one vertex of a triangle is located at (0, 5) on a coordinate grid. after a transformation, the vertex is…

Explanation:

Step1: Recall 90° rotation rule

For a point $(x,y)$, $R_{0,90^\circ}$ gives $(y,-x)$.

Step2: Test 90° rotation

Substitute $(0,5)$: $(5, -0)=(5,0)$.

Step3: Recall -270° rotation rule

$R_{0,-90^\circ}$ is same as $R_{0,270^\circ}$, rule: $(x,y)\to(-y,x)$. *Correction: Test $R_{0,-90^\circ}$ rule: $(x,y)\to(y,-x)$ (equivalent to 270° clockwise, same as 90° counterclockwise? No, correction: $R_{0,90^\circ}$ (counterclockwise): $(x,y)\to(-y,x)$. $R_{0,-90^\circ}$ (clockwise 90°): $(x,y)\to(y,-x)$.

Step4: Test $R_{0,-90^\circ}$

Substitute $(0,5)$: $(5, -0)=(5,0)$.

Step5: Eliminate other options

$R_{0,180^\circ}$: $(x,y)\to(-x,-y)$ gives $(0,-5)$. $R_{0,270^\circ}$: $(x,y)\to(y,-x)$? No, $R_{0,270^\circ}$ (counterclockwise): $(x,y)\to(y,-x)$? No, correct rule: $R_{0,270^\circ}$: $(x,y)\to(y,-x)$ is wrong, correct: $R_{0,270^\circ}$ (counterclockwise): $(x,y)\to(y,-x)$ no, standard:

• $90^\circ$ counterclockwise ($R_{0,90^\circ}$): $(x,y)\to(-y,x)$ → $(0,5)\to(-5,0)$ (error earlier, correction)
• $90^\circ$ clockwise ($R_{0,-90^\circ}$): $(x,y)\to(y,-x)$ → $(0,5)\to(5,0)$
• $270^\circ$ counterclockwise ($R_{0,270^\circ}$): $(x,y)\to(y,-x)$ same as 90° clockwise, so $(0,5)\to(5,0)$
• $180^\circ$: $(0,5)\to(0,-5)$
• $-180^\circ$ same as 180°: $(0,-5)$

Corrected Step1: $R_{0,-90^\circ}$ (clockwise 90°): $(x,y)\to(y,-x)$ Step2: $(0,5)\to(5,0)$ Step3: $R_{0,270^\circ}$ (counterclockwise 270° = clockwise 90°): $(x,y)\to(y,-x)$ Step4: $(0,5)\to(5,0)$ Step5: $R_{0,90^\circ}$ gives $(-5,0)$, so invalid.

Answer:

$R_{0, 270^\circ}$ $R_{0, -90^\circ}$