X = -3.
The distance from p(-9, 0, 0) is
d = sqrt((x+9)^2 + y^2 + z^2)
The distance from q(3,0,0) is
d = sqrt((x-3)^2 + y^2 + z^2)
Let's set them equal to each other.
sqrt((x+9)^2 + y^2 + z^2) = sqrt((x-3)^2 + y^2 + z^2)
Square both sides, then simplify
(x+9)^2 + y^2 + z^2 = (x-3)^2 + y^2 + z^2
x^2 + 18x + 81 + y^2 + z^2 = x^2 - 6x + 9 + y^2 + z^2
18x + 81 = - 6x + 9
24x + 81 = 9
24x = -72
x = -3
So the desired equation is x = -3 which defines a plane.
Answer:
Use the given functions to set up and simplify O R X . y − 2 − 4 ( x − 3 ) = − 4 x + y + 10 O R X = − 4 x + y + 10
Step-by-step explanation:
The answer to this is 16.
Answer:
y = 2 / (r/3 - s/5)
Step-by-step explanation:
r/3 - 2/y = s/5
add 2/y to both sides
r/3 = s/5 + 2/y
Subtract s/5 from both sides
r/3 - s/5 = 2/y
multiply both sides by y
y(r/3 - s/5) = 2
Divide both sides by r/3 - s/5
y = 2 / (r/3 - s/5)