Question

For each of the following points, find the coordinates of the image point under a half-turn about the origin

Answer 1

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:

`\begin{gathered} x^(\prime)=x\cos \theta-y\sin \theta \\ y^(\prime)=x\sin \theta+y\cos \theta \end{gathered}`

Where x and y is the original point and x' and y' is the image. For 180°, we have:

`\begin{gathered} \sin 180\degree=0 \\ \cos 180\degree=-1 \end{gathered}`

So, the equations simplify to:

`\begin{gathered} x^(\prime)=-x \\ y^(\prime)=-y \end{gathered}`

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):

`\begin{gathered} x=0 \\ x^(\prime)=-0=0 \\ y=4 \\ y^(\prime)=-4 \end{gathered}`

So, (0,-4).

b. (-1,3):

`\begin{gathered} x=-1 \\ x^(\prime)=-(-1)=1 \\ y=3 \\ y^(\prime)=-3 \end{gathered}`

So, (1,-3)

c. (-x, -y): here, it is simply change the sign:

`\begin{gathered} x^(\prime)=-(-x)=x \\ y^(\prime)=-(-y)=y \end{gathered}`

So, (x,y).