Question

which rule represents the translation from the pre - image, (square abcd), to the image, (square abcd)? (t_{1,-2}(x,y)) (t_{1,2}(x,y)) (t_{-2,1}(x,y)) (t_{2,1}(x,y))

Explanation:

Step1: Analyze horizontal translation

Let's consider a point from the pre - image to the image. For example, if we assume a general point \((x,y)\) in the pre - image \(ABCD\).
The horizontal change: If we observe the movement of the figure from \(ABCD\) to \(A'B'C'D'\), we can see that the figure is shifted to the right. The number of units of horizontal shift: Counting the units on the \(x\) - axis, the shift in the \(x\) - direction is \(1\) unit. So the \(x\) - coordinate of a point \((x,y)\) in the pre - image changes to \(x + 1\) in the image.

Step2: Analyze vertical translation

The vertical change: The figure is shifted upwards. Counting the units on the \(y\) - axis, the shift in the \(y\) - direction is \(2\) units. So the \(y\) - coordinate of a point \((x,y)\) in the pre - image changes to \(y+2\) in the image.
In the translation notation \(T_{a,b}(x,y)=(x + a,y + b)\), where \(a\) is the horizontal shift and \(b\) is the vertical shift. Here \(a = 1\) (right shift) and \(b=2\) (upward shift). So the translation rule is \(T_{1,2}(x,y)\)

Answer:

\(T_{1,2}(x,y)\)