Question

which rule describes the composition of transformations that maps $\triangle jkl$ to $\triangle jkl$?\
$\bigcirc\\ r_{o,\\ 90\degree} \circ t_{0,\\ -2}(x, y)$\
$\bigcirc\\ t_{0,\\ -2} \circ r_{o,\\ 90\degree}(x, y)$\
$\bigcirc\\ r_{o,\\ 90\degree} \circ t_{-2,\\ 6}(x, y)$\
$\bigcirc\\ t_{-2,\\ 0} \circ r_{o,\\ 90\degree}(x, y)$

Explanation:

Step1: Identify original points

Original $\triangle JKL$ points: $J(1,-4)$, $K(4,-4)$, $L(1,-2)$

Step2: Test option 1: $R_{0,90^\circ} \circ T_{0,-2}(x,y)$

First apply $T_{0,-2}(x,y)=(x,y-2)$:
$J'(1,-6)$, $K'(4,-6)$, $L'(1,-4)$
Then $R_{0,90^\circ}(x,y)=(-y,x)$:
$J''(6,1)$, $K''(6,4)$, $L''(4,1)$ → Does not match $\triangle J''K''L''$

Step3: Test option 2: $T_{0,-2} \circ R_{0,90^\circ}(x,y)$

First apply $R_{0,90^\circ}(x,y)=(-y,x)$:
$J'(4,1)$, $K'(4,4)$, $L'(2,1)$
Then $T_{0,-2}(x,y)=(x,y-2)$:
$J''(4,-1)$, $K''(4,2)$, $L''(2,-1)$ → Does not match

Step4: Test option 3: $R_{0,90^\circ} \circ T_{-2,0}(x,y)$

First apply $T_{-2,0}(x,y)=(x-2,y)$:
$J'(-1,-4)$, $K'(2,-4)$, $L'(-1,-2)$
Then $R_{0,90^\circ}(x,y)=(-y,x)$:
$J''(4,-1)$, $K''(4,2)$, $L''(2,-1)$ → Does not match

Step5: Test option 4: $T_{-2,0} \circ R_{0,90^\circ}(x,y)$

First apply $R_{0,90^\circ}(x,y)=(-y,x)$:
$J'(4,1)$, $K'(4,4)$, $L'(2,1)$
Then $T_{-2,0}(x,y)=(x-2,y)$:
$J''(2,1)$, $K''(2,4)$, $L''(0,1)$ → Matches the plotted $\triangle J''K''L''$

Answer:

$T_{-2, 0} \circ R_{0, 90^\circ}(x, y)$