For a given point (2, 0), find the coordinates of the image point under a half-turn a

Gr8fu13

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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.


 
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.



What does half-turn mean?
 
So would the coordinates be (-2,0) after the half-turn about the origin?
 
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 (22+02=2\displaystyle \sqrt{2^{2} + 0^{2}} = 2), and we know that a circle will be in this form:

x(θ)=rcos(θ+α)\displaystyle x(\theta) = rcos(\theta + \alpha)
y(θ)=rsin(θ+α)\displaystyle y(\theta) = rsin(\theta + \alpha)

We know that r=2\displaystyle r= 2. To find α\displaystyle \alpha, we just need to solve for x(0)=2\displaystyle x(0) = 2:

x(0)=2cos(α)=2cos(α)=1\displaystyle x(0) = 2cos(\alpha) = 2 \to cos(\alpha) = 1, which is true at α=0\displaystyle \alpha = 0

Therefore:

x(θ)=2cos(θ)\displaystyle x(\theta) = 2cos(\theta)
y(θ)=2sin(θ)\displaystyle y(\theta) = 2sin(\theta)

Which satisfies (x(0),y(0))=(2,0)\displaystyle (x(0),y(0)) = (2,0)

A half turn would be (x(π),y(π))=(2cos(π),2sin(π))=(2,0)\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 (x,y)(x,y)\displaystyle (x,y) \to (-x,-y) while a y-axis flip will always transform (x,y)(x,y)\displaystyle (x,y) \to (-x,y). It is clear from this why both would yield the same result at point (x1,0)\displaystyle (x_1, 0).
 
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