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.
<span>x=6</span>, <span>x=−5</span> or <span>x=9</span>
Explanation:
<span><span>f<span>(x)</span></span>=<span>(x−6)</span><span>(x+5)</span><span>(x−9)</span></span>
If all of the linear factors are non-zero, then so is their product <span>f<span>(x)</span></span>.
If any of the linear factors is zero, then so is their product <span>f<span>(x)</span></span>.
