Question

3
abcd is rotated counterclockwise about the origin. by how many degrees was abcd rotated?
a. 90°
b. 180°
c. 270°
d. 360°

Explanation:

Step1: Analyze rotation properties

A 180° rotation about the origin maps a point \((x,y)\) to \((-x,-y)\). Let's check a vertex of ABCD, say A. From the graph, A is at \((2,2)\). After rotation, \(A'\) should be at \((-2,-2)\)? Wait, no, looking at the red square, \(A'\) is at \((2,-2)\)? Wait, no, let's check coordinates again. Wait, original ABCD: A is (2,2), B is (2,5), C is (5,5), D is (5,2)? Wait, no, the grid: y-axis, x-axis. Wait, original A: x=2, y=2 (since it's on y=2, x=2). Then \(A'\) is at (2,-2)? No, the red square: \(A'\) is (2,-2)? Wait, no, the red square's \(A'\) is (2,-2)? Wait, no, looking at the red square, \(A'\) is (2,-2), \(B'\) is (5,-2), \(C'\) is (5,-5), \(D'\) is (2,-5)? Wait, no, the original ABCD: A(2,2), B(2,5), C(5,5), D(5,2). After 180° rotation, (x,y)→(-x,-y). So A(2,2)→(-2,-2), but in the red square, \(A'\) is (2,-2)? Wait, maybe I misread the coordinates. Wait, maybe original A is (2,2), and \(A'\) is (2,-2)? No, that's a reflection over x-axis. Wait, no, 180° rotation: let's take point B(2,5). After 180° rotation, it should be (-2,-5). But in the red square, \(B'\) is (5,-2)? Wait, no, maybe I messed up the coordinates. Wait, the original square is in the first quadrant (positive x, positive y), and the rotated square is in the fourth quadrant? No, the red square is in the fourth quadrant? Wait, no, x positive, y negative. Wait, a 180° rotation: if you rotate a figure 180° counterclockwise, it's the same as rotating 180° clockwise, and the image is opposite in both x and y. Let's take point A(2,2). After 180° rotation, it should be (-2,-2), but in the red square, \(A'\) is (2,-2)? No, that can't be. Wait, maybe the original A is (2,2), and \(A'\) is (2,-2)? No, that's a 180°? Wait, no, 180° rotation would flip both x and y signs. Wait, maybe the coordinates are different. Wait, looking at the graph, original ABCD: A is at (2,2) (x=2, y=2), B at (2,5) (x=2, y=5), C at (5,5) (x=5, y=5), D at (5,2) (x=5, y=2). Then the red square: \(A'\) at (2,-2), \(B'\) at (5,-2), \(C'\) at (5,-5), \(D'\) at (2,-5). Wait, that's a 180° rotation? Wait, no, (2,2)→(2,-2) is reflection over x-axis, but (2,5)→(5,-2) is not. Wait, maybe I made a mistake. Wait, no, a 180° rotation: the figure is rotated so that it's upside down and reversed. Let's check the direction. Original ABCD: A(2,2), B(2,5), C(5,5), D(5,2) (a square with sides vertical and horizontal). After 180° rotation, the square should have vertices at (-2,-2), (-2,-5), (-5,-5), (-5,-2), but in the graph, the red square is at (2,-2), (5,-2), (5,-5), (2,-5). Wait, that's a 180° rotation? Wait, no, the x-coordinates are same positive, y-coordinates negative. Wait, maybe the original square is at (2,2), (2,5), (5,5), (5,2), and the rotated square is (2,-2), (5,-2), (5,-5), (2,-5). So each point (x,y)→(x,-y) is reflection over x-axis, but that's 180°? No, reflection over x-axis is 180°? No, reflection over x-axis is 180°? Wait, no, rotation 180° counterclockwise is same as rotation 180° clockwise, and it's equivalent to reflecting over both x and y axes. So (x,y)→(-x,-y). But in the red square, the x-coordinates are same as original, y-coordinates are negative. Wait, maybe the original square is at (2,2), (2,5), (5,5), (5,2), and the rotated square is (2,-2), (5,-2), (5,-5), (2,-5). So the x-coordinates are same, y-coordinates are -y. That would be a 180° rotation? Wait, no, that's a 180° rotation? Wait, no, if you rotate 180° counterclockwise, the figure should be in the opposite quadrant. Wait, original is in first quadrant (x>0, y>0), 180° rotation should be in third quadrant (x<0, y<0), but the red square is in fourth quadrant (x>0, y<0). Wait, maybe I misread the coordinates. Wait, maybe original A is (2,2), and \(A'\) is (-2,-2), but in the graph, the red square is at (2,-2), (5,-2), (5,-5), (2,-5). So maybe the original square is (2,2), (2,5), (5,5), (5,2), and the rotated square is (2,-2), (5,-2), (5,-5), (2,-5). So the transformation is (x,y)→(x,-y), which is a 180° rotation? Wait, no, (x,y)→(x,-y) is reflection over x-axis, which is 180°? No, reflection over x-axis is 180°? Wait, no, rotation 180° counterclockwise: let's take vector from origin to A(2,2), rotating 180° counterclockwise would point to (-2,-2). But in the red square, the vector is (2,-2), which is 180° from (2,2)? No, (2,2) and (2,-2) are 180° apart? Wait, the angle between (2,2) and (2,-2) is 180°? Yes, because they are colinear with the y-axis, opposite directions. Wait, maybe the square is axis-aligned, so rotating 180° counterclockwise would flip both x and y signs, but in this case, the x-signs are same, y-signs are flipped. Wait, maybe the original square is (2,2), (2,5), (5,5), (5,2), and the rotated square is (2,-2), (5,-2), (5,-5), (2,-5). So the x-coordinates are same, y-coordinates are -y. So the rotation is 180°? Wait, no, if you rotate 180° counterclockwise, the figure should be rotated so that the top becomes bottom and left becomes right. In the original square, A is bottom-left, B is top-left, C is top-right, D is bottom-right. After 180° rotation, A should be top-right, B top-left? No, wait, let's list the vertices:

Original ABCD:
A: (2, 2) (bottom-left of square)
B: (2, 5) (top-left)
C: (5, 5) (top-right)
D: (5, 2) (bottom-right)

After 180° rotation counterclockwise, the new vertices should be:
A' should be opposite of A, so from origin, A is (2,2), so 180° rotation is (-2,-2), but in the red square, A' is (2,-2). Wait, maybe the square is centered at (3.5, 3.5)? No, the square has side length 3 (from x=2 to x=5, y=2 to y=5). So center at (3.5, 3.5). Rotating 180° about the origin would move the center to (-3.5, -3.5), but the red square's center is (3.5, -3.5). So that's a 180° rotation about the center? No, the question is rotation about the origin. Wait, maybe I'm overcomplicating. Let's use the property: a 180° rotation makes the figure look upside down and reversed. Looking at the original (blue) square and the rotated (red) square, they are opposite in both x and y directions? Wait, no, the blue square is in the first quadrant (positive y), red in fourth (positive x, negative y). Wait, no, 180° rotation about origin: the blue square's vertices are (2,2), (2,5), (5,5), (5,2). The red square's vertices: let's see, \(A'\) is (2,-2), \(B'\) is (5,-2), \(C'\) is (5,-5), \(D'\) is (2,-5). So each vertex (x,y) becomes (x,-y). Wait, that's a reflection over the x-axis, which is equivalent to a 180° rotation? No, reflection over x-axis is 180°? No, reflection over x-axis is 180°? Wait, no, rotation 180° counterclockwise is the same as rotation 180° clockwise, and it's the same as reflecting over both x and y axes. So (x,y)→(-x,-y). But in the red square, (x,y)→(x,-y), which is reflection over x-axis, not 180° rotation. Wait, maybe the original square is (2,2), (2,5), (5,5), (5,2), and the rotated square is (2,-2), (5,-2), (5,-5), (2,-5). So the x-coordinates are same, y-coordinates are -y. So the rotation is 180°? Wait, no, if you rotate 180° counterclockwise, the figure should be in the third quadrant (negative x, negative y), but here it's in fourth (positive x, negative y). So maybe the answer is 180°? Wait, the options are 90, 180, 270, 360. Let's check 90° rotation: 90° counterclockwise rotation maps (x,y) to (-y,x). So A(2,2)→(-2,2), which would be in second quadrant, but red square is in fourth, so not 90°. 270° counterclockwise is (y,-x), so A(2,2)→(2,-2), which is what we see for \(A'\) (2,-2). Wait, 270° counterclockwise rotation: (x,y)→(y,-x). So A(2,2)→(2,-2), B(2,5)→(5,-2), C(5,5)→(5,-5), D(5,2)→(2,-5). Which matches the red square's vertices: \(A'\)(2,-2), \(B'\)(5,-2), \(C'\)(5,-5), \(D'\)(2,-5). Yes! So 270° counterclockwise rotation. Wait, no, wait: 90° counterclockwise is (-y,x), 180° is (-x,-y), 270° counterclockwise is (y,-x). So (2,2)→(2,-2) (since y=2, -x=-2? Wait, no: 270° counterclockwise rotation formula is (x,y)→(y, -x). So x=2, y=2: (2, -2). Yes! So A(2,2)→(2,-2) which is \(A'\). B(2,5)→(5,-2) which is \(B'\). C(5,5)→(5,-5) which is \(C'\). D(5,2)→(2,-5) which is \(D'\). Perfect match. So the rotation is 270°? Wait, no, wait, 270° counterclockwise is same as 90° clockwise. Wait, but the question says counterclockwise. Wait, but according to the coordinates, 270° counterclockwise rotation gives the red square. Wait, but let's check again. Wait, original A(2,2), after 270° counterclockwise: (y, -x) = (2, -2). Correct. B(2,5): (5, -2). Correct. C(5,5): (5, -5). Correct. D(5,2): (2, -5). Correct. So that's 270°? But wait, the red square is below the x-axis, same x-range, y-range negative. So yes, 270° counterclockwise. Wait, but earlier I thought 180°, but that was a mistake. So the correct rotation is 270°? Wait, no, wait, 90° counterclockwise: (x,y)→(-y,x). So A(2,2)→(-2,2) (second quadrant), not matching. 180°: (x,y)→(-x,-y). A(2,2)→(-2,-2) (third quadrant), not matching. 270°: (x,y)→(y,-x). A(2,2)→(2,-2) (fourth quadrant), which matches. 360°: same as original, which it's not. So the answer should be 270°? Wait, but the options are A.90, B.180, C.270, D.360. So C is 270°. Wait, but let's check the graph again. Original square: top is B(2,5), right is C(5,5), bottom D(5,2), left A(2,2). After rotation, the red square: top is \(B'\)(5,-2), right is \(C'\)(5,-5), bottom \(D'\)(2,-5), left \(A'\)(2,-2). So the order of the vertices: original ABCD is A-B-C-D (left-top-right-bottom). After rotation, \(A'\)-\(B'\)-\(C'\)-\(D'\) is left-top-right-bottom? No, \(A'\) is (2,-2), \(B'\) is (5,-2) (right of \(A'\)), \(C'\) is (5,-5) (below \(B'\)), \(D'\) is (2,-5) (left of \(C'\)). So the order is A'-B'-C'-D', which is left-top-right-bottom, same as original, but rotated 270° counterclockwise. Yes, that makes sense. So the rotation is 270° counterclockwise, so the answer is C.270°. Wait, but earlier I thought 180°, but that was a mistake. So the correct step is to use the rotation formula for 270° counterclockwise: (x,y)→(y, -x), which matches the coordinates.

Step2: Confirm with vertex coordinates

Take vertex A(2,2) (blue square). After 270° counterclockwise rotation about origin, the new coordinates are (y, -x) = (2, -2), which is \(A'\) (red square). Similarly, B(2,5)→(5, -2) = \(B'\), C(5,5)→(5, -5) = \(C'\), D(5,2)→(2, -5) = \(D'\). All vertices match the red square's coordinates, confirming a 270° counterclockwise rotation.

Answer:

C. 270°