A half turn about the origin is the same thing as say a rotation about the origin of 180°. To calculate the point of each image, we can do it in two ways. Let me show it:
Say we have a counterclockwise rotation of θ. We can find the image of a point by using these equations:
[tex]\begin{gathered} x^{\prime}=x\cos \theta-y\sin \theta \\ y^{\prime}=x\sin \theta+y\cos \theta \end{gathered}[/tex]Where x and y is the original point and x' and y' is the image. For 180°, we have:
[tex]\begin{gathered} \sin 180\degree=0 \\ \cos 180\degree=-1 \end{gathered}[/tex]So, the equations simplify to:
[tex]\begin{gathered} x^{\prime}=-x \\ y^{\prime}=-y \end{gathered}[/tex]The second way is by just seeing that a 180° turn would get you to opposite quadrant, changing the signs of each coordinate. So, both methods are equivalent.
Now, for the points:
a. (0,4):
[tex]\begin{gathered} x=0 \\ x^{\prime}=-0=0 \\ y=4 \\ y^{\prime}=-4 \end{gathered}[/tex]So, (0,-4).
b. (-1,3):
[tex]\begin{gathered} x=-1 \\ x^{\prime}=-(-1)=1 \\ y=3 \\ y^{\prime}=-3 \end{gathered}[/tex]So, (1,-3)
c. (-x, -y): here, it is simply change the sign:
[tex]\begin{gathered} x^{\prime}=-(-x)=x \\ y^{\prime}=-(-y)=y \end{gathered}[/tex]So, (x,y).
