Question

the graph below shows a transformation from pre - image abcd to image ( abcd ) of a reflection across the line ( y = x ) across the line ( y=-x ) rotation of ( 90^{circ} ) clockwise about the point b of ( 180^{circ} ) about the point b translation of vector ( <10,2> )

Explanation:

Step1: Analyze the transformation

We check each transformation option.

• Reflection:
• Reflection across \(y = x\): If \((x,y)\) is reflected across \(y=x\), the image is \((y,x)\).
• Reflection across \(y=-x\): If \((x,y)\) is reflected across \(y =-x\), the image is \((-y,-x)\).
• Rotation:
• Rotation of \(90^{\circ}\) clockwise about a point \(P(x_0,y_0)\): The formula for rotating a point \((x,y)\) \(90^{\circ}\) clockwise about \((x_0,y_0)\) is \((x',y')=(x_0+(y - y_0),y_0-(x - x_0))\).
• Rotation of \(180^{\circ}\) about a point \(P(x_0,y_0)\): The formula for rotating a point \((x,y)\) \(180^{\circ}\) about \((x_0,y_0)\) is \((x',y')=(2x_0 - x,2y_0 - y)\).
• Translation:
• Translation by vector \(\langle a,b

angle\): If a point \((x,y)\) is translated by \(\langle a,b
angle\), the image is \((x + a,y + b)\).

Let's take a point, say \(A(-3,1)\). After transformation, \(A'\) is \((3,-1)\).
If we consider rotation of \(180^{\circ}\) about the origin \((0,0)\), using the formula \((x',y')=(2\times0-(-3),2\times0 - 1)=(3,-1)\). But we check about point \(B\).
Let \(B=(0,0)\) (from the graph). For a point \(A(-3,1)\), using the rotation of \(180^{\circ}\) about \(B(0,0)\) formula \((x',y')=(2\times0-(-3),2\times0 - 1)=(3,-1)\). For \(C(-1,4)\), its image \(C'(1,-4)\) using \((x',y')=(2\times0-(-1),2\times0 - 4)=(1,-4)\)

Answer:

rotation of \(180^{\circ}\) about the point \(B\)