Question

△abc is rotated around the origin to form △abc. which rotation is shown in the coordinate plane? 90° clockwise, 180° counterclockwise, 90° counterclockwise, 180° clockwise

Explanation:

Step1: Identify coordinates of points

$A(-4,-3), B(-1,-3), C(-1,-5)$; $A'(2,-4), B'(2,-1), C'(4,-1)$

Step2: Test 180° rotation rule

180° rotation rule: $(x,y)\to(-x,-y)$
For $A(-4,-3)$: $-(-4)=4, -(-3)=3$ → not $A'(2,-4)$

Step3: Test 90° clockwise rule

90° clockwise rule: $(x,y)\to(y,-x)$
For $A(-4,-3)$: $(y,-x)=(-3,4)$ → not $A'(2,-4)$

Step4: Test 90° counterclockwise rule

90° counterclockwise rule: $(x,y)\to(-y,x)$
For $A(-4,-3)$: $-(-3)=3, x=-4$ → $(3,-4)$ (not match)
Wait, correct 180°: Wait, recheck: 180° rotation (clockwise/counterclockwise same) swaps sign of both coordinates. Wait, $A(-4,-3)$ → $(4,3)$ no. Wait, 90° clockwise: $(x,y)\to(y,-x)$: $A(-4,-3)\to(-3,4)$ no. 90° counterclockwise: $(x,y)\to(-y,x)$: $A(-4,-3)\to(3,-4)$ no. Wait, 180°: $A(-4,-3)$ → $(4,3)$ no, but $A'(2,-4)$: Wait, no, wait the graph: $A$ is $(-4,-3)$, $A'$ is $(2,-4)$? No, wait grid: x-axis from -5 to 5, y-axis from -5 to 5. $A$ is at (-4,-3), $A'$ is at (2,-4)? No, wait no: 180° rotation: $(x,y)\to(-x,-y)$. $A(-4,-3)$ → $(4,3)$ no, but $A'$ is (2,-4)? No, wait I misread: $A$ is (-4,-2), $B$ is (-1,-2), $C$ is (-1,-4). Then $A'$ is (2,-2), $B'$ is (2,1), $C'$ is (4,1). No, wait 180°: $(-4,-2)\to(4,2)$ no. Wait 90° counterclockwise: $(x,y)\to(-y,x)$. $A(-4,-2)\to(2,-4)$ which matches $A'(2,-4)$. $B(-1,-2)\to(2,-1)$ which matches $B'(2,-1)$. $C(-1,-4)\to(4,-1)$ which matches $C'(4,-1)$.

Step5: Confirm 90° counterclockwise

All points match 90° counterclockwise rule.

Answer:

90° counterclockwise