Question

triangle def is rotated (270^circ) counterclockwise about the origin. the result is (\triangle def), as shown below.
(a) the arrows below show that the coordinates on the left are mapped to the coordinates on the right. fill in the blanks to give the coordinates after the rotation.
original coordinates (\to) final coordinates
(d(-5, -4) \to d(square, square))
(e(2, -2) \to e(square, square))
(f(-3, -6) \to f(square, square))
(b) choose the general rule below that describes the rotation mapping (\triangle def) to (\triangle def).
(circ (x,y) \to (x, -y)) (quad) (\bullet (x,y) \to (-y, x))
(circ (x,y) \to (-x, y)) (quad) (circ (x,y) \to (-y, -x))
(circ (x,y) \to (-x, -y)) (quad) (circ (x,y) \to (y, -x))
(circ (x,y) \to (y, x))

Explanation:

Step1: Recall 270° CCW rotation rule

A 270° counterclockwise rotation about the origin follows the rule: $(x, y) \to (y, -x)$

Step2: Calculate $D'$ coordinates

Substitute $D(-5, -4)$ into the rule:
$D' = (-4, -(-5)) = (-4, 5)$

Step3: Calculate $E'$ coordinates

Substitute $E(2, -2)$ into the rule:
$E' = (-2, -(2)) = (2, 2)$

Step4: Calculate $F'$ coordinates

Substitute $F(-3, -6)$ into the rule:
$F' = (-6, -(-3)) = (-6, 3)$

Step5: Identify general rotation rule

The standard rule for 270° counterclockwise rotation about the origin is $(x, y) \to (y, -x)$

Answer:

(a) $D'(-4, 5)$, $E'(2, 2)$, $F'(-6, 3)$
(b) $\boldsymbol{(x, y) \to (y, -x)}$