It is (-1, 6).
Also, if the rotation is 180 degrees, then clockwise or anticlockwise are irrelevant.
It is (-1, 6).
Conventionally positive angles are measured anticlockwise, by 180° is a half turn regardless of direction. It depends where the centre of rotation is, so where would you like the image to be? If the centre is at, say, (4, 3) then the image will be at (4, 3) regardless of the angle of rotation. If the centre is at, say, (4, 4) then the image will be at (4, 5) If the centre is at, say, the origin, ie (0, 0) then the image will be at (-4, -3).
Rotating it about the origin 180° (either way, it's half a turn) will transform a point with coordinates (x, y) to that with coordinates (-x, -y) Thus (2, 5) → (-2, -5)
The centre of rotation, the angle of rotation and, unless the angle is 180 degrees, the direction of rotation.
180 rotation
A 180 degree rotation?
The rule for a rotation by 180° about the origin is (x,y)→(−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.
Because 180 degrees clockwise is the same as 180 degrees counterclockwise.
Fomula(work with both clockwise/counterclockwise):(-x,-y)
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.
What is the image of point (3, 5) if the rotation is
The answer will depend on where the centre of rotation is. Since that it not specified, the image could by anywhere.
Conventionally positive angles are measured anticlockwise, by 180° is a half turn regardless of direction. It depends where the centre of rotation is, so where would you like the image to be? If the centre is at, say, (4, 3) then the image will be at (4, 3) regardless of the angle of rotation. If the centre is at, say, (4, 4) then the image will be at (4, 5) If the centre is at, say, the origin, ie (0, 0) then the image will be at (-4, -3).
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.
All rotations, other than those of 180 degrees should be further qualified as being clockwise or counter-clockwise. This one is not and I am assuming that the direction of rotation is the same as measurement of polar angles. Also, a rotation is not properly defined unless the centre of rotation is specified. I am assuming that the centre of rotation is the origin. Without these two assumptions any point in the plane can be the image. With the assumptions, for which there is no valid reason, the image is (3, -4).
If the point (3,5) is rotated 180 degrees, it becomes (-3,-5).
(-4,-3) anything with a 180 degree rotation regardless of being postive or negative is always negative the numbers in parenthesis.