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I have a doubt.

My teacher has posted a question that Is it always/sometimes/never true that the line y=x is at 45 degrees to the x-axis?

I know that the slope of the line y=x is always 1, and the line must make 45 degrees with x-axis.

I am a bit confused when I am plotting the line y=x at different X-scales (see Figures 1 and 2 attached). I do not visually see the line at 45 degrees to the x-axis in Figure 2.

What am I missing?

Which answer would be correct and why? (a) It is always true (b) It is sometimes true because when we change the scale of x or y axis, graph will distort and not be at 45 degrees.

See images attached. Fig 1 with equal x and y axis Fig 2 with different axis

Misha Lavrov
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Jeet Kun
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  • "I am a bit confused when I am plotting the line y=x at different X-scales" that's the problem. You are drawing it to a different scale. You are not allowed to do that. The basic premise of analytic cartesian geometry is that you universe is always cartesian space where the axes are perpendicular and all axes are on the same uniform scale. – fleablood Dec 02 '20 at 21:45
  • Here's a question. If $u = 10x$ does the line $y =u$ have an angle of $45^\circ$? Does the question even make sense? – fleablood Dec 03 '20 at 03:19

5 Answers5

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Yes, you can give different scales to your x and y axes, and then when you draw the line y=x, it will not be at 45 degrees to the x-axis. However, what someone usually means when they say that the line is at 45 degrees, what they almost always mean is that the line would be at 45 degrees if the x and y axes were scaled equally (and at right angles to each other). It is a convention. And it is a useful convention, because it allows us to uniformly represent slopes as the tangents of angles.

Math Keeps Me Busy
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  • I disagree that this is just a "convention". As with coordinates, angles must be defined in the context of a certain coordinate system. It does not matter how you render a particular plot on paper - what matters is the abstract values themselves. The line y=x always intersects the line y=0 at 45 degrees. It doesn't matter whether you scale the x-axis at 2x, or use a log-scale, or set the plots bounds so that it doesn't contain the origin at all - the lines still intersect at 45 degrees. Scaling just means you lose certain analogues with the rendered lines. – MooseBoys Mar 08 '22 at 19:23
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I'm going to say it's a meaningless question.

If you skew the graph by having the x axis and y axis be on different scales, or by depicting the x, y axis so they are not perpendicular, Or having the scale of the y axis be linear but the x axis be logrithmic or geometric or exponential (in which case $y=x$ won't even be a line) then no, the angle of that drawing of the function is not at $45^\circ$).

But those are skewed interpretations of "the" graph and just interpretive drawings. The are not "the" graph.

We define the "graph" of $y=x$ as: The set of all $(x,y)$ points in $\mathbb R^2$ where $(x,y)$ satisfy the equation $y=x$. But that's not a drawing... thats a bunch of points. What does it mean to say they are a "line" or that they have an angle? Well, it is presumed that we are depicting them on "THE" Cartesian plane and THE Cartesian plane has perpendicular $x$, $y$ axes on the same scale. And on that plane with that scale, and those axis, yes, there is only one instance of a graph $y= x$ and it is at an $45^\circ$ angle.

Graphing it on a skewed axis.... is simply not on the cartesian plane. They are another interpretation.

But it is vague as to what we actually "mean" when we say "$y=x$ is always at a right angle". Does that mean on the universal Cartesian plane? In which case the answer is "YES". Or does it mean any possible depiction? In which case the answer is "NO".

fleablood
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  • Well said, and hopefully this is exactly the sort of discussion the teacher will be having with the class – robert timmer-arends Dec 03 '20 at 02:17
  • The problem is that if $\Pi$ is the euclidean plane(2D affine space) the coordinate isomorphism $c_B: \Pi \to \matbb{R}^2$ is an isometry(so it preserves angles, lenght,...) iff $B$ is orthonormal. When you choose strange basis you get the discordance between the metric concepts in the plane and the metric concepts in the numerical affine space. – Kandinskij Dec 03 '20 at 14:19
  • That may be, but the question of "is $y=x$ always at an angle of $45^\circ$" is one of what does that mean. – fleablood Dec 03 '20 at 16:22
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A more advanced approach: we can show $y = x$ has a $45^{\circ}$ angle by using polar coordinates.

Let $x = r \cos \theta$ and $y = r \sin \theta.$ As $y = x$, $r \sin \theta = r \cos \theta$, but $r$ cancels out and we have $\cos \theta = \sin \theta$. Divide by $\cos \theta$ to get $\tan \theta = 1$, and thus $\theta = \arctan (1)$ or $45^{\circ}$.

bjcolby15
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  • Yeah.... That's not the level the OP is at. It presupposes the the OP accepts there is a cartesian plane (and scale) that is accepted as universal. If the OP could accept that s/he never would have asked the question or gotten thrown be drawing it other scale.... Here's a question. If $u = 10x$ does the line $y =u$ have an angle of $45^\circ$? Does the question even make sense? – fleablood Dec 03 '20 at 03:18
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It is always at $45^{\circ}$. It may not "look" like $45^{\circ}$, but that's because you're "cheating" and measuring the angle with the original coordinate system.

Then again, if you measure the coordinates of a point $(x,y)$ on the line $y=x$ in the scaled system using the original coordinate system, then clearly $y\neq x$.

So is $y=x$ or not? Clearly it is.

This is like drawing a circle on a piece of paper, then placing the paper flat on a table and squatting down near the level of the table and looking at the circle nearly edge-on. It looks squished and no longer looks like a circle but instead looks like a flattened ellipse. But of course, it is a circle.

MPW
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  • This assumes that the 1:1 coordinate system is in someway priviliged. Let's put it in another way. Let's assume you see a circle. Then that circle starts to travel at almost the speed of light, and by relativistic effect it appears as an ellipse to you. That figure is a circle or is an ellipse? – Kandinskij Dec 03 '20 at 14:31
  • My point is that the measurement must be made in the coordinate system in which it was created in order for the comparison to be meaningful. The circle remark just shows how a scaled system might arise. Superimposing a different coordinate system changes how the object is measured -- the line would no longer be the line $y=x$ in a different coordinate system. So, the point is that "the line at 45 degrees" and "the line $y=x$" are the same -- you can't measure one of these w.r.t. one coordinate system and the other w.r.t. a different coordinate system and then claim they are different – MPW Dec 03 '20 at 15:15
  • We have no information about the coordinate system. The question doesn't specify that we are in the Cartesian plane(that is simply a 2D euclidean affine space equipped with an orthonormal basis). It says just that we are in a 2D euclidean space(a plane). It doesn't specify what kind of basis is fixed(it could be anything). So if $y$ and $x$ are affine coordinates then they describe a different object if the affine basis changes. Even if you don't want to consider $x$ and $y$ as affine coordinates and work in the numerical affine space $\mathbb{R}^2$(so x and y are simply the couple entries). – Kandinskij Dec 03 '20 at 15:29
  • (Continuing the precedent comment) You should specify the meaning of the object:"x-axis" to talk about the angle of the line with "x-axis". So you have again to specify an affine basis. It's true that in this space we have a canonical choice for the basis, but the question doesn't specify that we make this choice. – Kandinskij Dec 03 '20 at 15:30
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Technically speaking the euclidean plane is a 2D affine euclidean space equipped with the standard product. The two axis(with their respective unit of measures) are simply a geometric representation of an orthogonal affine basis $(O,u,v)$ that we fix for the plane). Now $\mathcal{R}:y=x$ is a cartesian representation of an affine subspace in the affine basis $(O,u,v)$ (Where clearly $y$ is the second affine coordinate and $x$ the first). Formally an angle between two vectors $a,b$ is: $$\arccos(\frac{\langle a,b \rangle}{||a||||b||})$$ (Where with $a,b$ i mean two vectors and not two coordinate vectors, this would be the same if the basis was orthonormal). The angle between $\mathcal{R}$ and the "x-axis", is simply the angle between the coordinate vector $(1,1)$(which corresponds to the vector $v+u$) and the vector $u$. So:

$$\arccos(\frac{\langle v+u,u \rangle}{||u||||v||})$$

Since $v,u$ are orthogonal this becomes:

$$\arccos(\frac{||u||}{||v||})$$

That clearly depends on $\frac{||u||}{||v||}$ which is the scale factor of the two axes. So in general NO, that straight line doesn't always form a $45°$ angle with x-axis.

Kandinskij
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