For every point A = (x,y) in your figure, a 180 degree counterclockwise rotation about the origin will result in a point A' = (x', y') where: x' = x * cos(180) - y * sin(180) y' = x * sin(180) + y * cos(180)
Happy-fun time fact: This is equivalent to using a rotation matrix from Linear Algebra!
Because a rotation is an isometry, you only have to rotate each vertex of a polygon, and then connect the respective rotated vertices to get the rotated polygon.
You can rotate a closed curve as well, but you must figure out a way to rotate the infinite number of points in the curve. We are able to do this with straight lines above due to the property of isometries, which preserves distances between points.
The simplest fraction for is . Write the upper bound as a fraction with the same denominator:
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Hence the range for would be:
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If the denominator of is also , then the range for its numerator (call it ) would be . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than .
To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)
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At this point, the difference between the numerators is now . That allows a number ( in this case) to fit between the bounds. However, can't be written as finite decimals.
Try multiplying the numerator and the denominator by a different number.
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It is important to note that some expressions for can be simplified. For example, because of the common factor .