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Question 1191633: Find the image of point (5, 3) when it is rotated 1/4 of a turn counter-clockwise around the point (-1,4) .
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
The center of rotation is (-1,4)
We need to translate this center of rotation to the origin (0,0)
Apply the translation rule  \to (x+1,y-4)) so that (-1,4) moves to (0,0)
This will move (5,3) to (6,-1)
Another way to look at it: imagine that the xy axis is allowed to move but the points (5,3) and (-1,4) stay fixed.
If you moved the xy axis to have (-1,4) placed at the intersection of the axes, then it will give the illuison of movement described earlier.
Now that we have the center of rotation be the origin, we apply this 90 degree counter-clockwise rule:
 \to (-y,x))
This means (6,-1) becomes (1,6) after the counter-clockwise quarter turn.
After the rotation is done, we have to undo the translation we initially did. This will get us back to the correct frame of reference so to speak.
The translation we did was
the inverse or opposite translation would be
 \to (x-1,y+4))
That moves the point (1,6) to (0,10) which is the final answer.
Visual Confirmation (I used GeoGebra to make the graph):
Answer: (0,10)
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Here is a far less formal method for answering the question.
The displacement (I will avoid the use of the formal word "vector") from the center of the rotation (-1,4) to the given point (5,3) is "6 right, down 1".
When the point is rotated 1/4 turn counterclockwise about (-1,4), the "6 right" becomes "up 6" and the "down 1" becomes "right 1".
So the image of the point (5,3) under the given rotation is (-1+1,4+6) = (0,10).
ANSWER: (0,10)
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