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Given three points A(1, 2) , B(7, 4), C(5, 7). Find the points P(x,y) such that PB = 2 PA, and PC = 3 PA ?
1 Answer
- PopeLv 77 years ago
There are no solutions. The two conditions describe circles, which do not intersect.
PB = 2PA
√[(x - 7)² + (y - 4)²] = 2√[(x - 1)² + (y - 2)²]
(x - 7)² + (y - 4)² = 4[(x - 1)² + (y - 2)²]
x² + y² - 14x - 8y + 65 = 4x² + 4y² - 8x - 16y + 20
3x² + 3y² + 6x - 8y - 45 = 0
PC = 3PA
√[(x - 5)² + (y - 7)²] = 3√[(x - 1)² + (y - 2)²]
(x - 5)² + (y - 7)² = 9[(x - 1)² + (y - 2)²]
x² + y² - 10x - 14y + 74 = 9x² + 9y² - 18x - 36y + 45
8x² + 8y² - 8x - 22y - 29 = 0
So there you have the two circles. If you can find a point of intersection, then that represents a solution. But like I said, there are no points of intersection.
