Question

redicting results of rigid transformations: tutorial
in this graph, which transformation can produce
quadrilateral ( abcd ) from quadrilateral ( abcd )?

• a reflection across the line ( y = x )
• a reflection across the line ( y = 0 )
• a rotation ( 90^{circ} ) clockwise about the origin
• a rotation ( 180^{circ} ) counterclockwise about the origin

Explanation:

Step1: Recall reflection across \(y = 0\) (x - axis) rule

The rule for reflection across the \(x\) - axis (\(y = 0\)) is \((x,y)\to(x, - y)\).

Step2: Check the vertices

For example, if \(B=(1,1)\), after reflection across \(y = 0\), \(B'=(1,-1)\). If \(D=(1,2)\), after reflection across \(y = 0\), \(D'=(1, - 2)\). If \(C=(2,1)\), after reflection across \(y = 0\), \(C'=(2,-1)\). And \(A=(0,0)\), \(A'=(0,0)\) (invariant under reflection across \(y = 0\)).

Step3: Analyze other options

• Reflection across \(y=x\) has the rule \((x,y)\to(y,x)\). For \(B=(1,1)\), it would map to itself, but for \(D=(1,2)\) it would map to \((2,1)\) which is not the case.
• Rotation \(90^{\circ}\) clockwise about the origin has the rule \((x,y)\to(y,-x)\). For \(B=(1,1)\), it would map to \((1, - 1)\) (coincides in \(x\) - coordinate but the general shape of the figure: a rotation \(90^{\circ}\) clockwise would change the orientation in a different way than just flipping over the \(x\) - axis).
• Rotation \(180^{\circ}\) counter - clockwise about the origin has the rule \((x,y)\to(-x,-y)\). For \(B=(1,1)\), it would map to \((-1,-1)\) which is not the case.

Answer:

a reflection across the line \(y = 0\)