Because 180 degrees clockwise is the same as 180 degrees counterclockwise.
A 180° rotation is half a rotation and it doesn't matter if it is clockwise of counter clockwise. When rotating 180° about the origin, the x-coordinate and y-coordinates change sign Thus (1, -6) → (-1, 6) after rotating 180° around the origin.
180 degrees
360 degrees would be one full rotation. 180 degrees would be a half rotation. 360+180=540 So it would be a rotation and a half.
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.
Because 180 degrees clockwise is the same as 180 degrees counterclockwise.
Fomula(work with both clockwise/counterclockwise):(-x,-y)
A 180° rotation is half a rotation and it doesn't matter if it is clockwise of counter clockwise. When rotating 180° about the origin, the x-coordinate and y-coordinates change sign Thus (1, -6) → (-1, 6) after rotating 180° around the origin.
180 degrees
A 180 degree rotation between front and back is normal for US coins.
The rule for a rotation by 180° about the origin is (x,y)→(−x,−y) .
It is (-1, 6).Also, if the rotation is 180 degrees, then clockwise or anticlockwise are irrelevant.It is (-1, 6).
(-e, -h)
180 degrees is half a rotation so probably a half.
(x, y)-----> (-x, -y)
360 degrees would be one full rotation. 180 degrees would be a half rotation. 360+180=540 So it would be a rotation and a half.
To find the image of ABC for a 180-degree counterclockwise rotation about point P, we would reflect each point of the triangle across the line passing through P. The resulting image of ABC would be a congruent triangle with its vertices in opposite positions relative to the original triangle.