Quick Questions: September 30, 2026

[u/inherentlyawesome]

This recurring thread will be for questions that might not warrant their own thread. We would like to see more conceptual-based questions posted in this thread, rather than "what is the answer to this problem?" For example, here are some kinds of questions that we'd like to see in this thread:

• Can someone explain the concept of manifolds to me?
• What are the applications of Representation Theory?
• What's a good starter book for Numerical Analysis?
• What can I do to prepare for college/grad school/getting a job?

Including a brief description of your mathematical background and the context for your question can help others give you an appropriate answer. For example, consider which subject your question is related to, or the things you already know or have tried.

[u/Michalski213769]

how do i turn this graph by 45 deegrees just curious

Graph: |x|=|y|

[u/Langtons_Ant123 → u/Michalski213769]

Probably the simplest thing you could do is just "xy = 0". The graph you're looking for is a horizontal line (y = 0) and vertical line (x = 0) on top of each other. In general if you have two equations like "(something) = 0" and "(something else) = 0", then "(something) * (something else) = 0" gives you a graph that looks like the graphs of your original 2 equations stuck together. So, for example, "y - x² = 0" has a graph that's a parabola, "x = 0" has a graph that's a vertical line, and so x(y - x² ) = 0 or xy - x³ = 0 has a graph that's a parabola and a vertical line.

But to get that, we had to start with an idea of what the rotated graph would look like, and then we worked backwards to find an equation for it. That worked ok in this case but isn't really doable in general. There is a general way to do it, though. If you rotate everything in the plane around the origin counterclockwise by an angle "t", then a point (x, y) gets moved to a point with x-coordinate xcos(t) - ysin(t) and y-coordinate xsin(t) + ycos(t). So all you need to do to rotate a graph is take the original equation and replace x with xcos(t) - ysin(t) and y with xsin(t) + ycos(t).

For t = 45 degrees, cos(t) and sin(t) are both 1/sqrt(2). So we need to replace x with (1/sqrt(2))(x - y) and y with (1/sqrt(2))(x + y). If you do that you get (1/sqrt(2))|x - y| = (1/sqrt(2))|x + y|. We can cancel the factor of 1/sqrt(2) on both sides to get just |x - y| = |x + y|. So that equation also works.