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Coordinate geometry

  • Thread starter denian
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  • #1
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the vertices of a triangle are 0, and the points A (a, 0) and B (0,a) where a>0.
the point P moves in the plane OAB such that OP square + AP square + BP square = k square, where k is a positive constant. show that k mst be bigger than (2/3)a X square root of 3


i tried. but cant even know how to start. any hints?
 

Answers and Replies

  • #2
HallsofIvy
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Let P be (x,y). The OP square= x2+ y2, AP square= (x-a)2+ y2, and BP square= x2+ (y-a)2.
Now, what is OP square + AP square+ BP square. Keep in mind that x2 and y2 must be positive.
 

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