Wikipedia:Reference desk/Archives/Mathematics/2009 March 28
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March 28
editcoordinate axes
edithow 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 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 ( ) and the y-axis's slope is undefined ( ). 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 and , respectively). They're obviously not orthogonal, but I think they're still a basis of R2. After all, for any vector, . 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)
- 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 and , respectively). They're obviously not orthogonal, but I think they're still a basis of R2. After all, for any vector, . I could easily be wrong though. --Evan ¤ Seeds 08:21, 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)
- Using that definition, I would say that if the axes are perpendicular (like they usually are), the x-axis has a slope of zero ( ) and the y-axis's slope is undefined ( ). But can't you have non-perpendicular axes? --Evan ¤ Seeds 03:42, 28 March 2009 (UTC)
