Question

quadrilateral abcd is the image of quadrilateral abcd under a rotation 90° counterclockwise about the origin followed by a translation. write the rules for the rotation and translation. rotation: (x, y) ↦ (□, □) translation: (x, y) ↦ (□, □)

Explanation:

Step1: Determine Rotation Rule

A 90° counterclockwise rotation about the origin has the rule \((x, y) \to (-y, x)\). Let's verify with a point, say \(A(4, 0)\). Applying rotation: \((0, 4)\)? Wait, no, wait. Wait, original point \(A\) is \((4, 0)\). Wait, maybe better to take point \(B(0, 3)\). Rotating 90° counterclockwise: \((-3, 0)\). Wait, let's check the image after rotation (before translation). Wait, the final image is \(A'\) at \((0, 0)\). Wait, maybe first find the rotation of a point, then translation. Let's take point \(A(4, 0)\). After 90° counterclockwise rotation, the rule is \((x, y) \to (-y, x)\), so \(A(4, 0)\) becomes \((-0, 4) = (0, 4)\). Then, to get to \(A'(0, 0)\), we need to translate down 4 units? Wait, no, maybe another point. Let's take point \(C(0, 5)\). Rotating 90° counterclockwise: \((-5, 0)\). Then, to get to \(C'(-5, -4)\), we need to translate down 4 units? Wait, no, \(C'\) is at \((-5, -4)\). Wait, maybe the rotation is \((x, y) \to (-y, x)\), then translation. Let's check point \(A(4, 0)\): rotation gives \((0, 4)\), then translation to \((0, 0)\): so translation is \((x, y) \to (x, y - 4)\)? Wait, no, \(0 - 4 = -4\)? No, \(4 - 4 = 0\)? Wait, maybe I messed up. Wait, the rotation rule for 90° counterclockwise is \((x, y) \to (-y, x)\). Let's confirm with a standard rotation: yes, 90° counterclockwise about origin: \((x, y) \mapsto (-y, x)\).

Step2: Determine Translation Rule

Now, let's take point \(A(4, 0)\). After rotation, it should be \((-0, 4) = (0, 4)\). Then, the translated point \(A'\) is \((0, 0)\). So the translation from \((0, 4)\) to \((0, 0)\) is a vertical shift down by 4 units, or \((x, y) \to (x, y - 4)\). Let's check another point: \(B(0, 3)\). Rotation: \((-3, 0)\). Then translation: \((-3, 0 - 4) = (-3, -4)\), which is \(B'(-3, -4)\)? Wait, no, \(B'\) is at \((-3, -4)\)? Wait, the graph shows \(B'\) at \((-3, -4)\)? Wait, the purple points: \(C'\) is at \((-5, -4)\), \(B'\) at \((-3, -4)\), \(D'\) at \((-5, 1)\)? Wait, no, the grid: \(D'\) is at \((-5, 1)\), \(C'\) at \((-5, -4)\), \(B'\) at \((-3, -4)\), \(A'\) at \((0, 0)\). Wait, maybe my initial point selection is wrong. Let's take point \(D(5, 5)\). Rotating 90° counterclockwise: \((-5, 5)\). Then, to get to \(D'(-5, 1)\), we need to translate down 4 units: \((-5, 5 - 4) = (-5, 1)\), which matches \(D'\). Perfect! So \(D(5, 5)\) rotated 90° counterclockwise is \((-5, 5)\), then translated down 4 units (y - 4) gives \((-5, 1)\), which is \(D'\). Let's check point \(C(0, 5)\): rotated 90° counterclockwise is \((-5, 0)\), then translated down 4 units: \((-5, 0 - 4) = (-5, -4)\), which is \(C'\). Point \(B(0, 3)\): rotated 90° counterclockwise is \((-3, 0)\), then translated down 4 units: \((-3, 0 - 4) = (-3, -4)\), which is \(B'\). Point \(A(4, 0)\): rotated 90° counterclockwise is \((-0, 4) = (0, 4)\), then translated down 4 units: \((0, 4 - 4) = (0, 0)\), which is \(A'\). Perfect! So the rotation rule is \((x, y) \to (-y, x)\) (90° counterclockwise about origin), and the translation rule is \((x, y) \to (x, y - 4)\) (translate 4 units down, or 0 in x, -4 in y).

Answer:

Rotation: \((x, y) \to \boldsymbol{(-y, x)}\)
Translation: \((x, y) \to \boldsymbol{(x, y - 4)}\)