susan will rotate triangle abc 180° about one of the labeled points. susan says, he image will not touch the pre-image at all.\ which points could be her center of rotation? select each correct answer. ☐ point c ☐ point d ☐ point e ☐ point f

Question

susan will rotate triangle abc 180° about one of the labeled points. susan says, 	he image will not touch the pre-image at all.\ which points could be her center of rotation? select each correct answer. ☐ point c ☐ point d ☐ point e ☐ point f

Accepted Answer

# Brief Explanations:
- **Point C**: Rotating about point C, point C stays (since it's the center), so the image touches the pre - image at C. So not a valid center.
- **Point D**: Point D is inside the triangle. A 180° rotation about an interior point will result in the image overlapping or touching the pre - image (as the distance from D to vertices will map to points inside or on the triangle). So not a valid center.
- **Point E**: Point E is outside the triangle but relatively close. However, when we rotate 180° about E, some parts of the image may still touch the pre - image? Wait, no, let's think again. Wait, actually, point E is below the base BC. But maybe not. Wait, no, the key is for the image to not touch at all.
- **Point F**: Point F is far outside the triangle. When we rotate triangle ABC 180° about point F, the image of the triangle will be on the other side of F, and since F is far away, the image will not touch the pre - image. Also, let's re - evaluate point E. Wait, maybe my initial thought was wrong. Wait, let's recall the property of 180° rotation: for a point (x,y) rotated 180° about (h,k), the image is (2h - x, 2k - y). So if the center is outside the triangle and far enough, the image will not touch. Point F is far to the right, so rotating about F, the image of ABC will be on the left side of F, away from the original triangle. Point E: if we rotate about E, which is below BC, the image of A will be below E, and the image of B and C will be above E? Wait, no, 180° rotation about E: let's say E is (h,k), then A'(2h - Ax, 2k - Ay), B'(2h - Bx, 2k - By), C'(2h - Cx, 2k - Cy). If E is below BC, the image of the triangle may still intersect with the original? Wait, maybe the correct answers are point E and point F? Wait, no, let's look at the diagram again. The triangle is ABC, with D inside, E below BC, F to the right of C. When rotating about F: the original triangle is on the left of F, the image will be on the left of F? No, 180° rotation about F: the image is on the opposite side of F from the original triangle. So original is to the left of F, image is to the right? Wait, no, 180° rotation: the distance from F to each vertex is the same as from F to the image vertex, but in the opposite direction. So if F is to the right of C, then the image of C will be to the left of F, at a distance equal to FC from F. Similarly for B and A. Wait, maybe I messed up. Let's think of a simple case: if you rotate a triangle 180° about a point outside the triangle and not on any extension of its sides, and far enough, the image won't touch. So point F is far right, so the image will be far left of F, away from the original triangle. Point E: if E is below BC, the image of the triangle will be above E (since 180° rotation flips the y - coordinate relative to E). But the original triangle is above E (since E is below BC). So the image would be above E, but the original is also above E? Wait, no, E is below BC, so the original triangle is above E. Rotating 180° about E: the image of the triangle will be below E (since for each point (x,y) in the triangle, the image is (2h - x, 2k - y), where (h,k) is E. If E is (h,k) with k less than the y - coordinates of B and C, then 2k - y will be less than k (since y>k), so the image points will be below E, while the original is above E. So the image is below E and the original is above E, so they don't touch. Wait, so maybe both E and F? But wait, the problem says "the image will not touch the pre - image at all". Let's check each point:

- **Point C**: As C is a vertex, rotating about C, C is fixed, so image touches pre - image at C. Eliminate.
- **Point D**: Inside the triangle, so the image will overlap with the pre - image (since rotating 180° about an interior point, the image is inside or on the triangle). Eliminate.
- **Point E**: Outside the triangle, below the base. The image after 180° rotation will be on the opposite side of E from the original triangle, so no touch.
- **Point F**: Outside the triangle, to the right. The image after 180° rotation will be on the opposite side of F from the original triangle, so no touch.

Wait, but maybe my analysis of point E is wrong. Let's take coordinates as an example. Suppose E is at (0,0), B is at (1,1), C is at (3,1), A is at (2,5). Then rotating 180° about E: B' is (- 1,-1), C' is (- 3,-1), A' is (- 2,-5). The original triangle is in the first quadrant (y>0), the image is in the third quadrant (y < 0), so they don't touch. For point F, say F is at (10,0), B is (1,1), C is (3,1), A is (2,5). Rotating 180° about F: B' is (19,-1), C' is (17,-1), A' is (18,-5). The original triangle is at x from 1 - 3, y from 1 - 5. The image is at x from 17 - 19, y from - 5 to - 1. No touch. For point D (inside), say D is (2,2). Rotating 180° about D: B' is (3,3), C' is (1,3), A' is (2,-1). The original triangle has A at (2,5), B at (1,1), C at (3,1). The image point A' is (2,-1), B' is (3,3), C' is (1,3). The original triangle has B and C at y = 1, x = 1 and 3. The image B' and C' are at y = 3, x = 3 and 1, which is inside the original triangle's vertical range (x from 1 - 3) and above B and C. So the image touches the pre - image? Maybe. And point C: rotating about C, C is fixed, so image touches pre - image at C. So the correct points are E and F. Wait, but the options are point C, D, E, F. So the correct answers are point E and point F.

# Answer:
E. point E, F. point F

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