True or False: The line y=x is always at 45 degrees to the x-axis?

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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".

Answer 4

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 $45^{\circ}$.

Answer 5

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.