Respuesta :
Distance from P to the x-axis = 2x distance from P to the yz-plane
Distance to the x-axis of a point P=(x,y,z) is (y^2+z^2)^1/2
Distance to the yz-plane of a point P=(x,y,z) is x
So your equation is:
(y^2+z^2)^1/2 = 2x
=> y^2 + z^2 = 4x^2
=> y^2 + z^2 - 4x^2 = 0
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Distance to the x-axis of a point P=(x,y,z) is (y^2+z^2)^1/2
Distance to the yz-plane of a point P=(x,y,z) is x
So your equation is:
(y^2+z^2)^1/2 = 2x
=> y^2 + z^2 = 4x^2
=> y^2 + z^2 - 4x^2 = 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
The distance between two points, is the number of units between them.
The equation that represents the distance is [tex]\mathbf{ -25x^2 + y^2+z^2 = 0}[/tex]
The distance between a point and the x-axis is represented as:
[tex]\mathbf{D = (y^2+z^2)^{\frac 12}}[/tex]
From the question, we have:
[tex]\mathbf{D = 5x}[/tex]
Equate both expressions for D
[tex]\mathbf{5x = (y^2+z^2)^{\frac 12}}[/tex]
Square both sides
[tex]\mathbf{25x^2 = y^2+z^2}[/tex]
Equate to 0
[tex]\mathbf{ y^2+z^2 -25x^2 = 0}[/tex]
Rewrite as:
[tex]\mathbf{ -25x^2 + y^2+z^2 = 0}[/tex]
Hence, the equation that represents the distance is [tex]\mathbf{ -25x^2 + y^2+z^2 = 0}[/tex]
Read more about distance at:
https://brainly.com/question/10900823
