Question

Describe a transformation that carries the triangle with vertices (0,0), (13,0), and (3, 2) onto the triangle with vertices (0, 0), (12, 5), and (2, 3). Where does your transformation send the point (6, 0)?

Answer 1

@mathmate

Answer 2

I drew the triangles. It seems like as if I just need to rotate the original triangle by a certain number of degrees to get to its image

Answer 3

I just don't know what number to rotate it by.

Answer 4

Have you done transformation by matrices?

Answer 5

No.

Answer 6

This one the angle theta is defined by sin(\(\theta\))=5/13, as you can probably work out from the geometry of the graph.

Answer 7

I am sorry but we are not learning trig either

Answer 8

I understand you have not done the following before, but this is the classical way of rotating an angle which is not a multiple of 90 degrees.
The transformation matrix is written
\[\left[\begin{matrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{matrix}\right]\]
If we multiply the point (13,0) by the above matrix, we get the point (12,5) as we were given. Same for the other points.

The problem is, we need to find a way to do the transformation without using this.

Answer 9

Yes what other way can we do this?

Answer 10

Well, let the points be ABC and A'B'C' in the order given.
Then (6,0) lies in the segment AB, at a distance of 6 from A and 7 from B right?

Answer 11

I don't understand...

Answer 12

Is this the graph you drew?
I have added the letters ABC and A'B'C' to the vertices.

Answer 13

OK. I understand what you're saying. But what is the relevance of (6,0)?

Answer 14

Very relevant if we cannot use the matrix method of rotation!
I'd like you to draw the point D(6,0) on the diagram.

Answer 15

@calculusxy
Hmmm, still there?

Answer 16

Yeah

Answer 17

Somewhere like that?

Answer 18

Excellent! We're moving ahead!
Now can you calculate and mark on the diagram the distances AD and DB, please?

Answer 19

ok