Question

P is a variable point which moves such that 3PA = 2PB. If A = (-2, 2, 3) and
B=(13,-3, 13), prove that P satisfies the equation
x² + y^2 +2? + 28x-12 y +10z – 247 = 0.​

Answer 1

Answer:

x²  + y² + z² + 28x - 12y + 10z - 247 = 0

Step-by-step explanation:

Let say P ( x , y , z)

3PA = 2PB

Squaring both sides

=> 9 (PA)² = 4 (PB)²

(PA)² = (x + 2)² + (y - 2)² + (z - 3)²

(PB)² = (x - 13)² + (y + 3)² + (z - 13)²

=> 9 ( (x + 2)² + (y - 2)² + (z - 3)²)  = 4 ((x - 13)² + (y + 3)² + (z - 13)²)

=> 9 ( x² + 4 + 4x + y² + 4 - 4y + z² + 9 - 6z)  = 4 (x² + 169 - 26x + y² + 9 + 6y + z² + 169 - 26z)

=> 9 ( x²  + y² + z² + 4x  - 4y - 6z + 17 )  = 4 (x² + y² + z² - 26x+ 6y - 26z + 347 )

=> 5 (x²  + y² + z²) + 36x + 104x  -36y -24y  -54z + 104z + 153 - 1388 = 0

=> 5 (x²  + y² + z²) + 140x  -60y  + 50z  - 1235 = 0

Dividing by 5 both sides

=> x²  + y² + z² + 28x - 12y + 10z - 247 = 0

QED

Proved