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If a (3a, 0) and B (0, 3b) (a, B are not equal to 0) are two fixed points and P is the moving point on the straight line BX + ay = AB, then the trajectory equation of the center of gravity of the triangle ABP is obtained
Let P (m, n)
Then the coordinates of the center of gravity of the triangle are
Abscissa x = (3a + 0 + m) / 3
The ordinate is y = (0 + 3B + n) / 3
So m = 3x-3a, n = 3y-3b
P is on a straight line
bm+an=ab
therefore
b(3x-3a)+a(3y-3b)=ab
3bx-3ab+3ay-3ab=ab
3bx+3ay-7ab=0
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