My answer that i have gotten is 280
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.
The answer would be 17.8 hope this help you
I’m sorry, I wish I knew this one and I would help
Answer:
Slope is 3
Y- intercept is -4
Doesn't pass through zero.
Linear
Give me options and I can give a better answer :)
