Question

the figure is rotated counterclockwise. which rotation could have taken place?
○ a 45° rotation
○ a 90° rotation
○ a 135° rotation
○ a 225° rotation

Explanation:

Step1: Identify original points

Original triangle points: $S(0,0)$, $T(-4,4)$, $R(-2,1)$

Step2: Identify image points

Image triangle points: $S'(4,0)$, $T'(4,3)$, $R'(1,4)$

Step3: Test 90° rotation rule

Counterclockwise 90° rule: $(x,y)\to(-y,x)$
Apply to $S(0,0)$: $(0,0)\to(0,0)$ (does not match $S'(4,0)$)

Step4: Test 180° (not option) & 270° (not option), test 225° rule

225° counterclockwise = 180°+45°, but simpler: check vector direction. Original $ST$ vector: $(-4,4)$, image $S'T'$ vector: $(0,3)$. Alternatively, check rotation of point $T(-4,4)$:
225° counterclockwise rotation formula:
$$x' = x\cos225^\circ - y\sin225^\circ$$
$$y' = x\sin225^\circ + y\cos225^\circ$$
$\cos225^\circ=-\frac{\sqrt{2}}{2}$, $\sin225^\circ=-\frac{\sqrt{2}}{2}$
For $T(-4,4)$:
$x' = -4\times(-\frac{\sqrt{2}}{2}) - 4\times(-\frac{\sqrt{2}}{2}) = 2\sqrt{2}+2\sqrt{2}=4\sqrt{2}\approx5.66$ (does not match $T'(4,3)$)

Step5: Test 135° rotation rule

135° counterclockwise rotation formula:
$$x' = x\cos135^\circ - y\sin135^\circ$$
$$y' = x\sin135^\circ + y\cos135^\circ$$
$\cos135^\circ=-\frac{\sqrt{2}}{2}$, $\sin135^\circ=\frac{\sqrt{2}}{2}$
For $S(0,0)$: $x'=0$, $y'=0$ (does not match $S'(4,0)$)

Step6: Test 90° rotation about origin correction

Wait, recheck: original triangle is in 2nd quadrant, image in 1st. 90° counterclockwise from 2nd quadrant goes to 3rd, 180° to 4th, 270° to 1st. 270° counterclockwise = 90° clockwise, rule: $(x,y)\to(y,-x)$
For $S(0,0)$: $(0,0)\to(0,0)$ (no, but shift? No, rotation about origin. Wait, check point $T(-4,4)$: 225° counterclockwise is not, but 180°+45° no. Wait, check distance from origin:
$ST$ length: $\sqrt{(-4-0)^2+(4-0)^2}=\sqrt{32}=4\sqrt{2}$
$S'T'$ length: $\sqrt{(4-4)^2+(3-0)^2}=3$ (no, wait, $S'(4,0)$, $T'(4,3)$: length 3. Original $ST$ length 4√2≈5.66, no. Wait, rotation preserves length. So check $SR$ length: $\sqrt{(-2-0)^2+(1-0)^2}=\sqrt{5}$
$S'R'$ length: $\sqrt{(1-4)^2+(4-0)^2}=\sqrt{9+16}=5$ (no, not same). Wait, rotation about point (2,-2)? No, options are rotation angles. Wait, no, original triangle: $S(0,0)$, $T(-4,4)$, $R(-2,1)$. Image: $S'(4,0)$, $T'(4,3)$, $R'(1,4)$. Wait, 90° counterclockwise about (2,2):
For $S(0,0)$: translate to (-2,-2), rotate 90°: (2,-2), translate back: (4,0) which matches $S'(4,0)$
For $T(-4,4)$: translate to (-6,2), rotate 90°: (-2,-6), translate back: (0,-4) no. Wait, 180° no. Wait, 225° counterclockwise about origin: $T(-4,4)$ becomes $(-4)(-\frac{\sqrt{2}}{2}) -4(-\frac{\sqrt{2}}{2})= 2\sqrt{2}+2\sqrt{2}=4\sqrt{2}\approx5.66$, $y'=(-4)(-\frac{\sqrt{2}}{2}) +4(-\frac{\sqrt{2}}{2})=2\sqrt{2}-2\sqrt{2}=0$ no. Wait, 90° clockwise (270° counterclockwise) rule: $(x,y)\to(y,-x)$
For $T(-4,4)$: $(4,4)$ which matches $T'(4,3)$? No. Wait, wait, I misread points: original $T$ is at (-4,4)? No, original $T$ is at (-4,4)? No, grid: x-axis left is negative, right positive. Original $S$ is at (0,0), $T$ is at (-4,4), $R$ is at (-2,1). Image $S'$ is at (4,0), $T'$ is at (4,3), $R'$ is at (1,4). Wait, the length of $ST$ is 4√2, $S'T'$ is 3, which is not same. Wait no, I misread $T'$: $T'$ is at (3,4)? No, x=3, y=4? No, grid: x-axis 4 is at $S'$, $T'$ is at (3,4). Oh! I misread $T'$ as (4,3), it's (3,4). Now:
$S'(4,0)$, $T'(3,4)$, $R'(1,1)$? No, no, original $R$ is at (-2,1), image $R'$ is at (1,4). Now, $ST$ vector: (-4,4), $S'T'$ vector: (-1,4). Wait, 90° counterclockwise rotation: (x,y)→(-y,x). For $T(-4,4)$: (-4,-4) no. 135° rotation:
$x' = -4\times(-\frac{\sqrt{2}}{2}) -4\times(\frac{\sqrt{2}}{2})=2\sqrt{2}-2\sqrt{2}=0$, $y'=-4\times(\frac{\sqrt{2}}{2}) +4\times(-\frac{\sqrt{2}}{2})=-2\sqrt{2}-2\sqrt{2}=-4\sqrt{2}$ no. Wait, 225° rotation:
$x' = -4\times(-\frac{\sqrt{2}}{2}) -4\times(-\frac{\sqrt{2}}{2})=2\sqrt{2}+2\sqrt{2}=4\sqrt{2}\approx5.66$, $y'=-4\times(-\frac{\sqrt{2}}{2}) +4\times(-\frac{\sqrt{2}}{2})=2\sqrt{2}-2\sqrt{2}=0$ no. Wait, 90° clockwise: (x,y)→(y,-x). For $T(-4,4)$: (4,4) which matches $T'(3,4)$? No. Wait, the triangle is rotated about the origin, original triangle is in the second quadrant, image in first. 90° counterclockwise would take second quadrant to third, 180° to fourth, 270° to first. 270° counterclockwise is 90° clockwise, rule (x,y)→(y,-x). For $S(0,0)$: (0,0) no, but $S'$ is (4,0). Oh! I misread $S$: original $S$ is at (-5,0), not (0,0). Yes! Original $S$ is at (-5,0), $T$ at (-4,4), $R$ at (-2,1). Image $S'$ at (5,0), $T'$ at (4,3), $R'$ at (1,4). Now, 180° rotation: (x,y)→(-x,-y). For $S(-5,0)$: (5,0) which matches $S'(5,0)$. For $T(-4,4)$: (4,-4) no, $T'$ is (4,3). No. 90° counterclockwise: (x,y)→(-y,x). For $S(-5,0)$: (0,-5) no. 135° rotation:
$\cos135=-\frac{\sqrt{2}}{2}$, $\sin135=\frac{\sqrt{2}}{2}$
$S(-5,0)$: $x'=-5\times(-\frac{\sqrt{2}}{2}) -0\times(\frac{\sqrt{2}}{2})=\frac{5\sqrt{2}}{2}\approx3.54$, $y'=-5\times(\frac{\sqrt{2}}{2}) +0\times(-\frac{\sqrt{2}}{2})=-\frac{5\sqrt{2}}{2}\approx-3.54$ no.
225° counterclockwise:
$\cos225=-\frac{\sqrt{2}}{2}$, $\sin225=-\frac{\sqrt{2}}{2}$
$S(-5,0)$: $x'=-5\times(-\frac{\sqrt{2}}{2}) -0\times(-\frac{\sqrt{2}}{2})=\frac{5\sqrt{2}}{2}\approx3.54$, $y'=-5\times(-\frac{\sqrt{2}}{2}) +0\times(-\frac{\sqrt{2}}{2})=\frac{5\sqrt{2}}{2}\approx3.54$ no.
Wait, 90° clockwise (270° counterclockwise): (x,y)→(y,-x). For $S(-5,0)$: (0,5) no. Wait, the image is $S'(4,0)$, original $S$ is at (0,0). Original triangle: $S(0,0)$, $T(-4,4)$, $R(-2,1)$. Image: $S'(4,0)$, $T'(4,3)$, $R'(1,4)$. The rotation is about (2,0). 90° counterclockwise about (2,0):
Translate point by (-2,0): $S(0,0)\to(-2,0)$, rotate 90°: (0,-2), translate back: (2,-2) no. 180° about (2,0): $S(0,0)\to(4,0)$ which matches $S'(4,0)$. $T(-4,4)\to(8,-4)$ no. $R(-2,1)\to(6,-1)$ no.
Wait, the only option that makes sense is 90° rotation? No, wait, the triangle's orientation: original triangle has right angle at T? No, original $ST$ is vertical? No, $S(0,0)$, $T(-4,4)$ is a line with slope -1. Image $S'(4,0)$, $T'(4,3)$ is vertical line. Slope of $ST$ is -1, slope of $S'T'$ is undefined (vertical). The angle between slope -1 (135° from x-axis) and vertical (90° from x-axis) is 45°? No, 135-90=45. Wait, no, counterclockwise rotation from original to image: the angle between $ST$ and $S'T'$ is 135°. Wait, no, original $ST$ vector is (-4,4), image $S'T'$ vector is (0,3). The dot product: $(-4)(0)+(4)(3)=12$. The magnitude of $ST$ is $\sqrt{32}=4\sqrt{2}$, magnitude of $S'T'$ is 3. $\cos\theta=\frac{12}{4\sqrt{2}\times3}=\frac{1}{\sqrt{2}}$, so $\theta=45°$. So counterclockwise rotation of 45°? But that would mean the answer is a 45° rotation. Wait, yes! The dot product formula gives the angle between the vectors, which is the rotation angle.
$\cos\theta=\frac{\vec{ST}\cdot\vec{S'T'}}{|\vec{ST}||\vec{S'T'}|}=\frac{(-4)(0)+(4)(3)}{4\sqrt{2}\times3}=\frac{12}{12\sqrt{2}}=\frac{1}{\sqrt{2}}$, so $\theta=45°$.

Answer:

a 45° rotation