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In my notes, the following is written: The most general equation of a line in $\mathbb{R^2}$ is $ax+by=c,$ Where $a,b,c \in \mathbb{R}$ and $a,b$ not both $=0.$

I understand this, is it just something trivial, and a simple demonstration of a linear equation?

My question(s) is/are:

Are all linear equations in $\mathbb{R^2}$ simply straight lines, and linear equations in $\mathbb{R^3}$ represent planes?

And suppose, in the example above, $a=b=0$. Would $c=0$ represent a point in the one dimensional plane?

Just an idea, probably wrong $\ddot\smile.$

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    $\begingroup$ You mean "the general equation for a line", not "the most general equation". $\endgroup$ – quasi Mar 16 '17 at 22:42
  • $\begingroup$ @quasi my lecturer wrote 'the' most general equation...', but I would understand it as the standard/general equation of a line? $\endgroup$ – Gurjinder Mar 16 '17 at 22:45
  • $\begingroup$ For your other question, recall that for an equation with at most two vaiables $x,y$, the graph of the equation, by definition, is the set of all points $(x,y)$ which satisfy the equation. Hence if $a = b = c = 0$, the graph is ...? And if $a=b=0$ but $c \ne 0$, the graph is ...? $\endgroup$ – quasi Mar 16 '17 at 22:46
  • $\begingroup$ Well, if you consider a system of equations it gets more general, and that would allow you to transform the space so to speak. But perhaps this is the only simple equation. Does that help? $\endgroup$ – theREALyumdub Mar 16 '17 at 22:46
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    $\begingroup$ @theREALyumdub mmm, not really, sorry, I'm quite new to the introduction of mathematical rigour and goemetric intuitions. What would you mean by 'transforming the space's? $\endgroup$ – Gurjinder Mar 16 '17 at 22:58
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Edit: I realized after posting that you might not know what the scalar product is yet. In this case I suggest you stop at the "Yes" in the first two answers below, because the answers rely heavily on the intuition provided by the scalar product. If you don't know what that is, the answers are probably more confusing than helpful. However I do suggest that either way you take a look at the third answer ("Nope"), because that one only requires stuff you definitely know.

Are all linear equations in $\mathbb R^2$ simply straight lines? Yes, in the sense that you specified above (nonzero $a$ and $b$). The reason is this: The equation can be rewritten (in my opinion more intuitively) as $$ ax+by=c \quad \Leftrightarrow \quad \Biggl\langle\begin{bmatrix}a\\b\end{bmatrix} \Bigg| \begin{bmatrix}x\\y\end{bmatrix} \Biggr\rangle = c $$ where $\langle \boldsymbol u| \boldsymbol v\rangle$ is the scalar product $\boldsymbol u^\top \boldsymbol v$. The right equation should always evoke the intuition of the variable point $[x,y]^\top$ being projected onto the constant reference $[a,b]^\top$. That is the key feature of the scalar product. If $\|[a,b]^\top\|=1$, then $c$ is exactly the length of that part in $[x,y]^\top$ that is parallel to $[a,b]^\top$. If $\|[a,b]\| = A \neq 1$, then $c/A$ is the parallel length.

Either way, the scalar product expression essentially says: "I hold for all points $[x,y]$ that have a constant parallel part with $[a,b]$." If you draw it on a piece of paper, that means that $[a,b]^\top$ is perpendicular to the line of all possible $[x,y]^\top$, which intersects $[a,b]^\top$ at a distance of $c/A$ (from the origin $[0,0]^\top$).

Do all linear equations linear equations in $\mathbb R^3$ represent planes? Yes, in the same sense as above. Consider: $$ ax+by+cz=d \quad \Leftrightarrow \quad \Biggl\langle\begin{bmatrix}a\\b\\c\end{bmatrix} \Biggl| \begin{bmatrix}x\\y\\z\end{bmatrix} \Biggr\rangle = d $$ which holds for all points $[x,y,z]^\top$ that have with $[a,b,c]^\top$ a parallel part of $d/\|[a,b,c]^\top\|$, and an arbitrary perpendicular part. Again, $[a,b,c]$ specifies the normal direction, and the "arbitrary perpendicular part" has two degrees of freedom, meaning: All points on a plane that is offset by $d/\|[a,b,c]^\top\|$ in direction $[a,b,c]^\top$.

In the example above, $a=b=0$. Would $c=0$ represent a point in the one dimensional plane? Nope, but we don't even need geometric intuitions for this one. Just look at the resulting equation: $$ 0\cdot x + 0 \cdot y = 0 $$ For which choice of $x$ and $y$ is it satisfied? The answer will give you the set of $[x,y]^\top$ that this equation defines. Do they all lie on the same line? Or does this particular equation provide a different set of $[x,y]^\top$ points altogether?

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  • $\begingroup$ Thank you for your time and effort. I have found this very helpful, and will look into the scalar product. $\endgroup$ – Gurjinder Mar 17 '17 at 8:31

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