Triangle ABC with vertices of A(6,2) , B(2,5) , and C(2,2) is reflected over the x-axis to triangle \(A'B'C'\) . This triangle is reflected over the y-axis to triangle \(A''B''C''\) . What are the coordinates of point C ?
Yes, Point C is at (-2,-2) when reflected over the x-axis and then over the y-axis.
You don't have to graph this to figure out the answer. Just know the relationships (although I prefer graphing than remembering the relationships.
A reflection over the x-axis has the following effect:
\((x,y)--->(x,-y)\)
A reflection over the y-axis has the following effect:
\((x,y)--->(-x,y)\)
Let's try this with point C at (2,2):
First, you reflect over the x-axis. This arrow notation says that the x-coordinate does not change but the y-coordinate changes to -1*y-coordinate:
\((2,2)--->(2,-2)\),so \(C'(2,-2)\)
Take C' and apply the effect of reflecting over the y-axis:
\((2,-2)--->(-2,-2)\),so \(C''(-2,-2)\)
Hopefully, this makes sense. This is simply another method of solving. If you want to stick to graphing, that is OK.