#### WOODENkitttykat

##### New member

- Joined
- Jun 10, 2019

- Messages
- 3

The beginning of coordinate system moved to point \(\displaystyle O'(-1,2)\) and then rotated it over angle \(\displaystyle \alpha\) that \(\displaystyle tg\alpha = \frac{5}{12}\). Designate the point \(\displaystyle M\)coordinates in the old coordinate system if it coordinates in the new one are \(\displaystyle (2,-3)\).

So my proposition is:

I have to designate the \(\displaystyle \sin\) and \(\displaystyle \cos\)

\(\displaystyle \sin\alpha = \frac{5}{13}\) and \(\displaystyle \cos\alpha = \frac{12}{13}\).

I use the equation for rotation:

\(\displaystyle x' = x \cos\alpha - y \sin\alpha\)

\(\displaystyle y' = x \sin\alpha + y \cos\alpha\)

So

\(\displaystyle 2 = x \cdot \frac{12}{13} - y \cdot \frac{5}{13} \)

\(\displaystyle -3 = x \cdot \frac{5}{13} + y \cdot \frac{12}{13}\)

I designate \(\displaystyle x\) and \(\displaystyle y\) - coordinates of the point before the rotation.

Thats where I suspect my thinking is wrong because the results are too complicated.

But after that having \(\displaystyle x\) and \(\displaystyle y\) I use the equation for translation.

\(\displaystyle x' = x + a\)

\(\displaystyle y' = y + b\)

\(\displaystyle x'\) is earlier designated \(\displaystyle x\), as well as \(\displaystyle y'\). And \(\displaystyle a = 2, b = -3\).

I designate \(\displaystyle x\) and \(\displaystyle y\) from equation and thats the old coordinates of the point \(\displaystyle M\).

If anyone could tell me if this solution is correct? I would be very grateful.