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Find the equation of the set of point which are equidistant from the point (1, 2,3) and (3, 2,-1) |
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Answer» Let\xa0A(1,2,3)andB(3,2,-1)\xa0are the given points and\xa0P(x,y,z)\xa0is the point that is equidistant from these two points.Then,\xa0PA=√(x-1)2+(y-2)2+(z-3)2PA=√x2+y2+z2-2x-4y-6z+14PB=√(x-3)2+(y-2)2+(z+1)2PB=√x2+y2+z2-6x-4y+2z+14As, we are given,\xa0PA=PB∴√x2+y2+z2-2x-4y-6z+14=√x2+y2+z2-6x-4y+2z+14⇒x2+y2+z2-2x-4y-6z+14=x2+y2+z2-6x-4y+2z+14⇒4x=8z⇒x-2z=0, which is the required equation. X-2z=0 |
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