Question

The point in the xy plane which is equidistant from the points (5,0,6) ,(0,-3,2) and (4,5,0) is

Answer 1

Answer:

point is ( 75/14 , -13/14 , 0)

Step-by-step explanation:

Let say Point in the xy plane is ( x , y , 0)

is Equidistant  from the points (5,0,6) ,(0,-3,2) and (4,5,0) is

Distance² from (5,0,6) = ( x - 5)² + (y-0)² + (0 - 6)²

= x² + y² - 10x + 61     Eq1

Distance² from (0,-3,2) = ( x - 0)² + (y-(-3))² + (0 - 2)²

= x² + y² +6y + 13     Eq 2

Distance² from (4,5,0) = ( x - 4)² + (y-5)² + (0 - 0)²

= x² + y² - 8x -10y + 41    Eq3

Equating eq 1 & eq 2

x² + y² - 10x + 61  = x² + y² +6y + 13

=> 6y + 10x = 48

=> 3y + 5x = 24   Eq 4

Equating eq 1 & Eq 3

x² + y² - 10x + 61  = x² + y² - 8x -10y + 41

=> 2x - 10y = 20   Eq 5

10* eq 4 + 3 * eq 5

50x + 6x = 240 + 60

=> 56x = 300

=> x  = 75/14

 y = -13/14

point is ( 75/14 , -13/14 , 0)

The point in the xy plane which is equidistant from the points (5,0,6) ,(0,-3,2) and (4,5,0) is  ( 75/14 , -13/14 , 0)