# Thread: Translation and then a rotation of a triangle

1. ## Translation and then a rotation of a triangle

I have a triangle and the vertices were A(3,−1), B(−2,1) and C(2,3). I made a translation on a triangle (in another part of the paper) and now its new
vertices are A(5,−2), B(0,0) and C(4,2).

Next, I must rotate the triangle so that the line BA lies along the positive x-axis.

Here is the questions:

(3) Let rθ be the rotation that completes the required transformation, where θ lies in the interval (−π,π]. Find the exact values of tan θ,cos θand sin θ, and hence write down a formal definition of rθ using two-line notation. (There is no need to work out the value of the angle θ.)

- This is where I have the problems. I worked out as follows, but that is a lot of typing numbers so the answers I found are:

cos(θ)= 5*sqrt(29)/29

sin(θ)= 2*sqrt(29)/29

tan(θ)=2/5

-- Is this correct?

If so, I then tried to make the formal definition and I arrived with (but I can not show the right arrow shape!):

r(θ): R*2-->R*2
(x,y) |--> (x cos θ - y sin θ, x sin θ + y cos θ)

So I replace all of this with the values at the above paragraph (a lot of typing of the surds, so I don't be putting it all here the working out). We do not show matrices as we are not covering them yet in class, so I must always write with surds.

I finish with the (x,y) values of

(25*sqrt(29)/29 + 45*sqrt(29)/29, 10*sqrt(29)/29 - 10*sqrt(29)/29)

This makes coordinates ( (sqrt(29), 0).

-- But this does not look correct to me. Particular on the drawing I made. Please advise if I am doing this correct.

(4) Find the coordinates of the images of A' and C' under the rotation rθ. Give your answers as exact values. This means surds.

- But I just don't know how to do this question. I wrote a lot of workings and find different answers each time and they don't look right on the drawing that I made of the triangle in the x axis.

-- Can you please help me as I am spending so many hours on this today and yesterday and I go not very far!

(5) Write down a formal definition of the composite transformation: that is the result of the translation followed by the rotation above.

- I can not make this definition without getting the above question 2 well done.

Brigitte

2. Hello Brigitte
Originally Posted by dolkam
I have a triangle and the vertices were A(3,−1), B(−2,1) and C(2,3). I made a translation on a triangle (in another part of the paper) and now its new
vertices are A(5,−2), B(0,0) and C(4,2).

Next, I must rotate the triangle so that the line BA lies along the positive x-axis.

Here is the questions:

(3) Let rθ be the rotation that completes the required transformation, where θ lies in the interval (−π,π]. Find the exact values of tan θ,cos θand sin θ, and hence write down a formal definition of rθ using two-line notation. (There is no need to work out the value of the angle θ.)

- This is where I have the problems. I worked out as follows, but that is a lot of typing numbers so the answers I found are:

cos(θ)= 5*sqrt(29)/29

sin(θ)= 2*sqrt(29)/29

tan(θ)=2/5

-- Is this correct?
Yes. Good work!

If so, I then tried to make the formal definition and I arrived with (but I can not show the right arrow shape!):

r(θ): R*2-->R*2
(x,y) |--> (x cos θ - y sin θ, x sin θ + y cos θ)
Correct. But you have then used the coordinates of $A'$ in the next part of your working:
So I replace all of this with the values at the above paragraph (a lot of typing of the surds, so I don't be putting it all here the working out). We do not show matrices as we are not covering them yet in class, so I must always write with surds.

I finish with the (x,y) values of

(25*sqrt(29)/29 + 45*sqrt(29)/29, 10*sqrt(29)/29 - 10*sqrt(29)/29)

This makes coordinates ( (sqrt(29), 0).

-- But this does not look correct to me. Particular on the drawing I made. Please advise if I am doing this correct.
The formal definition will simply leave $x$ and $y$ alone, and just use the values of $\cos \theta$ and $\sin\theta$ that you have found:
$(x,y)\mapsto\Big(\frac{5\sqrt{29}}{29}x-\frac{2\sqrt{29}}{29}y,\frac{2\sqrt{29}}{29}x+\fra c{5\sqrt{29}}{29}y\Big)$
(4) Find the coordinates of the images of A' and C' under the rotation rθ. Give your answers as exact values. This means surds.

- But I just don't know how to do this question. I wrote a lot of workings and find different answers each time and they don't look right on the drawing that I made of the triangle in the x axis.

-- Can you please help me as I am spending so many hours on this today and yesterday and I go not very far!
You have already worked out the image of A': $(\sqrt{29},0)$, by putting $x = 5, y = -2$ in the formula above. Now use the coordinates of $C'$ $(4,2)$ in the same way to get the image of $C'$.
(5) Write down a formal definition of the composite transformation: that is the result of the translation followed by the rotation above.

- I can not make this definition without getting the above question 2 well done.

Brigitte
The translation is:
$(x,y)\mapsto(x+2,y-1)$
So when we follow this by the rotation, we get:
$(x,y)\mapsto\Big(\frac{5\sqrt{29}}{29}(x+2)-\frac{2\sqrt{29}}{29}(y-1),...\;?\Big)$
You complete it!

3. Thank you so much Grandad.

Now have I worked the coordinates of C' and have all the answers that I am required.

It is great that I have my answers but I don't really understand what I am doing really!

Can you please explain in a little words how those formulas know that I am wanting to put the BA line on the x-axis and to rotate the triangle from the origin! It is to me a mystery. It is probably obvious to you but to me not!

Only if you have the time. They do not require this answer for my assignment. I am the interested!

Thank you again.
Brigitte

4. ## Rotations and Matrices

Hello Brigitte
Originally Posted by dolkam

Now have I worked the coordinates of C' and have all the answers that I am required.

It is great that I have my answers but I don't really understand what I am doing really!

Can you please explain in a little words how those formulas know that I am wanting to put the BA line on the x-axis and to rotate the triangle from the origin! It is to me a mystery. It is probably obvious to you but to me not!

Only if you have the time. They do not require this answer for my assignment. I am the interested!

Thank you again.
Brigitte
Look at the diagram I've attached. (I'm assuming that the original points were called $A,B,C$ and that $A',B',C'$ are their images after the initial translation.)

We have to rotate the triangle $A'B'C'$ so that $A'B'$ lies along the $x$-axis.

Now, obviously, $B'$ is already on the $x$-axis, since it lies at the origin. So we shall need to rotate the triangle about $B'$ - since this will leave $B'$ where it is. The angle through which the triangle needs to rotate, then, is $\angle A'B'D' = \theta$. And, using simple trigonometry on $\triangle A'B'D'$:
$\tan\theta = \frac{A'D'}{B'D'}=\frac25$

$\sin\theta = \frac{2}{\sqrt{29}}$

$\cos\theta = \frac{5}{\sqrt{29}}$

You obviously have been told that the transformation that will rotate about the origin through an angle $\theta$ anticlockise is:
$(x,y)\mapsto(x\cos\theta-y\sin\theta, x\sin\theta+y\sin\theta)$
So, plugging in the values of $\sin\theta$ and $\cos\theta$, there's your answer.

Matrix Explanation of this formula (Ignore this if you haven't covered matrix multiplication)

You mention matrices, but you say you haven't covered them yet, so you may not understand this next bit. However, they do give the easiest explanation of this formula, if you do understand how matrix multiplication works. The matrix
$\begin{pmatrix}a&b\\c&d\end{pmatrix}$
transforms the points $(1,0)$ and $(0,1)$ as follows:
$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix }1&0\\0&1\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end {pmatrix}$
Therefore:
$\begin{pmatrix}1\\0\end{pmatrix}\mapsto\begin{pmat rix}a\\c\end{pmatrix}$
and
$\begin{pmatrix}0\\1\end{pmatrix}\mapsto\begin{pmat rix}b\\d\end{pmatrix}$
Now after a rotatation through an angle $\theta$ anticlockwise, it's very easy to check that:
$\begin{pmatrix}1\\0\end{pmatrix}\mapsto\begin{pmat rix}\cos\theta\\\sin\theta\end{pmatrix}$
and
$\begin{pmatrix}0\\1\end{pmatrix}\mapsto\begin{pmat rix}-\sin\theta\\\cos\theta\end{pmatrix}$
Therefore, using what we've just said using $a, b, c, d$ in a matrix, the matrix that will carry out this rotation is
$\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$
So, finally if we use this matrix to transform any point $(x,y)$, we get:
$\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\beg in{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x\cos\ theta-y\sin\theta\\x\sin\theta+y\cos\theta\end{pmatrix}$
In other words:
$(x,y)\mapsto(x\cos\theta-y\sin\theta, x\sin\theta+y\sin\theta)$