Question

derive the equation of the locus of a point twice as far from ( -2, 3,4 ) as from ( 3 ,-1 , -2)

Answer 1

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 = $\dfrac{\sf (x + 2)^2 + (y - 3)^2 + (z - 4)^2}{\sf 4}$

This equation describes the locus you're looking for :)

Cheers!

Answer 2

Answer:hello

Step-by-step explanation:

Answer