Question

o rotate figure abcd 90° counterclockwise about the origin. then translate the image 4 units to the left to form congruent figure efgh. o reflect figure abcd across the x - axis. then rotate the image 90° counterclockwise about the origin to form congruent figure efgh. o rotate figure abcd 90° clockwise about the origin. then reflect the image across the y - axis to form congruent figure efgh. o translate figure abcd 5 units down. then rotate the image 90° clockwise about the origin to form congruent figure efgh.

Explanation:

Step1: Analyze each transformation option

• Option 1 (Translate 5 units down then translate): Translating \(ABCD\) 5 units down will not align it with \(EFGH\) as the orientation and position relationship is more complex than just vertical translation.
• Option 2 (Rotate \(90^{\circ}\) clockwise about origin then reflect):
• When we rotate a point \((x,y)\) \(90^{\circ}\) clockwise about the origin, the transformation rule is \((x,y)\to(y, - x)\).
• Then reflecting a point \((x,y)\) across the \(y -\)axis has the rule \((x,y)\to(-x,y)\).
• Let's assume a point \(A(x_1,y_1)\) in \(ABCD\). After rotation \(90^{\circ}\) clockwise \(A_1=(y_1,-x_1)\), and after reflection across the \(y -\)axis \(A_2=(-y_1,-x_1)\) which can match the position of a point in \(EFGH\) considering the grid - based coordinates.
• Option 3 (Reflect across \(x -\)axis then rotate \(90^{\circ}\) counter - clockwise): Reflecting a point \((x,y)\) across the \(x -\)axis gives \((x, - y)\), and rotating \((x, - y)\) \(90^{\circ}\) counter - clockwise (rule \((x,y)\to(-y,x)\)) gives \((y,x)\) which does not match the transformation needed to get from \(ABCD\) to \(EFGH\).
• Option 4 (Rotate \(90^{\circ}\) counter - clockwise then translate 4 units left): Rotating \(90^{\circ}\) counter - clockwise (rule \((x,y)\to(-y,x)\)) and then translating 4 units left (rule \((x,y)\to(x - 4,y)\)) will not map \(ABCD\) to \(EFGH\) as per the grid - based positions.

Answer:

Rotate figure \(ABCD\) \(90^{\circ}\) clockwise about the origin. Then reflect the image across the \(y -\)axis to form congruent figure \(EFGH\).