Question

parallelogram klmn is the image of parallelogram klmn under a translation followe rotation 90° counterclockwise about the origin. write the rules for the translation and rotation. translation: (x, y) ↦ (□, □) rotation: (x, y) ↦ (□, □)

Explanation:

Step1: Determine the translation rule

Let's take a point from the original parallelogram \(KLMN\) and its corresponding point after translation (before rotation). For example, take point \(M(2,2)\).
The translation moves the \(x -\)coordinate and \(y -\)coordinate.
The formula for translation is \((x,y)\to(x + a,y + b)\).
If we assume the translation first, and then rotation. Let's check the movement of \(M\) (before rotation) to \(M'\) (after rotation). But first, we can also use the fact that translation is a slide.
Looking at the \(x -\)coordinates: For example, if we consider the horizontal movement. If we take \(K(-6,6)\), after translation (before rotation) and then rotation. But another way: The distance between \(K\) and \(K'\) (considering the combined transformation, but separating translation first).
The translation rule: \((x,y)\to(x + 4,y-8)\). Because if we take \(K(-6,6)\), \(x=-6\), \(y = 6\). After translation \(x=-6 + 4=-2\), \(y=6-8=-2\) (before rotation).

Step2: Determine the rotation rule

The rule for a \(90^{\circ}\) counter - clockwise rotation about the origin is \((x,y)\to(-y,x)\).
Let's verify. Suppose we have a point \((x,y)\) after translation \((x',y')=(x + 4,y-8)\).
If we take \(M(2,2)\), after translation \(x'=2 + 4=6\), \(y'=2-8=-6\). Then after \(90^{\circ}\) counter - clockwise rotation \((x'',y'')=(-y',x')=(6,6)\) (but wait, no, let's use the general formula.
The rotation formula about the origin for \(90^{\circ}\) counter - clockwise: \((x,y)\to(-y,x)\)

Answer:

Translation: \((x,y)\to(x + 4,y-8)\)
Rotation: \((x,y)\to(-y,x)\)