A quadrilateral with vertices G(−10,1)
, E(−6,−4)
, O(2,0)
, and M(0,4)
is rotated about the origin 270 degrees (counterclockwise). Apply rotation mapping rules to find the image of M
.
To rotate a point counterclockwise about the origin, we can apply the following rotation matrix:
```
[x'] [cos(θ) -sin(θ)] [x]
[y'] = [sin(θ) cos(θ)] [y]
```
where (x, y) are the coordinates of the original point, (x', y') are coordinates of the rotated point, and θ is the angle of rotation.
In this case, the angle of rotation is 270 degrees, which is equivalent to -90 degrees. We need to convert it to radians for the trigonometric functions:
θ = -90 degrees = -π/2 radians
Now, let's apply the rotation matrix to point M(0, 4):
[x'] [cos(-π/2) -sin(-π/2)] [0]
[y'] = [sin(-π/2) cos(-π/2)] [4]
Simplifying the matrix multiplication:
[x'] [0 -(-1)] [0]
[y'] = [(-1) 0] [4]
[x'] [0 1] [0]
[y'] = [-1 0] [4]
[x'] [0*0 + 1*4] [0]
[y'] = [-1*0 + 0*4] [4]
[x'] [4] [0]
[y'] = [0] [4]
Therefore, the image of M is M'(4, 0).
To find the image of point M(-5,2) after rotating 270 degrees counterclockwise about the origin, we can use rotation mapping rules.
The rotation mapping rule for counterclockwise rotation by 270 degrees is (x, y) -> (-y, x).
Applying this rule to point M, we have:
M' = (-2, 0)
Therefore, the image of point M after rotating 270 degrees counterclockwise about the origin is M'(-2, 0).
To find the image of point M after a 270-degree counterclockwise rotation about the origin, we can use rotation mapping rules.
The rotation mapping rule states that to rotate a point (x, y) counterclockwise by an angle θ about the origin, the new coordinates (x', y') can be found using the following equations:
x' = x * cos(θ) - y * sin(θ)
y' = x * sin(θ) + y * cos(θ)
In this case, θ = 270 degrees.
Let's apply the rotation mapping rules to find the image of point M.
Original coordinates:
M(0, 4)
Using the rotation mapping rules:
x' = 0 * cos(270°) - 4 * sin(270°)
y' = 0 * sin(270°) + 4 * cos(270°)
Calculating:
x' = 0 - 4 * (-1)
y' = 0 + 4 * 0
Simplifying:
x' = 4
y' = 0
Therefore, the image of the point M(0, 4) after a 270-degree counterclockwise rotation about the origin is M'(4, 0).