For a given point (2, 0), find the coordinates of the image point under a half-turn about the origin.
I am not sure exactly what I need to do here. I already inserted a graph with the X and Y axis with the (2,0) marked. I don't know where to go from there. Please help.
If would imagines a circle with center (0,0) that goes through point (2,0), a half turn about the origin would be moving 180° (or π rad) around that circle from the given point.
First, let's parametrize our circle. Since it goes through point (2,0), we know the radius is 2 (\(\displaystyle \sqrt{2^{2} + 0^{2}} = 2\)), and we know that a circle will be in this form:
\(\displaystyle x(\theta) = rcos(\theta + \alpha)\)
\(\displaystyle y(\theta) = rsin(\theta + \alpha)\)
We know that \(\displaystyle r= 2\). To find \(\displaystyle \alpha\), we just need to solve for \(\displaystyle x(0) = 2\):
\(\displaystyle x(0) = 2cos(\alpha) = 2 \to cos(\alpha) = 1\), which is true at \(\displaystyle \alpha = 0\)
Therefore:
\(\displaystyle x(\theta) = 2cos(\theta)\)
\(\displaystyle y(\theta) = 2sin(\theta)\)
Which satisfies \(\displaystyle (x(0),y(0)) = (2,0)\)
A half turn would be \(\displaystyle (x(\pi),y(\pi)) = (2cos(\pi),2sin(\pi)) = (-2,0)\)
The turn is about the origin - not about y-axis.
A skew or flip is about an axis; a turn (aka rotation) is around a point. To rotate around a point, the same distance is kept, but the direction (or angle) relative to the point changes. In this case, and only because the point is
on the x-axis, flipping about the y-axis yields the same result as rotating one half turn around the origin. A half turn will always transform \(\displaystyle (x,y) \to (-x,-y)\) while a y-axis flip will always transform \(\displaystyle (x,y) \to (-x,y)\). It is clear from this why both would yield the same result at point \(\displaystyle (x_1, 0)\).
