1)
Suppose you rotate triangle ABC clockwise 90º. If ABC has vertices A (-2, 1), B (-1, 4), and C (2, 2), vertex C' is
Suppose you rotate triangle ABC clockwise 90º. If ABC has vertices A (-2, 1), B (-1, 4), and C (2, 2), vertex B' is (? , 1). (Hint: Note the direction of rotation.)
Use a rotation matrix to rotate figure DEFGH counterclockwise 90º. If the figure has coordinates D (1, 3), E (3, 2), F (1, -1), G (-3, -2), and H (-2, 2), the coordinates of E' are (-2, ?).
Suppose you rotate triangle ABC counterclockwise 90º. If ABC has vertices A (-2, 1), B (-1, 4), and C (2, 2), vertex B' is ( ? , -1).
+238 （1）the anwser of number 1 can be any point ,because you didnt tell me rotate around which point
(rotate around point A ,C'(-1,-3)
rotate around Point B C'(-3,1）
rotate around point C , the martix of point Cno change (2,2)
rotate around the origin , C'(-2,-2)
(2)file:///C:/Users/Public/Pictures/Sample%20Pictures/New%20folder%20(2)/133.htm
In numbe 2 , I guess the image rotate around point (-1,1) clockwise 90 degrees , so the marix of point B' is (2,1)
(3)the figure rotate around point H（-2,2）counterclockwise 90 degrees,then the marix of point E is (-2,7)
(4) the figure rotate around point C (2,2) clockwise 90 degrees ,then the martix of point B' is (0,-1)
+238 （1）the anwser of number 1 can be any point ,because you didnt tell me rotate around which point
(rotate around point A ,C'(-1,-3)
rotate around Point B C'(-3,1）
rotate around point C , the martix of point Cno change (2,2)
rotate around the origin , C'(-2,-2)
(2)file:///C:/Users/Public/Pictures/Sample%20Pictures/New%20folder%20(2)/133.htm
In numbe 2 , I guess the image rotate around point (-1,1) clockwise 90 degrees , so the marix of point B' is (2,1)
(3)the figure rotate around point H（-2,2）counterclockwise 90 degrees,then the marix of point E is (-2,7)
(4) the figure rotate around point C (2,2) clockwise 90 degrees ,then the martix of point B' is (0,-1)
+23252 If you use matrices: (Sorry, but I don't know how to make real matrix symbols here.)
counter-clockwise matrix: | cosθ -sinθ | clockwise matrix: | cosθ sinθ | | sinθ cosθ | | -sinθ cosθ |
1) Point (2,2) 90° clockwise: | cos 90° sin 90° | X | 2 | | -sin 90° cos 90° | | 2 |
2) Point (-1,4) 90° clockwise: | cos 90° sin 90° | X | -1 | | -sin 90° cos 90° | | 4 |
3) Point (3,2) 90° counter-clockwise: | cos 90° -sin 90° | X | 3 | | sin 90° cos 90° | | 2 |
4) Point (-1.4) 90° counter-clockwise: | cos 90° -sin 90° | X | -1 | | sin 90° cos 90° | | 4 |
If you have trouble with the matrices, please ask.
+129899 I believe these points are meant to be rotated around the origin....
1. For the first one....C is at (2,2)......rotating a point in the first quadrant 90 degrees clockwise reverses the coordinates and changes the sign on the original x coordinate.....so....(2, 2) becomes (2, -2)......
2. Rotating a point in the second quadrant 90 degrees clockwise reverses the coordinates and changes the sign on the original x coordinate . So, B= (-1,4) and B' = (4, 1)
3. Here, we're just doing the opposite of what we did in (2). E = (3, 2), so E' = (-2, 3)
4. Rotating a point in the second quadrant 90 degrees counter-clockwise reverses the coordinates and changes the sign on the original y coordinate. So B = (-1, 4) and B' = (-4, -1).
