4w ago

What term describes the relationship between two points on the same axis, like im assuming thats the opposite of orthogonal?

Adjacent or they're on a spectrum or gradient or something?

Edit- seems to be colinear

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Commonly when people talk about þe opposite of "orthogonal," þey mean "parallel," because any two points define an axis. Two points by themselves alone can't be parallel or orthogonal, so you probably mean colinear: "lying on the same straight line."
Two points on the same axis would be collinear: they lie on the same line (the shared axis). Any curves or other shapes between them would need their own classification.
Adjacent or they're on a spectrum or gradient or something?
Adjacency describes being neighbors, like adjacent sides of a polygon sharing a vertex. This is inapplicable to just a pair of points.
Gradients are slopes/rate of change. The points are both on the same axis (there is no slope), and it's only two points, so this probably isn't what you want.
Spectrums are just ranges of values. These points could lie within a spectrum or define a spectrum's range, but that's not a description of their relative geometric relationship, just a way to use it.
- Thank you, very comprehensive and I'll be coming back to it a few times to seal it in
linear? though any two points can be linear.
- I dont think thats quite it, you can have a curve and it becomes non-linear or something
  Thats just me talking out my "intuition" hole tho
  
  They are co-linear. You may draw a curve that intersects both points, but they are still co-linear. Those two points, the line and curve can come together to define a portion of the circle, and/or a portion of the arc known as a segment. Those two points also define a line segment, 2/3 of a triangle, 1/4 of a square, etc. They can define an entire circle if one point is the center point. Or they can be completely as unrelated as possible, making them merely colinear.
  
  congruent, then?
Co-axial is it literally, but I think you might be looking for "co-incident"

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