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.

Question

M'(_,_)

Answer 1

To rotate a point \( (x, y) \) counterclockwise about the origin by an angle \( \theta \), you can use the following rotation formula:

\[ \begin{pmatrix} x' \ y' \end{pmatrix}

\begin{pmatrix} x \cos \theta - y \sin \theta \ x \sin \theta + y \cos \theta \end{pmatrix} \]

For a rotation of 270 degrees counterclockwise, we can first convert this to radians:

\[ 270^\circ = \frac{3\pi}{2} \text{ radians} \]

The cosine and sine values for \( 270^\circ \) are:

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

Now, let's apply the rotation to point \( M(0, 4) \):

Using the rotation formulas:

\[ x' = x \cos(270^\circ) - y \sin(270^\circ) \ y' = x \sin(270^\circ) + y \cos(270^\circ) \]

Plugging in the coordinates of point \( M \):

\[ x' = 0 \cdot 0 - 4 \cdot (-1) = 0 + 4 = 4 \]

\[ y' = 0 \cdot (-1) + 4 \cdot 0 = 0 + 0 = 0 \]

Thus, the image of point \( M \) after a 270-degree counterclockwise rotation about the origin is:

\[ M'(4, 0) \]

So, the final answer is:

\[ M'(4, 0) \]