Designate the old point coordinates (having the new one) after translation and rotation

WOODENkitttykat

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Hello! I'm struggling with this math problem:

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. :)
 

Dr.Peterson

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The beginning of coordinate system moved to point O′(−1,2) and then rotated it over angle α that \(\displaystyle tg\alpha = \frac{5}{12}\). Designate the point M coordinates in the old coordinate system if it coordinates in the new one are (2,−3).
I think you are saying that the origin, O(0, 0), is moved to O'(-1, 2), so that the plane is translated 1 unit left and 2 units up; and that the angle α is assumed to be in the first quadrant, not the third.

If anyone could tell me if this solution is correct? I would be very grateful.
I don't think you've shown your solution, just a description of your method of solution. The method does sound appropriate, except that I think you meant a = -1, b = 2.

I hope you would then check your answer by taking it through the indicated transformations and seeing that it works.

I will tell you that my own first idea for solving this is to reverse each transformation: first reverse the rotation by rotating (2, -3) by the same angle but clockwise; then reverse the translation by moving 1 right and 2 down. The best thing to do may be to try both methods: yours is more formal and practices following definitions, while mine is more informal and focuses on the meaning.
 

WOODENkitttykat

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I think you are saying that the origin, O(0, 0), is moved to O'(-1, 2), so that the plane is translated 1 unit left and 2 units up; and that the angle α is assumed to be in the first quadrant, not the third.
What does knowledge that it is the first quadrant give me?

I will tell you that my own first idea for solving this is to reverse each transformation: first reverse the rotation by rotating (2, -3) by the same angle but clockwise; then reverse the translation by moving 1 right and 2 down. The best thing to do may be to try both methods: yours is more formal and practices following definitions, while mine is more informal and focuses on the meaning.
Does reversing it means that i change the signs like so:

\(\displaystyle 2 = x \cdot (-\frac{12}{13}) + y \cdot \frac{5}{13}\)
\(\displaystyle -3 = x \cdot (-\frac{5}{3}) - y \cdot (-\frac{12}{13})\)

I don't think you've shown your solution, just a description of your method of solution. The method does sound appropriate, except that I think you meant a = -1, b = 2.
Of course, I meant that. My mistake.
 

Dr.Peterson

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What does knowledge that it is the first quadrant give me?
It tells you which of two different angles with the same tangent you are referring to! As it stands, α could be either 22.62° or 202.62°. I'm just guessing that the former is intended.

Does reversing it means that i change the signs like so:

2=x⋅(−1213)+y⋅513\displaystyle 2 = x \cdot (-\frac{12}{13}) + y \cdot \frac{5}{13}
−3=x⋅(−53)−y⋅(−1213)
No. You would reverse the angle (using -α, which changes only the sine) in order to undo the transformation (getting the original point from the final point):

\(\displaystyle x = 2 \cdot (\frac{12}{13}) - (-3) \cdot (-\frac{5}{13})\)​
\(\displaystyle y = 2 \cdot (-\frac{5}{3}) + (-3) \cdot (-\frac{12}{13})\)​

This should give the same result you get by solving your system of equations for x and y. It just takes a little less work.
 

WOODENkitttykat

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OK. So either like this:

\(\displaystyle 2 = \frac{12}{13}x + \frac{5}{13}y\)
\(\displaystyle -3 = -\frac{5}{13}x + \frac{12}{13}y\)

I'm adding it:

\(\displaystyle -1 = \frac{7}{13}x + \frac{17}{13}y\)
\(\displaystyle -\frac{17}{13}y = \frac{7}{13}x + 1\)
\(\displaystyle y = \frac{7}{13}x \cdot (-\frac{13}{17}) - \frac{13}{17}\)
\(\displaystyle y = -\frac{7}{17}x - \frac{13}{17}\)


\(\displaystyle 2 = \frac{12}{13}x + \frac{5}{13}(-\frac{7}{17}x - \frac{13}{17})\)
\(\displaystyle 2 = \frac{12}{13}x - \frac{35}{221} - {5}{17}\)
\(\displaystyle \frac{442}{221} + \frac{65}{221} = \frac{204}{221}x - \frac{35}{221}x\)
\(\displaystyle \frac{507}{221} = \frac{169}{221}x\)
\(\displaystyle \frac{507}{221} \cdot \frac{221}{169} = x\)
\(\displaystyle \frac{507}{169} = x\)
\(\displaystyle x = 3\)


\(\displaystyle 2 = \frac{12}{13} \cdot 3 + \frac{5}{13}y\)
\(\displaystyle \frac{26}{13} = \frac{36}{13} + \frac{5}{13}y\)
\(\displaystyle -\frac{10}{13} = \frac{5}{13}y\)
\(\displaystyle -\frac{10}{13} \cdot \frac{13}{5} = y\)
\(\displaystyle y = -2\)

Or like this:

No. You would reverse the angle (using -α, which changes only the sine) in order to undo the transformation (getting the original point from the final point):

\(\displaystyle x = 2 \cdot (\frac{12}{13}) - (-3) \cdot (-\frac{5}{13})\)​
\(\displaystyle y = 2 \cdot (-\frac{5}{3}) + (-3) \cdot (-\frac{12}{13})\)​

This should give the same result you get by solving your system of equations for x and y. It just takes a little less work.
\(\displaystyle x = \frac{24}{13} - \frac{15}{13} = \frac{9}{13}\)
\(\displaystyle y = -\frac{10}{13} + \frac{36}{13} = \frac{26}{13} = 2\)

So it doesn't give me the same result. But maybe I didn't understand you correctly.

So that gives me the point coordinates before rotation.

then reverse the translation by moving 1 right and 2 down.
\(\displaystyle 3 = x - 1\)
\(\displaystyle -2 = y - (-2)\)

\(\displaystyle x = 4\)
\(\displaystyle y = -4\)

OR

\(\displaystyle \frac{9}{13} = x - 1\)
\(\displaystyle 2 = y + 2\)

\(\displaystyle x = \frac{22}{13}\)
\(\displaystyle y = 0\)

So the points \(\displaystyle M\) coordinates are \(\displaystyle (4,-4)\) or \(\displaystyle (\frac{22}{13},0)\)

I'm sure I didn't understand you correcty. But is one of this version right?
 

Dr.Peterson

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First, you can make the work your way a little easier by clearing fractions at the start, multiplying everything by 13. But your work on that part is correct, and the work on the second part is correct your way. (The second part, of course, gives the wrong result by using "my" wrong values.)

Second, I made a couple typos, one of which you fixed (changing 3 to 13), and the other was less obvious (both 12/13 should have been positive).

Third, I think one or both of us is being confused by the fact that the problem is about moving the coordinate system, not points, which in itself reverses things. We'll get the same answer if I don't change the sign of the angle for my method; I'll have to give that some more thought than I can at the moment.

Rather than me trying to straighten that out, how about if you take your answer, (4, -4), and check whether the transformations described result in assigning that point the coordinates (-2, 3) in the final coordinate system. (Do that both by the formal methods you've learned, and by graphing.) Then either correct your method, if necessary, or move on an think about what transformations would reverse the process (if you want to pursue my approach).
 

Dr.Peterson

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I have now both solved the problem correctly (as I read it), and checked my answer. The answer is one that hasn't been written here yet.

The key is that you used the wrong equations for the rotation, assuming the problem really is about rotating the axes rather than the point. See https://en.wikipedia.org/wiki/Rotation_of_axes, which shows that your equations are for the inverse transformation.

So if I'm right, I gave some wrong advice based on trusting your equations, and some based on my own misreading of the problem. Or maybe I'm wrong now. Please state the problem exactly as given to you, if you didn't; even if you translated it from another language, give us the original (and a direct translation).
 
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