Wikipedia:Reference desk/Archives/Mathematics/2009 March 28#coordinate axes

how do we define the slopes of the coordinate axes??? or

do we simply say that he slope of x-axis is zero and that of the y-axis is infinity?? —Preceding unsigned comment added by 117.197.116.50 (talk) 02:16, 28 March 2009 (UTC)

:Slope is defined as $m\; =\; \backslash frac\{\backslash Delta\; y\}\{\backslash Delta\; x\}.$ 122.107.207.98 (talk) 03:19, 28 March 2009 (UTC)

::Using that definition, I would say that if the axes are perpendicular (like they usually are), the x-axis has a slope of zero ($\backslash Delta\; y\; =\; 0$) and the y-axis's slope is undefined ($\backslash Delta\; x\; =\; 0$). But can't you have non-perpendicular axes? --Evan ¤ Seeds 03:42, 28 March 2009 (UTC)

:::The formulae, m₁ * m₂ = -1 for perpendicular lines having slopes m₁ and m₂, does not hold if either m₁ or m₂ is undefined. I agree with a part of the previous answer. Non-perpendicular axis would not form a basis so it is usually appropriate to deal with perpendicular axis. --PST 05:59, 28 March 2009 (UTC)

::::Wouldn't they though? It certainly wouldn't be an orthogonal basis, obviously, but you could still have a basis. Consider the vectors (0, 1) and (1, 1) (let's say $\backslash vec\{x\}$ and $\backslash vec\{y\}$, respectively). They're obviously not orthogonal, but I think they're still a basis of R². After all, for any vector, $z\; =\; (a,\; b)\; =\; a\backslash vec\{y\}\; +\; (b-a)\backslash vec\{x\}$. I could easily be wrong though. --Evan ¤ Seeds 08:21, 28 March 2009 (UTC)

:::::They're a perfectly good basis, yes. Of course, when dealing with an inner product space the natural notion is 'orthonormal basis'. Algebraist 13:36, 28 March 2009 (UTC)