# Thread: Rotation of a triangle

1. ## Rotation of a triangle

A triangle has vertices at the points
A(6, 9), B(6, 3) and C(9,6).
Suppose that the triangle is to be moved so that
B is at the origin
and
BC lies along the positive x-axis. One isometry that achieves this
transformation is the composite of a translation followed by a
rotation. (You may find it helpful to sketch the triangle.)
(i) Determine the translation that moves
B to the origin, giving your
ta,b. Write down a formal definition of this
translation in two-line notation.
2
(ii) Find the coordinates of the images
A! of A and C! of C under the
translation in part (a)(i).

(iii) 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 θ and 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.)

(iv) Find the coordinates of the images of
A! and C! under the
rotation

(v) Write down a formal definition of the composite transformation,
that is, the result of the translation in part (a)(i) followed by the
rotation in part (a)(iii).

I've been stuck on this all day! I have t-6,3 for part (i) and found A to be (-12,6) and C to be (3,-9) for part (ii).
Other than that I'm totally lost!
Any help would be really appreciated, thanks.

2. ## Re: Rotation of a triangle

Hey alexlbrown59

Do you have to put your answers in matrix form (where you find the matrices and apply them to the vectors themselves)?

3. ## Re: Rotation of a triangle

I haven't covered matrix form yet, so no I don't think so.
Can you still help?

5. ## Re: Rotation of a triangle

I've made slight progress...I have sin theta = 3sqrt10/10, cos theta = sqrt10/10 and tan theta = 3
Am I on the right lines?

6. ## Re: Rotation of a triangle

Let's work out the translation, first (since we have to APPLY it first). We need to move point B = (6,3) to the origin O = (0,0).

Since we're going to move "everything at once", let's just subtract 6 from every x-coordinate, and 3 from every y-coordinate. That should do the trick.

Hence: $$t_{-6,-3}(x,y) = (x-6,y-3)$$

or:

$$\begin{array}{l}x' = x - 6\\y' = y - 3 \end{array}$$

Before we do the rotation, let's think about what we're going to do: we need to put point C on the x-axis.

Now after the translation, C is at point (9 - 6, -6 - 3) = (3,-9).

We can form a right triangle with hypotenuse OC. The hypotenuse has length:

$$\sqrt{3^2 + (-9)^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10}$$.

This tells us:

$$\cos(-\theta) = \frac{1}{\sqrt{10}},\ \sin(-\theta) = \frac{-3}{\sqrt{10}}$$

and therefore that:

$$\cos\theta = \frac{1}{\sqrt{10}},\ \sin\theta = \frac{3}{\sqrt{10}}$$

Hence:

$$\tan\theta = 3$$.

We can write the rotation as:

$$r_{\theta}(x',y') = (x'\cos\theta - y'\sin\theta,x'\sin\theta + y'\cos\theta)$$

or:

$$\begin{array}{l}x'' = x'\cos\theta - y'\sin\theta\\y'' = x'\sin\theta + y'\cos\theta \end{array}$$

so, explicitly, this becomes:

$$r_{\theta}(x',y') = \left(\frac{x'- 3y'}{\sqrt{10}},\frac{3x'+y'}{\sqrt{10}}\right)$$

The composition is thus:

$$r_{\theta} \circ t_{-6,-3}(x,y) = r_{\theta}(t_{-6,-3}(x,y))$$

which you should be perfectly capable of evaluating.

7. ## Re: Rotation of a triangle

Thank you so much Deveno!
I had tan theta = 3, sin theta = 3sqrt10/10 (which is the same as 3/sqrt10) and cos theta = sqrt10/10 (which is the same as 1/sqrt10) so you've proven me right!
Thanks for taking the time to help