A quadrilateral with vertices G(−10,1)

Question

𝐺
(
−
10
,
1
)
, E(−6,−4)
𝐸
(
−
6
,
−
4
)
, O(2,0)
𝑂
(
2
,
0
)
, and M(0,4)
𝑀
(
0
,
4
)
is rotated about the origin 270 degrees (counterclockwise). Apply rotation mapping rules to find the image of M
𝑀
.(1 point)
M′
𝑀
′
(,)

Answer 1

To find the image of point \( M(0, 4) \) after a 270-degree counterclockwise rotation around the origin, we can use the rotation mapping rules.

The general rule for rotating a point \( (x, y) \) counterclockwise by \( \theta \) degrees about the origin is given by: \[ (x', y') = (x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta) \]

For a 270-degree rotation, we know:

• \( \cos(270^\circ) = 0 \)
• \( \sin(270^\circ) = -1 \)

Plugging in these values, the rotation transformation becomes: \[ (x', y') = (x \cdot 0 - y \cdot (-1), x \cdot (-1) + y \cdot 0) \] This simplifies to: \[ (x', y') = (y, -x) \]

Now we apply this to point \( M(0, 4) \):

• Here, \( x = 0 \) and \( y = 4 \). Substituting these values into the transformation: \[ (x', y') = (4, -0) = (4, 0) \]

Thus, the image of point \( M \) after a 270-degree counterclockwise rotation about the origin is: \[ M' = (4, 0) \]