Question

triangle rst is rotated ( 90^{circ} ) counterclockwise about the origin. the result is ( \triangle r^{prime} s^{prime} t^{prime} ), 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 (
ightarrow ) final coordinates ( r(-8,-2)
ightarrow r^{prime}(square, square) ) ( s(1,-4)
ightarrow s^{prime}(square, square) ) ( t(-2,-6)
ightarrow t^{prime}(square, square) ) (b) choose the general rule below that describes the rotation mapping ( \triangle r s t ) to ( \triangle r^{prime} s^{prime} t^{prime} ). ( (x, y)
ightarrow(-x,-y) ) ( (x, y)
ightarrow(x,-y) ) ( (x, y)
ightarrow(-x, y) ) ( (x, y)
ightarrow(-y,-x) ) ( (x, y)
ightarrow(y,-x) ) ( (x, y)
ightarrow(-y, x) ) ( (x, y)
ightarrow(y, x) )

Explanation:

Step1: Apply rotation rule

The rule for a \(90^{\circ}\) counter - clockwise rotation about the origin is \((x,y)\to(-y,x)\).
For point \(R(-8,-2)\):
Substitute \(x = - 8\) and \(y=-2\) into the rule \((x,y)\to(-y,x)\).
\(-y=-(-2) = 2\) and \(x=-8\), so \(R'=(2,-8)\).

Step2: Apply rotation rule for point \(S\)

For point \(S(1,-4)\):
Substitute \(x = 1\) and \(y = - 4\) into the rule \((x,y)\to(-y,x)\).
\(-y=-(-4)=4\) and \(x = 1\), so \(S'=(4,1)\).

Step3: Apply rotation rule for point \(T\)

For point \(T(-2,-6)\):
Substitute \(x=-2\) and \(y = - 6\) into the rule \((x,y)\to(-y,x)\).
\(-y=-(-6)=6\) and \(x=-2\), so \(T'=(6,-2)\).

Step4: Check the general rule

The general rule \((x,y)\to(-y,x)\) is the one that maps \(\triangle RST\) to \(\triangle R'S'T'\)

Answer:

(a) \(R(-8,-2)\to R'(2,-8)\), \(S(1,-4)\to S'(4,1)\), \(T(-2,-6)\to T'(6,-2)\)
(b) \((x,y)\to(-y,x)\)