Question

parallelogram ( f^{prime prime} g^{prime prime} h^{prime prime} j^{prime prime} ) is the final image after the rule ( r_{y - axis} circ t_{1,2}(x,y) ) was applied to parallelogram ( fghj ). what are the coordinates of vertex ( f ) of parallelogram ( fghj )? ( (-2,2) ) ( (-2,6) ) ( (-3,4) ) ( (-4,2) )

Explanation:

Step1: Analyze the transformation rule

The transformation is \(r_{y - axis}\circ T_{1,2}(x,y)\). First, the translation \(T_{1,2}(x,y)=(x + 1,y+2)\), then the reflection over the \(y -\)axis \(r_{y - axis}(x,y)=(-x,y)\). Let the original coordinates of \(F\) be \((x,y)\). After translation \(T_{1,2}\), the coordinates become \((x + 1,y + 2)\), and after reflection over the \(y -\)axis, the coordinates of \(F''\) are \((-(x + 1),y + 2)\).
From the graph, the coordinates of \(F''\) are \((2,4)\).

Step2: Solve for \(x\) and \(y\)

Set up the equations:
\(-(x + 1)=2\) and \(y + 2=4\)
For \(-(x + 1)=2\), we have \(x+1=-2\), so \(x=-3\)
For \(y + 2=4\), we have \(y=2\)

Answer:

\((-3,4)\)