Question 1168329: Write an indirect proof to prove Theorem 4-2: If there is a line and a point not on the line, then exactly one plane contains them.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **Theorem 4-2:** If there is a line and a point not on the line, then exactly one plane contains them.
**Indirect Proof (Proof by Contradiction):**
**1. Assume the Opposite:**
Assume that there is a line *l* and a point *P* not on the line *l*, and that *more than one plane* contains both *l* and *P*.
**2. Develop the Contradiction:**
* Let's call two of these planes that contain *l* and *P*, plane *A* and plane *B*.
* Since both plane *A* and plane *B* contain line *l*, and line *l* has at least two distinct points (let's call them *X* and *Y*), then both plane *A* and plane *B* contain points *X* and *Y*.
* We also know that both plane *A* and plane *B* contain point *P*.
* Therefore, plane *A* and plane *B* contain three non-collinear points *X*, *Y*, and *P*.
* However, a postulate of geometry (Postulate 4-1) states that through any three non-collinear points, there is exactly one plane.
* This means that plane *A* and plane *B* must be the same plane, which contradicts our initial assumption that they were different planes.
**3. State the Conclusion:**
Since our assumption that more than one plane contains the line *l* and the point *P* leads to a contradiction, our assumption must be false. Therefore, exactly one plane contains the line *l* and the point *P*.
**In summary:**
We started by assuming the opposite of what we wanted to prove (more than one plane). We then showed that this assumption led to a contradiction with a known postulate (only one plane through three non-collinear points). This forces us to reject our initial assumption, thereby proving the original theorem.
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