derive the equation of the locus of a point twice as far from (-2,3,4) as from ( 3,-1,-2)
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Hello!
✓ The locus of all points ‘ k ’ units away from (3, -1, -2) is the sphere.
(x - 3)² + (y + 1)² + (z + 2)² = k²
Then,
✓ The locus of all points ‘ 2k ’ units away from (-2, 3, 4) is the sphere.
(x + 2)² + (y - 3)² + (z - 4)² = 4k²
All points in the intersection of these two spheres are :-
• k units away from (3, -1, -2)
• 2k units away from (-2, 3, 4)
Dividing the 2nd eqn. by four :-
Thus,
They are twice as far from (-2, 3, 4) as from (3, -1, -2), as required.
(x - 3)² + (y + 1)² + (z + 2)² =
This equation describes the locus you're looking for :)
Cheers!
✓ The locus of all points ‘ k ’ units away from (3, -1, -2) is the sphere.
(x - 3)² + (y + 1)² + (z + 2)² = k²
Then,
✓ The locus of all points ‘ 2k ’ units away from (-2, 3, 4) is the sphere.
(x + 2)² + (y - 3)² + (z - 4)² = 4k²
All points in the intersection of these two spheres are :-
• k units away from (3, -1, -2)
• 2k units away from (-2, 3, 4)
Dividing the 2nd eqn. by four :-
Thus,
They are twice as far from (-2, 3, 4) as from (3, -1, -2), as required.
(x - 3)² + (y + 1)² + (z + 2)² =
This equation describes the locus you're looking for :)
Cheers!
Anonymous:
thank uhh so much @ihelper ...
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