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:

\[ $$\left(\begin{array}{c}{x}^{\prime }{y}^{\prime }\end{array}\right)$$

$$\left(\begin{array}{c}x\cos \theta -y\sin \theta x\sin \theta +y\cos \theta \end{array}\right)$$ \]

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:

$$x^{\prime }=x\cos ({270}^{\circ })-y\sin ({270}^{\circ })y^{\prime }=x\sin ({270}^{\circ })+y\cos ({270}^{\circ })$$

Plugging in the coordinates of point $M$:

$$x^{\prime }=0\cdot 0-4\cdot (-1)=0+4=4$$

$$y^{\prime }=0\cdot (-1)+4\cdot 0=0+0=0$$

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

$$M^{\prime }(4,0)$$

So, the final answer is:

$$M^{\prime }(4,0)$$