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)?

Re: Rotation of a triangle

Hi Chiro, thanks for replying!

I haven't covered matrix form yet, so no I don't think so.

Can you still help?

Re: Rotation of a triangle

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?

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.

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 :)