Question

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.

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:

\[
M'(4, 0)
\]