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:
1. x=5
2.x=-33/8
3.x=10
4.x=5
Step-by-step explanation:
I’m 100% sure it’s the 3rd one :)
<h3>✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽</h3>
➷ It would cross at (0, -1) or just -1
y = mx + c
The 'c' value in the equation is the y intercept
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➶ Hope This Helps You!
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