WOODENkitttykat
New member
- Joined
- Jun 10, 2019
- Messages
- 3
Hello! I'm struggling with this math problem:
The beginning of coordinate system moved to point [MATH]O'(-1,2)[/MATH] and then rotated it over angle [MATH]\alpha[/MATH] that [MATH]tg\alpha = \frac{5}{12}[/MATH]. Designate the point [MATH]M[/MATH]coordinates in the old coordinate system if it coordinates in the new one are [MATH](2,-3)[/MATH].
So my proposition is:
I have to designate the [MATH]\sin[/MATH] and [MATH]\cos[/MATH]
[MATH]\sin\alpha = \frac{5}{13}[/MATH] and [MATH]\cos\alpha = \frac{12}{13}[/MATH].
I use the equation for rotation:
[MATH]x' = x \cos\alpha - y \sin\alpha[/MATH][MATH]y' = x \sin\alpha + y \cos\alpha[/MATH]
So
[MATH]2 = x \cdot \frac{12}{13} - y \cdot \frac{5}{13} [/MATH][MATH]-3 = x \cdot \frac{5}{13} + y \cdot \frac{12}{13}[/MATH]
I designate [MATH]x[/MATH] and [MATH]y[/MATH] - 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 [MATH]x[/MATH] and [MATH]y[/MATH] I use the equation for translation.
[MATH]x' = x + a[/MATH][MATH]y' = y + b[/MATH]
[MATH]x'[/MATH] is earlier designated [MATH]x[/MATH], as well as [MATH]y'[/MATH]. And [MATH]a = 2, b = -3[/MATH].
I designate [MATH]x[/MATH] and [MATH]y[/MATH] from equation and thats the old coordinates of the point [MATH]M[/MATH].
If anyone could tell me if this solution is correct? I would be very grateful.
The beginning of coordinate system moved to point [MATH]O'(-1,2)[/MATH] and then rotated it over angle [MATH]\alpha[/MATH] that [MATH]tg\alpha = \frac{5}{12}[/MATH]. Designate the point [MATH]M[/MATH]coordinates in the old coordinate system if it coordinates in the new one are [MATH](2,-3)[/MATH].
So my proposition is:
I have to designate the [MATH]\sin[/MATH] and [MATH]\cos[/MATH]
[MATH]\sin\alpha = \frac{5}{13}[/MATH] and [MATH]\cos\alpha = \frac{12}{13}[/MATH].
I use the equation for rotation:
[MATH]x' = x \cos\alpha - y \sin\alpha[/MATH][MATH]y' = x \sin\alpha + y \cos\alpha[/MATH]
So
[MATH]2 = x \cdot \frac{12}{13} - y \cdot \frac{5}{13} [/MATH][MATH]-3 = x \cdot \frac{5}{13} + y \cdot \frac{12}{13}[/MATH]
I designate [MATH]x[/MATH] and [MATH]y[/MATH] - 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 [MATH]x[/MATH] and [MATH]y[/MATH] I use the equation for translation.
[MATH]x' = x + a[/MATH][MATH]y' = y + b[/MATH]
[MATH]x'[/MATH] is earlier designated [MATH]x[/MATH], as well as [MATH]y'[/MATH]. And [MATH]a = 2, b = -3[/MATH].
I designate [MATH]x[/MATH] and [MATH]y[/MATH] from equation and thats the old coordinates of the point [MATH]M[/MATH].
If anyone could tell me if this solution is correct? I would be very grateful.