still serves me well. I like the control it gives over thickness of lines and colours.

Transformations that leave the origin invariant can be represented by a matrix transformation.  So, in theory there must be one for this one but I'll have to do a bit of work to find it.  I'll come back to this later.

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

Bob

Administrator

Registered: 2010-06-20

Posts: 10,602

Re: Linear transformations

OK, got it.

Are you familiar with matrix multiplication?  If not then there are two useful pages here:

https://www.mathsisfun.com/algebra/matr … lying.html

https://www.mathsisfun.com/algebra/matr … lator.html

In 2D transformation geometry 2 by 1 vectors (coordinates switched row and column) can be transformed by multiplying by a 2 by 2 matrix.

As any such matrix transforms (0,0) to (0,0) this only works when the transform leaves the origin invariant.  The one for the question does as y=x goes through the origin.

I'll call the matrix

y=x is invariant so (x,x) maps onto (x,x)

This gives us two equations

a + b = 1 so b = 1-a         and c + d = 1 so d = 1-c

So the matrix becomes

Now to fix which stretch by considering a single point under the transformation.

I'll start on the line at (2,2) and go one right and one down to (3,1) and again to (4,0)

That's a good point to consider as there's a zero in the calculation.

It's image is at one right and one down and the same again ie at (6,-2)

So (4,0) maps onto (6,-2)

This leads to a = 1.5 and c = -0.5

So the matrix for this transformation is

So what does this do to our points?

QED

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