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 ?

tertre May 20, 2017

#2

#3**0 **

Yes, Point C is at (-2,-2) when reflected over the x-axis and then over the y-axis.

TheXSquaredFactor
May 20, 2017

#4

#5**0 **

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.

TheXSquaredFactor
May 20, 2017