Wikipedia:Reference desk/Archives/Mathematics/2009 March 28

= 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 $$m = \frac{\Delta y}{\Delta x}.$$ 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 ($$\Delta y = 0$$) and the y-axis's slope is undefined ($$\Delta x = 0$$). 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. -- PS T  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 $$\vec{x}$$ and $$\vec{y}$$, respectively). They're obviously not orthogonal, but I think they're still a basis of R2. After all, for any vector, $$z = (a, b) = a\vec{y} + (b-a)\vec{x}$$. 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)