# Thread: Coordinates of image of point

1. ## Coordinates of image of point

Hi All,

This is a transformation problem, that I need a little help with.

Point X(6, 5) is given a translation OP where O is (0, 0) and P is (4, 8). The result is then rotated 60 degrees clockwise about the origin O.

(a) State the coordinates of the final image.
(b) What will the final image of the point be if it is first rotated 60 degrees clockwise and then given the translation OP?

The translation is simple enough, X1 = (10, 13). But I haven't dealt with a transformation with rotations of other than multiples of 90 degrees.

Only thing I see at the moment is, that the 60 degrees and 2 congruent sides would make it an equilateral triangle. Any advice on how to proceed further would be great!

Thanks.

2. There is something called dot product which I'm sure will give you the answer here, but I'm not sure whether you know about it or not. The formula is:

$\displaystyle \vec{a} \cdot \vec{b} = |a||b| \cos\theta$

If you want to get further in formation about how to use it, just ask

Otherwise, solving it by sketching would be the way I'd take.

3. Hey @unknown008,

I haven't done vector multiplication yet. But I know the cosine rule(derived via Pythagoras theorem). You suggestion looks similar to it. Can you please clarify?

4. Originally Posted by mathguy80
Point X(6, 5) is given a translation OP where O is (0, 0) and P is (4, 8). The result is then rotated 60 degrees clockwise about the origin O.
(a) State the coordinates of the final image.
(b) What will the final image of the point be if it is first rotated 60 degrees clockwise and then given the translation OP?
Rotating the point $\displaystyle (x,y)\to(x',y')$ about the origin through an angle of $\displaystyle \theta$ is accomplished by this system:
$\displaystyle \begin{gathered} x' = x\cos (\theta ) - y\sin (\theta ) \hfill \\ y' = x\sin (\theta ) + y\cos (\theta ) \hfill \\ \end{gathered}$

In this question $\displaystyle \theta=60^o$.

5. Hey Plato,

Thanks, I used your method and the answer checks out. Now I need to understand how to get there! This looks like some derived Trigonometry identities. Can you point out what I areas I need to study? Thanks.

6. Originally Posted by mathguy80
I used your method and the answer checks out. Now I need to understand how to get there! This looks like some derived Trigonometry identities. Can you point out what I areas I need to study? Thanks.
It is done with matrix multiplication. It is standard in vector geometry.
$\displaystyle \left( {\begin{array}{*{20}c} {\cos (\theta )} & { - \sin (\theta )} \\ {\sin (\theta )} & {\cos (\theta )} \\ \end{array} } \right)\left[ {\begin{array}{*{20}c} x \\ y \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {x\cos (\theta ) - y\sin (\theta )} \\ {x\sin (\theta ) + y\cos (\theta )} \\ \end{array} } \right]$

7. Thanks. More work ahead for me.

8. It's only now that I get a notification that you answered

Okay, the way it is worked out is:

$\displaystyle \displaystyle \binom{x_1}{y_1} \cdot \binom{x_2}{y_2} = \sqrt{x_1\ ^2 + y_1\ ^2}\sqrt{x_2\ ^2 + y_2\ ^2} \cos\theta$

Simplify:

$\displaystyle x_1x_2 + y_1y_2 = \sqrt{x_{1}\ ^2 + y_{1}\ ^2}\sqrt{x_2\ ^2 + y_2\ ^2}\cos\theta$

Putting in the values gives you:

$\displaystyle 4x_2 + 8y_2 = \sqrt{4 ^2 + 8^2}\sqrt{x_2 \ ^2 + y_2\ ^2}\cos60$

(the vector from the origin to the point is (4 8)

$\displaystyle 4x_2 + 8y_2 = \sqrt{80}\sqrt{x_2\ ^2 + y_2\ ^2}\cos60$

And since you know that the point is rotated through the origin, the length from the origin to the point stays the same. Hence $\displaystyle \sqrt{x_2\ ^2 + y_2\ ^2} = \sqrt{80}$

$\displaystyle 4x_2 + 8y_2 = 80 \cos60$

$\displaystyle x_2 + 2y_2 = 10$

From here, you have two equations, which gives two coordinates.

$\displaystyle x_2\ ^2 + y_2\ ^2 = 80$

$\displaystyle x_2 + 2y_2 = 10$

You can solve for both and make a quick sketch to reject one answer.

x2 and y2 is the vector OP' and you can get the coordinates of P easily from there.

I'm not that familiar with matrix transformations, so... I'm using what I'm sure to solve problems.

9. Thanks Unknown008. I think I understand your idea, working towards simultaneous equations. I guess I need to study more to be able to do things like this by myself!