Question

22 a rhombus is graphed on the set of axes below. which transformation does not carry the rhombus onto itself? (1) a rotation of 180° about the origin (2) a rotation of 180° about point (1,0) (3) a reflection over the line ( y=\frac{1}{2}x-\frac{1}{2} ) (4) a reflection over the line ( y=-2x + 2 )

Explanation:

Step1: Analyze rotation about the origin

A rhombus is a centrally - symmetric figure. A rotation of \(180^{\circ}\) about the origin (the center of symmetry for many plane figures when properly placed) will map the rhombus onto itself.

Step2: Analyze rotation about \((1,0)\)

A rotation of \(180^{\circ}\) about a non - center - of - symmetry point (in this case, the point \((1,0)\) is not the center of the rhombus) will not map the rhombus onto itself.

Step3: Analyze reflection over \(y=\frac{1}{2}x - \frac{1}{2}\)

If the line \(y=\frac{1}{2}x-\frac{1}{2}\) is a line of symmetry of the rhombus (by checking the distances of corresponding points from the line and using the property of reflection: for any point \(P(x,y)\) on the rhombus, its image \(P'(x',y')\) after reflection over the line \(ax+by + c = 0\) satisfies the mid - point of \(PP'\) lies on the line and the line \(PP'\) is perpendicular to the given line). We can find that this reflection maps the rhombus onto itself.

Step4: Analyze reflection over \(y=-2x + 2\)

Using the properties of reflection (distance from points to the line and perpendicularity of the line joining a point and its image with the reflecting line), we can show that this reflection maps the rhombus onto itself.

Answer:

(2) a rotation of \(180^{\circ}\) about point \((1,0)\)