Question

triangle rst is rotated 180° counterclockwise about the origin. the result is △rst, 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 → final coordinates r(-6, 2) → r( , ) s(-4, -2) → s( , ) t(-1, 6) → t( , ) (b) choose the general rule below that describes the rotation mapping △rst to △rst. (x, y) → (-x, y) (x, y) → (-x, -y) (x, y) → (y, -x) (x, y) → (y, x) (x, y) → (-y, x) (x, y) → (x, -y) (x, y) → (-y, -x)

Explanation:

Step1: Recall rotation rule

When a point $(x,y)$ is rotated 180° counter - clockwise about the origin, the rule is $(x,y)\to(-x,-y)$.

Step2: Find $R'$ coordinates

For point $R(-6,2)$, applying the rule: $x=-6,y = 2$, then $-x = 6$ and $-y=-2$. So $R'=(6, - 2)$.

Step3: Find $S'$ coordinates

For point $S(-4,-2)$, with $x=-4,y=-2$, then $-x = 4$ and $-y = 2$. So $S'=(4,2)$.

Step4: Find $T'$ coordinates

For point $T(-1,6)$, where $x=-1,y = 6$, then $-x=1$ and $-y=-6$. So $T'=(1,-6)$.

Step5: Determine general rule

The general rule for a 180° counter - clockwise rotation about the origin is $(x,y)\to(-x,-y)$.

Answer:

(a) $R(-6,2)\to R'(6,-2)$; $S(-4,-2)\to S'(4,2)$; $T(-1,6)\to T'(1,-6)$
(b) $(x,y)\to(-x,-y)$