Question

carlos performed a transformation on trapezoid efgh to create efgh, as shown in the figure below: what transformation did carlos perform to create efgh? rotation of 270° clockwise about the origin reflection across the x - axis rotation of 90° clockwise about the origin reflection across the line of symmetry of the figure

Explanation:

Step1: Recall rotation rules

For a point \((x,y)\) rotated \(270^{\circ}\) clockwise about the origin, the new coordinates are \((y, -x)\).

Step2: Check coordinates

Let's assume a point \(E\) (say \(E(-8,-3)\)). After \(270^{\circ}\) clockwise rotation about the origin, using the rule \((x,y)\to(y, -x)\), we get \((-3,8)\) which matches the position of \(E'\) in the figure.

Step3: Eliminate other options

• Reflection across \(x -\)axis: \((x,y)\to(x,-y)\). This would not give the figure \(E'F'G'H'\) as shown.
• Rotation of \(90^{\circ}\) clockwise: \((x,y)\to(y,-x)\) (incorrect as per the figure's transformation).
• Reflection across line of symmetry (not applicable as no such line is evident in the general trapezoid transformation shown).

Answer:

Rotation of \(270^{\circ}\) clockwise about the origin.